CORNELL UNIVERSITY LIBRARY 924 062 612 258 Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924062612258 Production Note Cornell University Library produced this volume to replace the irreparably deteriorated original. It was scanned using Xerox software and equipment at 600 dots per inch resolution and compressed prior to storage using CCITT Group 4 compression. The digital data were used to create Cornell's replacement volume on paper that meets the ANSI Standard Z39. 48-1984. The production of this volume was supported in part by the Commission on Preservation and Access and the Xerox Corporation. 1992. CORNELL UNIVERSITY LIBRARY GIFT OF L.L.Lan^don lAATHH-iATIOS -ELEMENTS Geometry and Trigonometry FROM THE WORKS OF A. M. LEGENDRE ADAPTED TO THE COUESE OF MATHEMATICAL INSTETJOTIOM IN THE UNITED STATES By CHAELES DAVIES, LL.D. AITTHOR OF A FULL CODBSE OF UATHEMATICd EDITED BY J. HOWAED VAN AMEINGE, A.M., Ph.D. PR0FE8»:>R OF MATHEUATICS IN COLUMBIA COLLBOE NEW YORK •:• CINCINNATI •;. CHICAGO AMERICAN BOOK COMPANY 6 '\0[ FX4. 6-j DAVIES'S MATHEMATICAL SERIES For Elementary Schools Davies's Primary Arithmetic Davies's Intellectual Arithmetic Davies's First Book in Arithmetic Davies's Standard Arithmetic Davies's Practical Arithmetic Davies's Complete Arithmetic For Secondary Schools Davies's University Arithmetic Davies's New Elementary Algebra Davies's Bourdon's Algebra Davies's Elementary Geometry and Trigonometry Davies's Legendre's Geometry and Trigonometpy For Colleges and Advanced Students Davies's University Algebra Davies's Analytical Geometry Davies's Analytical Geometry and Calculus Davies's Descriptive Geometry Davies's Elements of Surveying Copyright, 1862, 1882, 1885, and i8go, by A. S. Barnes & Co. D. LEGENDRE GEOM. AND TRIG. W. P. 4 PREFACE 01" the various treatises on Elementary Q-eometry wliloh have appeared during the present century, that of It. Iiegendre stands pre-eminent. Its peculiar merits have ■won for it not only a, European reputa- tion, but have also caused it to be selected as the basis of many of the best -works on the subject that have been published In this country. In the original treatise of Le^endrei the propositions are not enun- ciated in general terms, but by means of the diagrams employed in their demonstration. This departure from the method of EucUd is much to be regretted. The propositions of Geometry are general truths, and ought to be stated in general terms, -without reference to particular diagrams. In the foUo-wing work, each proposition is first enunciated in general terms, and after-ward -with reference to a particular figure, that figure being taken to represent any one of the class to -which it belongs. By this arrangement, the difficulty experienced by beginners in comprehending abstract truths Is lessened, -without in any manner impairing the gener- ality of the truths evolved. The term solid, used not only by Iiegendre, but by many other authors, to denote a limited portion of space, seems calculated to Introduce the foreign idea of matter into a, science -which deals only -with the abstract properties and relations of figured space. The term volume has been introduced in Its place, under the belief that it corresponds more exactly to the idea intended. Many other departures have been made from the original text, the value and utility of -which have been made manifest in the practical tests to -which the -work has been subjected. In the present edition, numerous changes have been made, both in the Geometry and in the Trigonometry. The definitions have been care- fully revised— the demonstrations have been harmonized, and, in many instances, abbreviated— the principal object being to simplify the subject as much as possible, -without departing from the general plan. These changes are due to Professor Peck, of the Department of Pure Mathematics iV PREFACE. and Asironomy in Columbia College. For his aid, in giving to the woi* Its prflsont permanent form, I tender him my grateful acknowledgments. The edition of liceendre, referred to in the last paragraph, will not be altered la form or substance ; and yet, Q-eometry must be made a more practical science. To attain this object, without deranging a system so long used, and so generally approved, an Appendix has been prepared and added to Iiegrendre, embracing many Problems of Geometrical construc- tion, and many applications of Algebra to Geometry. It would be unjust to those giving instruction, to add to their dally labors, the additional one, of finding appropriate solutions to so many difficult problems : hence, a Key has been made for the use of Teachers, in which the best methods of construction and solution are fully given. CHARLES DAVIES. PlSHKILL-ON'HUDSON, ^ifffP, F^7S- NOTE. — The edition of lie^endre referred to in the foregoing preface was prepared by the late Professor Davles the year before his lamented death. The present edition is the result of a careful re-examination of the work, into which have been incorporated such emendations, in the way of greater clearness of expression or of proof, as could be made with- out altering it in form or substance. Practical exercises have been placed at the end of the several books, and comprise additional theorems, problems, and numerical exercises upon the principles of the Book or Books preceding. They will, it Is hoped, be found of service in accustoming students, early tu and throughout their course, to make for themselves practical application of geometric princi- ples, and constitute, in addition, a large body of review and test questions for the convenience of teachers. The Trigonometry has been carefully revised throughout, to simplify the discussions and to make the treatment conform in every particular to the latest and best methods. It is believed that In clearness and precision of definition, in general simplicity and rigor of demonstration, in orderly and logical development of the subject, and in compactness of form, Davies' Iiegrendre Is superior to any work of its grade for the general training of the logical powers of pupils, and for their instruction in the great body of elementary geometric truth. J. H. VAN AMEIN&E, Edii»r of Davits'' Course o/ Mathemmtit*. Columbia Collegb, N. Y., June^ 1885. CONTENTS GEOMETRY. TABM Introduction, .. 9 BOOK I. Definitions, 18 Propositions, 20 Exercises, 60 BOOK II. Batlos and Proportions 62 BOOK III The Circle, and the Measurement of Angles, 61 Problems relating to the Plrst and Third Books 84 Exercises, Bi BOOK IV. Proportions of Figures — Measurement of Areas, 97 Problems relating to the Fourth Book 183 Exercises 140 BOOK V. Eegular Polygons — Measurement of the Circle, , 142 Exercises, •■ 163 BOOK VI. Planes, and Polyedral Angles 165 Exercises, 187 BOOK VII. Polyedrons, 189 Exercises 221 VI Contents. BOOK VIII. PAGE Cylinder, Cone, and Sphere, 223 Exercises, 248 BOOK IX. Spherical Geometry, 250 Exercises, 277 PLANE TRIGONOMETRY. INTRODUCTION. Deflnltlon of Logarithms 3 Rules for Characterlstlos, 4 General Principles, 5 Table of Logarithms Manner of Using the Table 7 Multiplication by Logarithms, 11 Division by Logarithms, 12 Arithmetical Complement, 13 Raising to Powers by Logarithms, 15 Extraction of Roots by Logarithms, 16 PLANE TRIGONOMETRY. Plane Trigonometry Defined, 17 Functions of an Arc, 18—21 Table of Natural Sines 22 Table of Logarithmic Sines, 22 Use of the Table 24—27 Solution of Right-angled Triangles, 27—36 Solution of Oblique-angled Triangles ^} 86—49 Problems, BO ANALYTICAL TRIGONOMETRY. Analytical Trigonometry Defined 53 Definitions and General Principles, 53—56 Rules for Signs of the Functions, 66 CONTENTS. Vll PAGE Limiting Value of Circular functions 57 Eelatlons of Circular Functions, 59—61 Sanctions of Negative Arcs, 62—65 Particular Values of Certain Functions, 66 Formulas of Eelation between Functions and Arcs, 67—70 Functions of Double and Half Arcs 70—71 Additional Formulas, 71—73 Method of Computing a Table of Natural Sines, 74 SPHERICAL TRIGONOMETRY. Spberioal Trigonometry Defined 76 General Principles, 76 Formulas for Eigbt-angled Triangles, 77—80 Napier's Circular Parts, 80 Solution of Eight-angled Spherical Triangles, 84—88 Quadrantal Triangles 89 Formulas for Oblique-angled Triangles, 90— 98 Solution of Oblique-angled Triangles, 98—116 MENSURATION. Mensuration Defined 117 The Area of a Parallelogram 118 The Area of a Triangle 118 Formula for the Sine and Cosine of Half an Angle, 120 Area of a Trapezoid, 125 Area of a Quadrilateral, 126 Area of a Polygon 126 Area of a Eegular Polygon 127 To find the Circumference of a Circle 129 To find the Diameter of a Circle, 130 To find the Length of an Arc, 130 Area of a Circle, 131 Area of a Sector 131 Area of a Segment 132 Area, of a Circular Eing, 138 Vm CONTENTS. rAGS Area of the Stirface of a Prism, 184 Area of the Surface of a Pyramid, 134 ,Vrea of the lyustum of a Cone, 136 Area of the Svtrface of a Sphere, 136 Area of a Zone, 137 Area of a Spherical Polygon 137 Volume of a Prism, 138 Volume of a Pyramid 139 VolUD^ie of the Trustum of a Pyramid 139 Volume of a Sphere, 141 Volume of a "Wedge, 141 Volume of a Prlsmoid, 144 Volumes of Begular Polyedrons, 146 LOGARITHMIC TABLES. Logarithms of Numbers from 1 to 10,000 1—16 Sines and Tangents 17—62 ELEMENTS OF GEOMETRY. INTRODUCTION. DEFINITIONS OF TERMS. 1. Quantity is any thing whicli can be increased, dimin- ished, and measured. To measure a thing, is to find out ho"w many times it contains some other thing, of the same kind, taken as a stand- ard. The assumed standard is called the unit of measure. 2. In Q-EOMETBT, there are four species of quantity, viz. : Lines, Surfaces, Volumes, and Angles. These are called Geometrical Magnitudes. Since the unit of measure of a quantity is of the same kind as the quantity measured, there are four kinds of units of measure, viz. : Units of Length, Units of Surface, Units of Volume, and Units of Angular Measure. 3. Q-EOMETRY is that branch of Mathematics which treats of the properties, relations, and measurement of the Q-eo- metrical Magnitudes. 4. In Geometry, the quantities considered are generally represented by means of the straight line and curve. The operations to be performed upon the quantities, and the rela- tions between them, are indicated by signs, as in Analysis. 10 GEOMETRY, The following are the principal signs employed : The Sign of Addition, + , called plus : Thus, A + B, indicates that B is to be added to A. The Sign of Subtraction, — , called minus : Thus, A — B, indicates that B is to be subtracted from A. The Sign of Multiplication, x : Thus, A X B, indicates that A is to be multiplied by B. The Sign of Division, -i- : A Thus, A -T- B, or, ^ , indicates that A is to be divided by B. The Exponential Sign: Thus, A', indicates that A is to be taken three times as a factor, or raised to the third power. The Radical Sign, V : Thus, a/a, •v^B, indicate that the square root of A, and the cube root of B, are to be taken. When a compound quantity is to be operated upon as a single quantity, its parts are connected by a vinculum or by a parenthesis: Thus, A + B X C, indicates that the sum of A and B is to be multiplied by C ; and (A + B) -4- C, indi- cates that the sum of A and B is to be divided by C. A numbejr written before a quantity, shows how many times it is to be taken. Thus, 3 (A + B), indicates that the sum of A • and B is to be taken three times. The Sign of Equality, = : Thus, A = B + C, indicates that A is equal to the sum of B and C. INTRODUCTION. 11 The expression, A = B + C, is called an equation. The part on the left of the sign of equality is called the first member ; that on the right, the second member. The Sign of Inequality, < : Thus, VA < -v^B, indicates that the square root of A is less than the cube root of B. The opening of the sign is towards the greater quantity. The sign, . • . is used as an abbreviation of the word hence, or consequently. The symbols, 1°, 2°, etc., mean 1st, 2d, etc. 5. The general truths of Geometry are deduced by a course of logical reasoning, the premises being definitions and principles previously established. The course of reasoning employed in establishing any truth or principle is called a demonstration. 6. A Theorem is a truth requiring demonstration. 7. An Axiom is a self-evident truth. 8. A Problem is a question requiring solution. 9. A Postulate is a self-evident Problem. Theorems, Axioms, Problems, and Postulates, are all called Propositions. 10. A Lemma is an auxiliary proposition. 11. A Corollary is an obvious consequence of one or more propositions. 12. A Scholium is a remark made upon one or more propositions, with reference to their connection, their use, their extent, or their limitation. 12 GEOMETRY. 13. An Hypothesis is a supposition made, either in the statement of a proposition, or in the course of a demonstrar tion. 14. Magnitudes are equal to each other, when each con- tains the same unit an equal niimber of times. 15. Magnitudes are equal in all respects, when they may be so placed as to coincide throughout their whole extent; they are equal in all their parts when each part of one is equal to the corresponding part of the other, when taken either in the same or in the reverse order. ELEMENTS OF GEOMETRY. BOOK I. ELEMENTARY PRINCIPLES. DEFINITIONS. 1. Geometry is that branch of Mathematics which creats of the properties, relations, and measurements of the Q-eo- metrical Magnitudes. 2. A Point is that which has position, but not magni- tude. 3. A Line is that which has length, but neither breadth nor thickness. Lines are divided into two classes, straight and curved. 4. A Straight Line is one which does not change its direction at any point. 5. A Curved Line is one which changes its direction at every point. When the sense is obvious, to avoid repetition, the word line, alone, is commonly used for straight line; and the word curve, alone, for curved line. 6. A line made up of straight lines, not lying in the same direction, is called a broken line. 7. A Surface is that whjch has length and breadth without thickness. 14 I GEOMETRY. Surfaces are divided into two classes, plane and curved surfaces. 8. A Plane is a surface, such, that if any two of its points be joined by a straight line, that line wiU lie wholly in the surface. 9. A Curved Surface is a surface which is neither a plane nor composed of planes. 10. A Plane Angle is the amount of divergence of two straight lines lying in the same plane. Thus, the amount of divergence of the Unas AB and AC, is an angle. The lines AB and AC are called sides, and their com- mon point A, is called the vertex. An angle is designated by naming its sides, or sometimes by simply naming its vertex; thus, the above is called the angle BAC, or simply, the angle A. 11. When one straight line meets another, the two angles which they form ^^^ are called adjacent angles. Thus, the angles B ^ ABD and DBC are adjacent. 12. A Right Angle is formed by one straight line meeting another so as to make the adjacent angles equal. The first hne is then said to be perpendicular to the second. 13. An Oblique Angle is formed by one straight line meeting another so as to make the adjacent angles unequal. Oblique angles are subdivided into two classes, acute angles, and obtuse angles. 14. An Acute Angle is less than a right angle. BOOK I. 15 15. An Obtuse Angle is greater than a right angle. 16. Two straight lines are parallel, when they lie in the same plane and can not meet, how far soever, either way, both may be produced. They then have the same direction. 17. A Plane Figuee is a portion of a plane bounded by lines, either straight or curved. 18. A Polygon is a plane figure bounded by straight lines. The bounding lines are called iides of the polygon. The broken line, made up of all the sides of the polygon, is called the perimeter of the polygon. The angles formed by the sides are called angles of the polygon. 19. Polygons are classified according to the number of their sides or angles. A Polygon of three sides is called a triangle; one of four sides, a quadrilateral j one of five sides, a pentagon; one of six sides, a hexagon; one of seven sides, a heptor gon; one of eight sides, an octagon; one of ten sides, a decagon; one of twelve sides, a dodecagon, &c. 20. An Equilateeal Polygon is one whose sides are all equal. An EQUL4.NGULAK POLYGON is One whose angles are all equal. A Regulae Polygon is one which is both equilateral and equiangular. 21. Two polygons are mutually equilateral, when their sides, taken in the same order, are equal, each to each : that is, following their perimeters in the same direction, the first 16 GEOMETRY. side of the one is equal to the first side of the other, the second side of the one to the second side of the other, and so on. 22. Two polygons are mutually equiangular, when their angles, taken in the same order, are equal, each to each. 23. A Diagonal of a polygon is a straight line joining the vertices of two angles, not consecutive. 24. A Base of a polygon is any one of its sides on which the polygon is supposed to stand. 25. Triangles may be classified with reference to either their sides, or their angles. When classified with reference to their sides, there are two classes : scalene and isosceles. 1st. A Scalene Tela.ngle is one which has no two of its sides equal. 2d. An Isosceles Triangle is one which has two of its sides equal. When all of the sides are equal, the triangle is Equilateral. When classified with reference to their angles, there are two classes : right-angled and oblique-angled. 1st. A Right-angled Triangle is one that has one right angle. The side opposite the right angle is called the hypothe- nuse. 2d. An Oblique-angled Triangle is one whose angles are all oblique. BOOK I. 17 If one angle of an oblique-angled triangle is obtuse, the triangle is said to be obtuse-angled. If all of the angles are acute, the triangle is said to be acute-angled. 26. Quadrilaterals are classified -with reference to the rel- ative directions of their sides. There are then two classes; the first class embraces those which have no two sides par- allel ; the second class embraces those which have at least two sides parallel. Quadrilaterals of the first class, are called trapeziums. Quadrilaterals of the second class, are divided into two species : trapezoids and parallelograms. 27. A Trapezoid is a quadrilatoral which has only two of its sides parallel. 28. A Paballelogeam is a quadrilateral which has its opposite sides parallel, two and two. There are two varieties of parallelograms : rectangles and rhomboids. 1st. A Rectangle is a parallelogram whose angles are all right angles. A Square is an equilateral rectangle. 2d. A Rhomboid is a parallelogram whose angles are all oblique. A Rhombus is an equilateral rhomboid. 9 18 GEOMETRY. 29. Space is indefinite extension. 30. A Volume is a limited portion of space, combining the tliree dimensions of length, breadth, and thickness. AXIOMS. 1. Things which are equal to the same thing, are equal to each other. 2. If equals are added to equals, the sums are equal. 3. If equals are subtracted from equals, the remainders are equal. 4. If equals are added to unequals, the sums are un- equal. 5. If equals are subtracted from unequals, the remain- ders are unequal. 6. If equals are multiplied by equals, the products are equal. 7. If equals are divided by equals, the quotients are equal. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 10. AU right angles are equal. 11. Only one straight line can be drawn joining two given points. 12. The shortest distance from one point to another is measured on the straight line which joins them. 13. Through the same point, only one straight line can be drawn parallel to a given straight line. BOOK I. 19 POSTULATES. 1. A straight line can be drawn joining any two points. 2. A, straight line may be prolonged to any length. 3. If two straight lines are unequal, the length of the less may be laid off on the greater. 4. A straight line may be bisected; that is, divided into two equal parts. 5. An angle may be bisected. 6. A perpendicular may be drawn to a given straight Une, either from a point without, or from a point on the line. 7. A straight line may be drawn, making with a given straight line an angle equal to a given angle. 8. A straight line may be drawn through a given point, parallel to a given line. NOTE. In making references, the following abbreviations are employed, viz.: A. for Axiom; B. for Book; C. for Corollary; D. for Definition; I. for Introduction; P. for Proposition; Prob. for Problem; Post, for Postulate; and S. for Scholium. In referring to the same Book, the number of the Book i» not given; in referring to any other Book, the number of the Book it given. 20 GEOMETRY. PROPOSITION I. THEOREM. If a straight line meets another straight line, the sum of the adjacent angles is equal to two right angles. Let DC meet AB at C: then is the sum of the angles DCA and DCB equal to two right an- gles. At C, let CE be drawn perpen- dicular to AB (Post. 6) ; then, by- definition (D. 12), the angles ECA and ECB are both right angles, and consequently, their sum is equal to two right angles. The angle DCA is equal to the sum of the angles ECA and ECD (A. 9) ; hence. But, DCA + DCB = ECA + ECD + DCB ; ECD + DCB is equal to ECB (A. 9) ; hence, DCA + DCB = ECA + ECB. The sum pi the angles ECA and ECB, is equal to two right angles; consequently, its equal, that is, the sum of the angles DCA and DCB, must also be equal to two right angles ; which was to he proved. Cor. 1. If one of the angles DCA, DCB, is a right angle, the other must also be a right angle. Cor. 2. The sum of the an- gles BAC, CAD, DAE, EAF, formed about a given point on the same side of a straight line BF, is equal to two right angles. For, their B- eum is equal to the sum of the BOOK I. 21 angles EAB and EAF; -which, from the proposition just demonstrated, is equal to two right angles. DEFINITIONS. If two straight lines intersect each other, they form four angles about the point of intersection, which have received dififerfent names, with respect to each other. 1°. Adjacent Angles are those which lie on the same side of one line, and on opposite sides of the other ; thus, ACE and ECB, or ACE and ACD, are adjacent angles. 2°. Opposite, or Vertical Angles, are those which lie on opposite sides of both lines ; thus, ACE and DCB, or ACD and ECB, are opposite angles. From the proposition just demonstrated, the sum of any two adjacent angles is equal to two right angles. PROPOSITION n. THEOREM. If two straight lines intersect each other, the opposite or vertical angles are equal. Let AB and DE intersect at C : then are the opposite or vertical angles equal. The sum of the adjacent angles ACE and ACD, is equal to two right angles (P. I.) : the sum of the adjacent angles ACE and ECB, is also equal to two right angles. But things which are equal to the same thing, are equal to each other (A. 1) ; hence, 22 GEOMETRY. ACE + ACD = ACE + ECB ; Taking from both the common angle ACE (A. 3), there remains, ACD = ECB. In hke manner, we find, ACD + ACE = ACD + DCB ; and, taking away the common angle ACD, we have, ACE = DCB. Hence, the proposition is proved. Cor. 1. If one of the angles about C is a right angle, all of the others are right angles also. For, (P. I., C. 1), each of its adjacent angles is a right angle ; and from the proposition just demonstrated, its opposite angle is also a right A angle. Cor. 2. If one line DE, is per- pendicular to another AB, then is the second line AB perpendicTilar to the first DE. For, the angles DC A and DCB are right angles, by definition (D. 12); and from what has just been proved, the angles ACE and BCE are also right angles. Hence, the two lines are mutually perpen- dicular to each other. Cor. 3. The sum of all the angles ACB, BCD, DCE, ECF, FCA, that can be formed about a point, is equal to four right angles. B O K I. 23 For, if two lines are drawn through the point, mutually perpendicular to each other, the sum of the angles which they form is equal to four right angles, and it is also equal to the sum of the given angles (A. 9). Hence, the sum of the given angles is equal to four right angles. PROPOSITION III. THEOREM. If two straight lines have two points in common, they coin- cide throughout their whole extent, and form one and the same line. Let A and B be two points common to two lines : then the - E lines coincide throughout. A g C ""■■ Between A and B they must coincide (A. 11). Suppose, now, that they begin to separate at some point C, beyond AB, the one becoming ACE, and the other ACD. If the lines do separate at C, one or the other must change direction at this point; but this is contradictory to the definition of a straight line (D. 4) : hence, the supposition that they separate at any point is absurd. They must, therefore, coincide throughout; which was to ie proved. Cor. Two straight lines can intersect in only one point. Note. — The method of demonstration employed above, is called the reductio ad ahsurdum. It consists in assuming an hypothesis which is the contradictory of the proposition to be proved, and then continuing the reasoning until the assumed hypothesis is shown to be false. Its contradictory is thus proved to be true. This method of demonstration is often used in Geometry. 24 GEOMETEY. PROPOSITION IV. THEOREM. // a straight line meets two other straight lines at a com- mon point, making the sum of the contiguous angles equal to two right angles, the two lines met form one and the same straight line. Let DC meet AC and BC at C, making the sum of the angles DCA and DCB equal to two right angles: then is CB the prolonga- tion of AC. For, if not, suppose CE to be the prolongation of AC ; then is the sum of the angles DCA and DCE equal to two right angles (P. I.) : consequently, we have (A. 1), DCA + DCB = DCA + DCE ; Taking from both the common angle DCA, there re- mains DCB = DCE, which is impossible, since a part can not be equal to the whole (A. 8). Hence, CB must be the prolongation of AC ; which was to he proved. PROPOSITION V. THEOREM. // two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the triangles are equal in all respects. In the triangles ABC and DEF, let AB be equal to DE, BOOK I. 2& AC to DF, and the angle A to the angle D : then are the triangles equal in all respects. For, let ABC be ap- plied to DEF, in such a A P manner' that the angle A shall coincide with the angle D, the side AB taking the direction DE, and the side AC the direction DF. Then, because AB is equal to DE, the ver- tex B will coincide with the vertex E ; and because AC is equal to DF, the vertex C will coincide with the vertex F ; consequently, the side BC will coincide with the side EF (A. 11). The two triangles, therefore, coincide through- out, and are consequently equal in all respects (I., D. 15); which was to be proved. PROPOSITION VI. THEOREM. // two triangles have two angles and the included side of the one equal to two angles and the included side of the other, each to each, the triangles are equal in all respects. , In the triangles ABC and DEF, let the angle B be equal to the angle E, the angle C to the angle F, and the side BC to the side EF : then are the triangles equal in all re- spects. For, let ABC be applied to DEF in such a manner that the angle B shall coincide with the angle E, the side BC taking the direction EF, and the side BA the direc 26 GEOMETRY. tion ED. Then, because BC is equal to EF, the vertex C will coincide with the vertex F; and because the angle C is equal to the angle F, the side CA will take the direction FD. Now, the vertex A being at the same time on the lines ED and FD, it must be at their intersection D (P. m., C.) : hence, the triangles coincide throughout, and are therefore equal in all respects (I., D. 15) ; which was to be proved. PROPOSITION VII. THEOREM. The sum, of any two sides of a triangle is greater than the third side. Let ABC be a triangle : then will the sum of any two sides, as AB, BC, be greater than the third side AC. For, the distance from A to C, measured on any broken line AB, BC, is greater than C B the distance measured on the straight line AC (A. 12): hence, the sum of AB and BC is greater than AC ; which was to be proved. Cof. If from both members of the inequality, AC < AB + BC, we take away either of the sides AB, BC, as BC, for ex- ample, there remains (A. 5), AC — BC < AB ; that is, the difference between any two sides of a triangle is less than the third side. Scholium. In order that any three given lines may rep- BOOK I. 27 resent the sides of a triangle, the sum of any two must be greater than the third, and the difference of any two must be less than the third. PROPOSITION Vm. THEOREM. .// from any point within a triangle two straight lines are drawn to the extremities of any side, their, sum is less than that of the two remaining sides of the triangle. Let be any point within the triangle BAC, and let the lines OB, OC, be drawn to the extremities of any side, as BC : then the sum of BO and OC is less than the sum of the sides BA and AC. Prolong one of the lines, as BO, till it meets the side AC in D ; then, from Prop. VII., we have, OC < OD + DC ; adding BO to both members of this inequality, recollecting that the sum of BO and OD is equal to BD, we have (A. 4), BO + OC < BD + DC. From the triangle BAD, we have (P. Vn.), BD < BA + AD ; adding DC to both members of this inequality, recollecting that the sum of AD and DC is equal to AC, we have, BD + DC < BA + AC. But it was shown that BO + OC is less than BD + DC •; still more, then, is BO + OC less than BA + AC ; which was to be proved. 28 GEOMETRY. PROPOSITION IX. THEOREM. If two triangles have two sides of the one ecfual to two sides of the other, each to each, and the included angles unequal, the third sides are unequal ; and the greater side belongs to the triangle which has the greater included angle. m In the triangles BAC and DEF, let AB be equal to DE, AC to DF, and the angle A greater than the angle D : then is BC greater than EF. Let the line AG be drawn, making the angle CAG equal to the angle D (Post. 7) ; make AG equal to DE, and draw GC. Then the triangles AGC and DEF have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each ; consequently, GC is equal to EF (P. V.). Now, the point G may be without the triangle ABC, it may be on the side BC, or it may be within the triangle ABC. Each case will be considered separately. 1°. When G is without the triangle ABC. In the triangles GIC and AIB, we have, (P. VII.), Gl + IC > GC, G E^ and Bl + lA > AB; whence, by addition, recollecting that the sum of BI and IC is equal to BC, and the sum of Gl and lA, to GA, we have, AG + BC > AB + GC. BOOK I. Or, since AG = AB, and GC = EF, we have, AB + BC > AB + EF. Taking away the common part AB, there remains (A. 5), BC > EF. 2°. When G is on BC. In this, case, it is obvious that GC is less than BC ; or since GC = EF, we have, BC > EF. 29 3°. When G is within the triangle ABC. From Proposition VIII., we have, BA + BC > GA + GC ; or, since GA = BA, and GC = EF, we have, BA + BC > BA + EF. Taking away the common part AB, there remains, BC > EF. Hence, in each case, BC is greater than EF; which was to be proved. Conversely: If in two triangles ABC and DEF, the side AB is equal to the side DE, the side AC to DF, and BC greater than EF, then is the angle BAC greater than the angle EDF. For, if not, BAC must either be equal to, or less than, EDF. In the former case, BC would be equal to EF (P. v.), and in the latter case, BC would be less than EF ; either of which would contradict the hypothesis : hence, BAC must be greater than EDF. 30 GEOMETRY PROPOSITION X. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are equal in all respects. In the triangles ABC and DEF, let AB be equal to DE, AC to DF, and BC to EF: then are the triangles equal in all respects. For, since the sides AB, AC, are equal to DE, DF, each to each, if the angle A were greater than D, it ^ Q would follow, by the last Proposition, that the side BC would be greater than EF; and if the angle A were less than D, the side BC would be less than EF. But BC is equal to EF, by hypothesis; therefore, the angle A can neither be greater nor less than D : hence, it must be equal to it. The two triangles have, therefore, two sides and the included angle of the one equal to two sides and the included angle of the other, each to each ; and, conse- quently, they are equal in all respects (P. V.) ; which was to he proved. Scholium. In triangles, equal in aU respects, the equal sides lie opposite the equal angles ; and conversely. PROPOSITION XI. THEOREM. In an isosceles triangle the angles opposite the equal sides are equal. Let BAC be an isosceles triangle, having the side AB equal to the side AC: then the angle C is equal to the angle B. BOOK 1. 31 Join the vertex A and the middle point D of the base BC. Then, AB is equal to AC, by hypothesis, AD com- mon, and BD equal to DC, by con- struction : hence, the triangles BAD, and DAC, have the three sides of the one equal to those of the other, each to each ; therefore, by the last PropositiAa, the angle B is equal to the angle t; which was to he proved. Cor. 1. An equilateral triangle is equiangular. Cor. 2. The angle BAD is equal to DAC, and BDA to CDA: hence, the last two are right angles. Consequently, a straight line drawn from the vertex of an isosceles tri- angle to the middle of the base, bisects the angle at the vertex, and is perpendicular to the base. PROPOSITION Xri. THEOREM. If two angles of a triangle are equal, the sides opposite to them are also equal, and consequently, the triangle is isosceles. In the triangle ABC, let the angle ABC be equal to the angle ACB: then is AC equal to AB, and consequently, the triangle is isosceles. For, if AB and AC are not equal, suppose one of them, as AB, to be the greater. On this, take BD equal to AC (Post. 3), and draw DC. Then, in the triangles ABC, DBC, we have the side BD equal to AC, by construction, the side BC common, and the included angle ACB equal to the included angle DBC, by hypothesis : hence, the two triangles are equal 32 GEOMETKT. in all respects (P. V.). But this is impossible, because a part can not be equal to the whole (A. 8) : hence, the hypothesis that AB and AC are unequal, is false. They tAist, therefore, be equal ; which was to be proved. Cor. An equiangular triangle is equilateral. PROPOSITION Xm. THEOREM. In any triangle, the greater side is opposite the greater angle; and, conversely, the greater angle is opposite the greater side. In the triangle ABC, let the angle ACB be greater than the angle ABC : then the side AB is greater than the side AC. For, draw CD, making the angle BCD equal to the angle B (Post. 7) : then, in the triangle DCB, we have the angles DCB and DBC equal: hence, the opposite sides DB and DC are equal (P. XII.). In the triangle ACD, we have (P. VII.), AD 4- DC > AC ; or, since DC = DB, and ,AD + DB = AB, we have, AB > AC; which was to be proved. Conversely : Let AB be greater than AC : then the angle ACB is greater than the angle ABC. For, if ACB were less than ABC, the side AB would be less than the side AC, from what has just been proved ; if ACB were equal to ABC, the side AB would be equal to AC, by Prop. XII. ; but both conclusions contradict BOOK I. 33 the hypothesis : hence, ACB can neither be less than, nor equal to, ABC ; it must, therefore, be greater ; which was to be proved. PROPOSITION XIV. THEOREM. From a given point only one perpendicular can be draion to a given straight line. Let A be a given point, and AB a perpendicular to DE : then can no other perpendicular to DE be drawn from A. C- For, suppose a second perpendicTilar AC to be drawn. Prolong AB till BF is equal to AB, and draw CF. Then, the triangles ABC and FBC have AB equal to BF, by con- struction, CB common, and the included angles ABC and FBC equal, because both are right angles: hence, the angles ACB and FCB are equal (P. V.). But ACB is, by a hypothesis, a right angle : hence, FCB must also be a right angle, and consequently, the line ACF must be a straight line (P. IV.). But this is impossible (A. 11). The hypothesis that two perpendiciilars can be drawn is, therefore, absurd ; consequently, only one such perpendic- ular can be drawn ; which was to be proved. B If the given point is on the given line, the proposition is equally true. For, if from A two perpendiculars AB and AC could be drawn to DE, we should have BAE and CAE each equal to a right angle; and consequently, equal to each other; which is absurd (A. 8). D 34 GEOMETEY. PROPOSITION XV. THEOREM. If from a point without a straight line a perpendicular is let fall on the line, and, dbliqwe lines are drawn to dif- ferent points of it: 1°. The perpendicular is shorter than any oblique line. 2°. Any two oblique lines that meet the given line at points equally distant from the foot of the perpendicular, are equal. 3°. Of two oblique lines that meet the given line at points itnequally distant from the foot of the perpendicular, the one which meets it at the greater distance is the longer. Let A be a given point, DE a given straight line, AB a perpendicular to DE, and AD, AC, AE oblique lines, BC being equal to BE, and BD greater than BC. Then AB is less than any of the oblique lines, AC is equal to AE, and AD greater than AC. Prolong AB until BF is equal to AB, and draw FC, FD. 1°. In the triangles ABC, FBC, we have the side AB equal to BF, by construction, the side BC common, and the included angles ABC and FBC equal, because both are right angles: hence, FC is equal to AC (P. V.), But, AF is shorter than ACF (A. 12): hence, AB, the half of AF, is shorter than AC, the half of ACF ; which ' was to be proved. 2°. In the triangles ABC and ABE, we have the side BC equal to BE, by hypothesis, the side AB common, and the included angles ABC and ABE equal, because both are BOOK I. 35 right angles : hence, AC is equal to AE ; which was to be proved. 3°. It may be shown, as in the first case, that AD is equal to DF. Then, because the point C lies within the triangle ADF, the sum of the lines AD and DF is greater than the sum of the lines AC and CF (P. VIII.) : hence, AD, the half of ADF, is greater than AC, the half of ACF ; which was to be proved. Cor. 1. The perpendicular is the shortest distance from a point to a line. Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn ; for, if there could be more, there would be at , least two equal oblique lines on the same side of the perpendicular; which is im- possible. PROPOSITION XVI. THEOREM. If a perpendicular is drawn to a given straight line at its middle point: 1°. Any point of the perpendicular is equally distant from the extremities of the line: 2°. -^-~->^^ ^■^-'-—^ and EG ; then the triangles / \ / \ ACD and EOG have two sides ( A ) A ) and their included angle, in \/\yV/\/ the one, equal to two sides ^^^—^ f^^;.::^::;^ and their included angle, in the other, each to each. They are, therefore, equal in all respects ; consequently, AD is equal to EG. But, since the chords AD and EG are equal, the arcs AMD and ENG are also equal (P. IV.) ; which was to be proved. 76 GEOMKTRY. 2°. Let the arcs AMD and ENG be equal: then the angles ACD and EOG are equal. For, since the arcs AMD and ENG are equal, the chords AD and EG are equal (P. IV.) ; consequently, the triangles ACD and EOG have their sides equal, each to each ; they are, therefore, equal in all respects : hence, the angle ACD is equal to the angle EOG ; which was to be proved. PROPOSITIOlSr XVI. THEOREM. In equal circles, commensuraile angles at the centre are -proportional to their intercepted arcs. In the equal circles, whose centres are C and 0, let the angles ACB and DOE be commensurable ; that is, be exactly measured by a common unit : then are they pro- portional to the intercepted arcs AB and DE. Let the angle M be a common unit; and suppose, for example, that this unit is contained 7 times in the angle ACB, and 4 times in the angle DOE. Then, suppose ACB be divided into 7 angles, by the radii Cm, Cn, Cp, '&c. ; and DOE into 4 angles, by the radii Ox, Oy, and Oz, each equal to the unit M. BOOK III. 77 From the last proposition, the arcs Am, mn, &c., Dx, xy, &c., are equal to each other ; and hecause there are 7 of these arcs in AB, and 4 in DE, we shall have, arc AB : arc DE : : 7 : 4. But, hy hypothesis, we have, angle ACB : angle DOE : : 7 : 4 ; hence, from (B. II., P. IV.), we have, angle ACB : angle DOE : : arc AB : arc DE. If any other numbers than 7 and 4 had been used, the same proportion would have been found ; which was to be proved. Cor. If the intercepted arcs are commensurable, they are proportional to the corresponding angles at the centre, as may be shown by changing the order of the couplets in the above proportion. PROPOSITION XVn. THEOREM. In equal circles, incommensurable angles at the centre are proportional to their intercepted arcs. In the equal circles, . whose centres are C and 0, let ACB and FOH be incom- mensurable : then are they proportional to the arcs AB and FH. For, let the less angle FOH, be placed upon the greater angle ACB, so that it shall take the position ACD. Then, 78 GEOMETRY. if the proposition is not true, let us suppose that the angle ACB is to the angle FOH, or its equal ACD, as the arc AB is to an arc AO, greater than FH, or its equal AD ; whence, angle ACB : angle ACD arc AB : arc AO. Conceive the arc AB to be divided into equal parts, each less than DO : there "will be at least one point of division between D and ; let I be that point ; and draw CI. Then the arcs AB, Al, will be commensurable, and we shaH have (P. XVI.), angle ACB : angle ACI arc AB arc AI. Comparing the two proportions, we see that the antece- dents are the same in both : hence, the consequents are proportional (B. 11., P. IV., C.) ; hence. angle ACD : angle ACI arc AO arc Al. But, AO is greater than Al : hence, if this proportion is true, the angle ACD must be greater than the angle ACI. On the contrary, it is less: hence, the fourth term of the assumed proportion can not be greater than AD. In a similar manner, it may be shown that the fourth term can not be less than AD : hence, it must be equal to AD ; therefore, we have, angle ACB : angle ACD : : arc AB : arc AD ; which was to be proved. Cor. 1. The intercepted arcs are proportional to the cor- responding angles at the centre, as may be shown by BOOK III. 79 changing the order of the couplets in the preceding pro- portion. Cor. 2. In equal circles, angles at the centre are pro- portional to their intercepted arcs, and the reverse, whether they are conamensurable or incommensurable. Cor. 3. In equal circles, sectors are proportional to their angles, and also to their arcs. Scholium. Since the intercepted arcs are proportional to the corresponding angles at the centre, the arcs may be taken as the measures of the angles. That is, if a cir- cumference be described from the vertex of any angle, as a centre, and with a fixed radius, the arc intercepted between the sides of the angle may be taken as the measure of the angle. In Q-eometry, the right angle, which is measured by a quarter of a circumference, or a quadrant, is taken as a unit. If, therefore, any angle is measured by one half or two thirds of a quadrant, it is equal to one half or two thirds of a right angle. PROPOSITION XVIII. THEOREM. An inscribed angle is measured by half of the arc included between its sides. There may be three cases: the centre of the circle may lie on one of the sides of the angle ; it may lie within the angle ; or, it may lie without the angle. 1°. Let EAD be an inscribed angle, one of whose sides AE passes through the centre : then it is measured by half of the arc DE. 80 GEOMETRY. For, draw the radius CD. The external angle DCE, of the triangle DCA, is equal to the sum of the opposite interior angles CAD and CDA (B. I., P. XXV., C. 6). But, the triangle DCA being isosceles, the angles D and A are equal; therefore, a the angle DCE is double the angle DAE. Because DCE is at the centre, it is measured by the arc DE (P. XVII., S.) : hence, the angle DAE is measured by half of the arc DE ; which was to be proved. 2°. Let DAB be an inscribed angle, and let the centre lie within it: then the angle is measured by half of the arc BED. For, draw the diameter AE. Then, from what has just been proved, the angle DAE is measured by half of DE, and the angle EAB by half of EB : hence, BAD, which is the sum of EAB and DAE, is measured by half of the sum of DE and EB, or by haK of BED ; which was to be proved. 3°. Let BAD be an inscribed angle, and let the centre lie without it: then it is measured by half of the arc BD. For, draw the diameter AE. Then, from what precedes, the angle DAE is measured by half of DE, and the angle BAE by half of BE : hence,. BAD, which is the difference of BAE and DAE, is measured by half of the difference of BE and DE, or by half of the arc BD ; which was to be proved. BOOK III. 81 Cor. 1. All the angles BAC, BDC, BEC, inscribed in the same segment, are equal ; because they are each measured by half of the same arc BOC. Cor. 2. Any angle BAD, inscribed in a semicircle, is a right angle ; because it is measured by half the semi-circum- ference BOD, or by a quadrant (P. XVIL, S.). Cor. 3. Any angle BAC, inscribed in a segment greater than a semicircle, is acute ; for it is measured by half the arc BOC, less than a semi-circum- ference. Any angle BOC, inscribed in a seg- ment less than a semicircle, is obtuse ; for it is measured by half the arc BAC, greater than a semi-circumference. Cor. 4. The opposite angles A and C, of an inscribed quadrilateral ABCD, are together equal to two right angles; for the angle DAB is measured by half the arc DCB, the angle DCB by half the arc DAB : hence, the two angles, taken together, are measured by half the circumference: hence, their sum is equal to two right angles. 82 GEOMETRT. PROPOSITION XIX. THEOREM. Any angle farmed by two chords, which intersect, is meas- ured hy half the sum of the included ares. Let DEB be an angle formed by the intersection of the chords AB and CD : then it is measured by half the sum of the arcs AC and DB. For, draw AD : then, the angle DEB, being an exterior angle of the triangle DEA, is equal to the sum of the angles EDA and EAD (B. L, P. XXV., C. 6). But, the angle EDA is measured by half the arc AC, and EAD by half the arc DB (P. XVIII.) : hence, the angle DEB is measured by half the sum of the arcs AC and DB ; which was to be proved. PROPOSITION XX. THEOREM. The angle formed hy two secants, intersecting without the circumference, is measured hy half the difference of the included arcs. Let AB, AC, be two secants : then the angle BAC is measured by half the difference of the arcs BC and DF. Draw DE parallel to AC : the arc EC is equal to DF (P.' X.), and the angle BDE to the angle BAC (B. L, P. XX., C. 3). But BDE is measured by half the arc BE (P. XVIII.): hence, BAC is also measured by half the arc BE ; that is, by half the difference of BC and EC, or by half the difference of BC and DF; proved. which was to he BOOK III. 83 PROPOSITION XXI. THEOEEM. An angle formed by a tangent and a chord meeting it at the point of contact, is measured by half the included arc. Let BE be tangent to the circle AMC, and let AC be a chord drawn from the point of contact A : then BAG is measured by half of the arc AMC. For, draw the diameter AD. The angle BAD is a right angle (P. IX.), and is measured by half the semi-circumfer- ence AMD (P. XVIL, S.); the angle DAC is ' measured by half of the arc DC (P. XVIII.) : hence, the angle BAG, which is equal to the sum of the angles BAD and DAC, is measured by half the sum of the arcs AMD and DC, or by half of the arc AMC ; which was to be proved. The angle CAE, which is the difference of DAE and DAC, is measured by half the difference of the arcs DCA and DC, or by half the arc CA. PRACTICAL APPLICATIONS. PROBLEM I >CE To bisect a given straight line. Let AB be a given straight line. From A and. B, as centres, with a radius greater than one half of AB, describe arcs intersecting at E and F: -+- join E and F, by the straight line EF. Then EF bisects the given line AB. For, E and F are each equally distant from A and B ; and consequently, the line EF bisects AB (B; L, P. XVI., C). T >CF PROBLEM n. To erect a perpendicular to a given straight line, at a given point of that line. Let EF be a given line, and let A be a given point of that line. From A, lay off the equal distances AB and AC ; from B and C, as centres, with a radius greater than one half >;d BOOK III. 85 of BC, describe arcs intersecting at D ; draw the line AD : then AD is the perpendicular required. For, D and A are each equally distant from B and C ; consequently, DA is perpendicular to BC at the given point A (B. I., P. XVI., C). PROBLEM III. To draw a -perpendicular to a given straight line, from a given point without that line. Let FG be the given line, and A the given point. From A, as a centre, with a radius sufficiently great, describe an arc cut- ting FG in two points, B and D ; with B and D as centres, and a radius greater than one half of BD, describe arcs intersecting at E ; draw AE : then AE is the perpendicular required. For, A and E are each equally distant from B B--. ^ and D : hence, AE is perpendicular to BD (B. L, P. XVI., C). PROBLEM IV. At a point on a given straight line, to construct an angle eqi^al to a given angle. Let A be the given point, AB the given line, and IKL the given angle. From the vertex K as a cen- ter, with any radius Kl, describe the arc IL, terminating in the sides of the angle. From A as a centre, with a radius AB, equal to Kl, describe the 86 GEOMETRY. indefinite arc BO ; then, with a radius equal to the chord LI, from B as a centre, describe an arc cutting the arc BO in D ; draw AD : then BAD is equal to the angle K. l^ \q^ For the arcs BD, IL, have equal radii and equal chords: k^ hence, they are equal (P. IV.) ; therefore, the angles BAD, IKL, measured by them, are also equal (P. XV.). PROBLEM V. To bisect a given arc or a given angle. 1°. Let AEB be a given arc, and C its centre. Draw the chord AB ; through C, draw CD perpendicular to AB (Prob. III.) : then CD bisects the arc AEB (P. VI.). 2°. Let ACB be a given angle. With C as a centre, and any radius CB, describe the arc BA ; bisect it by the line CD, as just explained : then CD bisects the angle ACB. For, the arcs AE and EB are equal, from what was just shown; consequently, the angles ACE and ECB are also equal (P. XV.). Scholium. If each half of an arc or angle is bisected, the original arc or angle is divided into four equal parts ; and if each of these is bisected, the original arc or angle is divided into eight equal parts ; and so on. BOOK III. 87 PROBLEM VI. Through a given point, to draw a straight line parallel to a given straight line. Let A be a given point, and BC a given line. From the point A as a centre, with a radius AE, greater than the g_F E_^ shortest distance from A to BC, de- \ ,.'-"" scribe an indefinite arc .EO ; from E \, .,■■■-•"" as a centre, with the same radius, describe the arc AF ; lay off ED equal to AF, and draw AD : then AD is the parallel required. For, drawing AE, the angles AEF, EAD, are equal (P. XV.) ; therefore, the lines AD, EF are parallel (B. L, P. XIX., C. 1). PROBLEM Vn. Given,two angles of a triangle, to construct the third angle. Let A and B be given angles of a triangle. Draw a line DF, and at some point of it, as E, construct the angle FEH D E F equal to A, and HEC equal to B. A. yi. Then, CED is equal to the required \ angle. For, the sum of the three angles at E is equal to two right angles (B. I., P. I., 0. 2), as is also the sum of the three angles of a triangle (B. I., P. XXV.). Consequently, the third angle CED must be equal to the third angle of the triangle. 88 GEOMETET. PROBLEM Vin. Given, two sides and the included angle of a triangle, to construct the triangle. Let B and C denote the given sides, and A the given angle. Draw the indefinite line DF, and at D construct an angle FDE, equal to the angle A; on DF, lay off DH equal to the side w and on DE, lay- off DG equal to the sideO^; draw GH : then DGH is the required triangle (B. L, P. V.). PROBLiLM ES. Given, one side and two angles of a tri^angle, to construct the triangle. The two angles may be either both adjacent to the given side, or one may be adjacent and the other oppo- site to it. In the latter case, construct the third angle by Problem VII. We shall then, have two angles and their included side. Draw a straight line, and on it lay off q ,F DE equal to the given side ; at D con- struct an angle equal to one of the adjacent angles, and at E construct -an angle equal to the other adjacent angle ; produce the sides DF and EG till they intersect at H : then DEH is the triangle required (B. I., P. VI.). BOOE III. 89 PROBLEM X. Given, the three sides of a triangle, to construct the tri- angle. Let A, B, and C, be -the given sides. Draw DE, and make it equal to the side A; from D as a centre, with a radius equal to the side B, describe *' ' an arc ; from E as a centre, with a O 1 radius equal to the side C, describe an arc intersecting the former at F ; draw DF and EF : then DEF is the triangle required (B. I., P. X.). Scholium. In order that the construction may be pos- sible, any one of the given sides must be less than the sum of the two others, and greater than their difference (B. L, P. VII., S.). PROBLEM XI. Given, two sides of a triangle, and the angle opposite one of them, to construct the triangle. Let A and B be the given sides, and C the given angle. Draw an indefinite line DG, and Ai > ^ at some point of it, as D, construct B^ an angle GDE equal to the given angle ; on one side of this angle lay off the distance DE equal to the side B adjacent to the given angle ; from E as a centre, with a radius equal to the side opposite the given angle, describe an arc cutting the side DG at G : draw EG, Then DEG is the required triangle. 90 GEOMETKY. For, the sides DE and EG are equal to the given sides, and the angle D, opposite one of them, is equal to the given angle. Scholium. If the side opposite the given angle is greater than the other given side, there is but one solu- tion. If the giveri< angle is acute, and the side opposite the given angle is less than the other given side, and greater than the shortest distance from E to DG, there are two solutions, DEG and DEF. If the side opposite the given angle is equal to the short- est distance from E to DG, the arc will be tangent to DG, the angle opposite DE is a right angle, and there is but one solution. If the side opposite the given angle is shorter than the distance from E to DG, there is no solution. PROBLEM XII. Given, two adjacent sides of a parallelogram and their included angle, to construct the parallelogram. Let A and B be the given sides, and C the given angle. Draw the line DH, and at some point as D, construct the angle HDF equal to the angle C. Lay off DE equal to the side A, and DF equal to the ^^ side B ; draw FG parallel to DE, gl- and EG parallel to DF; then DFGE is the parallelogram required. BOOK III. 91 For, the opposite sides are parallel by construction ; and consequently, the figure is a parallelogram (D. 28); it is also formed with the given sides and given angle. PROBLEM Xin. To find the centre of a given oircwmference or -arc. Take any three points A, B, and C, on the circumference or arc, and join them by the chords AB, BC ; bisect these chords by the perpendiculars DE and FG : then their point of intersection, 0, is the centre required (P. VII.). Scholium. The same construc- tion enables us to pass a circumference through any three points not in a straight line. If the points are vertices of a triangle, the circle is circumscribed about it. PROBLEM XIV. Through a given point, to draw a tangent to a given circle. There may be two cases : the given point may lie on the circumference of the given circle, or it may lie with- out the given circle. 1°. Let C be the centre of the given circle, and A a point on the cir- cumference, through which the tangent is to be drawn. Draw the radius CA, and at A draw AD perpendicular to AC : then AD is the tangent required (P. IX.). 92 GEOMETRY. 2°. Let C be the centre of the given circle, and A a point without the circle, through "which the tangent is to be drawn. Draw the line AC ; bisect it at 0, and from as a centre, with a radius OC, describe the circumference ABCD ; join the point A with the points of intersection D and B : then both AD and AB are tangent to the given circle and there are two solutions. For, the angles ABC and ADC are right angles (P. XVIIL, C. 2) : hence, each of the lines AB and AD is per- pendicular to a radius at its extremity ; and consequently, they are tangent to the given circle (P. IX.). Corollary. The right-angled triangles ABC and ADC, have a common hypothenuse AC, and the side BC equal to DC ; and consequently, they are equal in all respects (B. I., P. XVII.) : hence, AB is equal • to AD, and the angle CAB is equal to the angle CAD. The tangents are therefore equal, and the line AC bisects the angle between them. PROBLEM XV. To inscribe a circle in a given triangle Let ABC be the given tri- angle. Bisect the angles A and B, by the lines AG and BO, meet- ing in the point (Prob. V.) ; from the point let fall the BOOK III. 93 perpendiculars OD, OE, OF, on the sides of the triangle: these perpendiculars are all equal. For, in the triangles BOD and BOE, the angles OBE and OBD are equal, by construction; the angles ODB and OEB are equal, because each is a right angle ; and conse- quently, the angles BOD and BOE are also equal (B. I., P. XXV., C. 2), and the side OB is common; and there- fore, the triangles are equal in all respects (B. I., P. VI.) : hence, OD is equal to OE. In like manner, it may be shown that OD is equal to OF. From as a centre, with a radius OD, describe a circle, and it wUl be the circle required. For, each side is perpendicular to a radius at its extremity, and is there- fore tangent to the circle. Corollary. The lines that bisect the three angles of a triangle all meet in one point. PROBLEM XVI. On a given straight line, to construct a segment that shall contain a given angle. Let AB be the given line. Produce AB towards D ; at B construct the angle DBE equal to the given angle ; draw BO perpendicular to BE, 94 GEOMETRY. and at the middle point G, of AB, draw GO perpendicular to AB; from their point of intersection 0, as a centre, with a radius OB, describe the arc AMB: then the seg- ment AMB is the segment required. For, the angle ABF, equal to EBD, is measured by hall of the arc AKB (P. XXL); and the inscribed angle AMB is measured by half of the same arc: hence, the angle AMB is equal to the angle EBD, and consequently, to the given angle. Note. — ^A quadrant or quarter of a circumference, as CD, is, for convenience, divided into 90 equal parts, each of which is called a degree. A degree is denoted by the symbol ° ; thus, 25° is read 25 degrees, etc. Since a quad- rant contains 90°, the whole circumfer- ence contains 360°. A right angle, as CAD, which is the unit of measure for angles, being measured by a quadrant (P. XVn., S.), is said to be an angle of 90°; an angle which is one third of a right angle is an angle of 30°; an angle of 120° is -W or f of a right angle, etc. BOOK III. 95 EXERCISES. 1. Draw a circumference of given radius through two given points. 2. Construct an equilateral triangle, having given one of its sides. 3. At a point on a given straight line, construct an angle of 30°. 4. Through a given point without a given line, draw a line forming with the given line an angle of 30°. 5. A line 8 feet long is met at one extremity by a second line, making with it an angle of 30°; find the centre of the circle of which the first line is a chord and the second a tangent. 6. How many degrees in an angle inscribed in an arc of 135°? 7. How many degrees in the angle formed by two secants meeting without the circle and including arcs of 60° and 110°? 8. At one extremity of a chord, which divides the cir- cumference into two arcs of 290° and 70° respectively, a tangent is drawn; how many degrees in each of the angles formed by the tangent and the chord? 9. Show that the sum of the alternate angles of an inscribed hexagon is equal to four right angles. 10. The sides of a triangle are 3, 5, and 7 feet; con- struct the triangle. 11. Show that the three perpendiculars erected at the middle points of the three sides of a triangle meet in a common point. 12. Construct an isosceles .triangle with a given base and a given vertical angle. 13. At a point on a given straight line, construct an angle of 45° 96 GEOMETBT. 14. Construct an isosceles triangle so that the base shall be a given line and the vertical angle a right angle. 15. Construct a triangle, having given one angle, one of its including sides, and the difiference of the two other sides. 16. From a given point. A, without a circle, draw two tangents, AB and AC, and at any point, D, in the included arc, draw a third -tangent and produce it to meet the two others; show that the three tan- gents form a triangle whose perimeter is constant. 17. On a straight line 5 feet long, con- struct a circular segment that shall contain an angle of 30°. 18. Show that parallel tangents to a circle include semi-circumferences between their points of contact. 19. Show that four circles can be drawn tangent tC' three intersecting straight lines. BOOK IV, MEASUREMENT AND RELATION OF POLYGONS. DEFINITIONS. 1. Similar Polygons are polygons whicli are mutually equiangular, and which have the sides about the equal angles, taken in the same order, proportional. 2. In similar polygons, the parts which are similarly placed in each, are called homologous. The corresponding angles are homologous angles, the corresponding sides are homologous sides, the corresponding diagonals are homologous diagonals, and so on. 3. Similar Arcs, Sectors, or Segments, in different cir- cles, are those which correspond to equal angles at the centre. Thus, if the angles A and are ^ equal, the arcs BFC and DGE are simi- lar, the sectors BAG and DOE are / v p/ »p similar, and the segments BFC and B^^ > C G DGE are similar. 4. The Altitude of a Triangle is the perpendicular distance from the vertex of any angle to the opposite side, or the opposite side pro- duced. The vertex of the angle from which the distance is measured, is called the vertex of the triangle, and the opposite side is called the base of the triangle. 98 GEOMETBY. 5. The Altitude of a Parallelogram is the perpen- dicular distance between two opposite sides. These sides are called bases j one the upper, and the other, the lower base. 6. The Altitude of a Trapezoid is the perpendicular / distance between its parallel sides. These sides are called bases; one the upper, and the other, the lower base. 7. The Area of a Surface is its numerical value expressed in terms of some other surface taken as a unit. The unit adopted is a square described on the linear unit as a side. PROPOSITION I. THEOREM. Parallelograms which have equal bases and equal altitudes, are equal. Let the parallelograms ABCD and EFGH have equal bases and equal altitudes: then the parallelograms are equal. For, let them be so placed that their lower bases shall coincide ; then, because they have the same altitude, their upper bases will be in the same line DG, parallel ,to AB. The triangles DAH and CBG, have the sides AD and BC equal, because they are opposite sides of the parallel- ogram AC (B. I., P. XXVm.) ; the sides AH and BG equal, because they are opposite sides of the parallelogram AG ; the angles DAH and CBG equal, because their sides are BOOK IV. 99 parallel and lie in the same direction (B. I., P. XXIV.) : hence, the triangles are equal (B. I., P. V.). If from the quadrilateral ABGD, we take away the tri- angle DAH, there will remain ^he parallelogram AG ; if from the same quadrilateral ABGD, we take away the tri- angle CBG, there will remain the parallelogram AC : hence, the parallelogram AC is equal to the parallelogram EG (A. 3); which ipas to he proved. PROPOSITION II. THEOREM. A triangle is equal to one half of n parallelogram having an equal idse and an equal altitude. Let the triangle ABC, ai^d the parallelogram ABFD, have equal bases and equal altitudes : then the triangle is equal to one half of the parallelogram. ' For, let them be so q ^ p ^ placed that the base of V 7 \-^^/ ^^-^/ the triangle shall coincide \ /^^"""^ \/ ^^^^^^^^f with the lower base of * ^ A B the parallelogram ; then, be- cause they have equal altitudes, the vertex of the triangle will lie in the upper base of the parallelogram, or in the prolongation of that base. From A, draw AE parallel to BC, forming the parallel- ogram ABCE. This parallelogram is equal to the parallel- ogram ABFD, from Proposition I. But the triangle ABC is equal to half of the parallelogram ABCE (B. I., P. XXVni., C. 1) I hence, it is equal to half of the parallelogram ABFD (A. 7) ; which was to be proved. Cor. Triangles having equal bases and equal, altitudes are equal, for they are halves of equal parallelograriis. 100 GEOMETRY. PROPOSITION m. THEOREM. Rectangles having equal altitudes, are proportional to their bases. There may be two cases : the bases may be commen- surable, or they may be incommensurable. 1°. Let ABCD and HEFK, be two rectangles whose alti- tudes AD and HK are equal, and whose bases AB and HE are commensurable : then the areas of the rectangles are proportional to their bases. D C K F A E H E Suppose that AB is to HE, as 7 is to 4. Conceive AB to be divided into 7 equal parts, and HE into 4 equal parts, and at the points of division, let perpendiculars be drawn to AB and HE. Then will ABCD be divided into 7, and HEFK into 4 rectangles, all of which are equal, because they have equal bases and equal altitudes (P. I.) : hence, we have, ABCD : HEFK : : 7 : 4. But we have, by hypothesis, AB : HE : : 7 : 4. From these proportions, we have (B. 11., P. TV.), ' ABCD : HEFK : : AB : HE. Had any other numbers than 7 and 4 been used, the same proportion would have been found; which was to be proved. D F K C A ; OB BOOK IV. 101 2°. Let the bases of the rectangles be incommensiira- hle : then the rectangles are proportional to their bases. For, place the rectangle HEFK upon the rectangle ABCD, so that it shall take the position AEFD. Then, if the rectangles are not proportional to their bases, let us suppose that ABCD : AEFD : : AB : AO ; in which AO is greater than AE. Divide AB into equal parts, each less than OE ; at least one point of division, as I, will fall between E and 0; at this point, draw IK perpendicular to AB. Then, because AB and Al are com- mensurable, we shall have, from what has just been shown, ABCD : AIKD : : AB : Al. The above proportions have their antecedents the sanae in each; hence (B. II.„ P. IV., 0.), AEFD : AIKD : : AO : Al. The rectangle AEFD is less than AIKD ; and if the above proportion were true, the line AO would be less than Al ; whereas, it is greater. The fourth term of the proportion, therefore, cannot be greater than AE. In like manner, it may be shown that it cannot be less than AE ; conse- quently, it must be equal to AE : hence, ABCD : AEFD : : AB : AE ; which was to be proved. Cor. If rectangles have equal bases, they are to each other as their altitudes. 102 GEOMETilT. PEOPOSITION IV. THEOREM. Any two rectangles are to each other as the products of their bases and altitudes. Let ABCD and AEGF be two rectangles: then ABCD is to AEGF, as ABxAD is to AExAF. For, place the rectangles so that A P the angles DAB and EAF shall be opposite or vertical ; then, produce ^ L IB the sides CD and GE till they meet in H. G The rectangles ABCD and AD HE have the same altitude AD : hence (P. III.), ABCD : ADHE : : AB : AE. The rectangles ADHE and AEGF have the same altitude AE : hence, ADHE : AEGF : : AD : AF. Multiplying these proportions, term by term (B. 11., P. XEL), and omitting the common factor ADHE (B, JL, P. Vn.), we have, ABCD : AEGF :: ABxAD : AExAF; which was to he proved. Cor. If we suppose AE and AF, each to be equal to the Linear imit, the rectangle AEGF is the superficial vmit, and we have, ABCD : 1 :: ABxAD : 1; ABCD = ABxAD: BOOK IV. 103 hence, the area of a rectangle is equal to the product of its base and altitude; that is, the number of superficial units in the rectangle, is equal to the product of the number of linear units in its base by the number of linear units in its altitude. The product of two lines is sometimes called the rectangle of the lines, because the product is equal to the area of a rectangle constructed with the lines as sides. PROPOSITION Y. THEOREM. The area of a parallelogram is equal to the product of its base and altitude. Let ABCD be a parallelogram, AB its base, and BE its altitude : then the area of ABCD is equal to AB x BE. For, construct the rectangle ABEF, having the same base and altitude : then will the rectangle be equal to the parallelogra,m (P. I.) ; but the area of the rectangle is equal to AB x BE : hence, the area of the parallelogram is also equal to AB x BE ; which was to be proved. Cor. Parallelograms are to each other as the products of their bases and altitudes. If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes. 104 GEOMETRY. PROPOSITION VI. THEOREM. The area of a triangle is eqaal to half the product of its base and altitude. Let ABC be a triangle, BC its base, and AD its altitude : then its area is equal to |BCxAD. For, from C, draw CE parallel to BA, and from A, draw AE parallel to BC. The area of the parallelogram BCEA is BCxAD (P. V.); but the triangle ABC is half of the parallel- ogram BCEA : hence, its area is equal to J^BC x AD ; which was to be proved. Cor. 1. Triangles are to each other, as the products of their bases and altitudes (B. II., P. VII.). If the altitudes are equal, they are to each other as their bases. If the bases are equal, they are to each other as their altitudes. Cor. 2. The area of a triangle is equal to half the product of its perimeter and the radius of the inscribed circle. For, let DEF be a circle in- scribed in the triangle ABC. Draw OD, OE, and OF, to the points of contact, and OA, OB, and OC, to the vertices. The area of OBC is equal to |0E X BC ; the area of OAC is equal to |0F x AC ; and the area of OAB is equal to |ODxAB; and since OD, OE, and OF, are equal, the area of the triangle ABC (A. 9), is equal to iOD (AB + BC -|- CA). BOOK IV. 105 PROPOSITION Vn. THEOREM. . The area of a trapezoid is equal to the product of its alti- tude and half the sum of its parallel sides. Let ABCD be a trapezoid, DE its altitude, and AB and DC its parallel sides : then its area is equal to DE x i (AB + DC). For, draw the diagonal AC, forming the triangles ABC and ACD. The alti- tude of each of these triangles is equal to DE. The area of ABC is equal to iAB X DE (P. VI.) ; the area of ACD is equal to ^DC x DE : hence, the area of the trapezoid, which is the sum of the triangles, is equal to the sum of |-AB x DE and ^DC x DE, or to DE X ^ (AB + DC) ; which was to be proved. Scholium. Through I, the middle point of BC, draw IH parallel to AB, and LI parallel to AD, meeting DC produced, at K. Then, since Al and HK are parallelograms, we have AL = HI = DK ; and therefore, HI = ^ (AL + DK). But since the triangles LIB and CIK are equal in all respects, LB = CK ; hence, AL + DK = AB + DC ; and we have HI = I (AB + DC) : hence, The area of a trapezoid is equal to its altitude multi- plied by the line which connects the middle points of its inclined sides. PROPOSITION Vm. THEOREM. The square described on the sum of two lines is' equal to the sum of the squares described on the lines, increased by twice the rectangle of the lines. 106 GEOMETBT. Let AB and BC be two lines, and AC their sum : then AC* = AB* H- BC* + 2AB X BC. On AC, constniot the square AD ; from B, draw BH parallel to AE ; lay off AF equal to AB, and from F, draw FG parallel to AC : then IG and IH are each equal to BC ; and IB and IF, to AB. The square ACDE is composed of four parts. The part ABIF is a square described on AB; the part IGDH is equal to a square described on BC; the part BCGI is equal to the rectangle of AB and BC ; and the part FIHE is also equal to the rectangle of AB and BC : hence, we have (A. 9), A? = AB^ + BC^ + 2AB X BC ; which was to be proved. Cor. If the lines AB and BC are equal, the four parts of the square on AC are also equal : hence, the square descrUted on a line is equal to four times the square described on half the line. PROPOSITION IX. THEOREM. The square described on the difference of two lines is equal to the sum of the squares described on the lines, dimin- ished by twice the rectangle of the lines. Let AB and BC be two lines, and AC their difference ; then AC* = AB' + BC* - 2AB X BC. On AB construct the square ABIF; from C draw CG parallel to Bl ; lay off CD equal to AC, and from D draw DK parallel and equal to BA ; complete the square EFLK ; BOOK IV. 107 then EK is equal to BC, and EFLK is equal to the square of BC. The whole figure ABILKE is equal to the sum of the squares described on AB and BC. The part CBIG is equal to the rectangle of AB and BG; the part DGLK is also equal to the rectangle of AB and BC. If from the whole figure ABILKE, the two parts CBIG and DGLK be taken, there will remain the part ACDE, which is equal to the square of AC: hence, AC' = AB' + BC* - 2AB x BC ; which was to be proved. L F G 1 E D A C B PROPOSITION X. THEOREM. The rectangle contained by the sum and difference of two lines, is equal to the difference of their squares. Let AB and BC be two lines, of which AB is the greater : then (AB 4- BC) (AB - BC) = AB* - BC'. On AB, construct the square ABIF; prolong AB, and make BK equal to BC ; then AK is equal to AB + BC ; from K, draw KL parallel to Bl, and make it equal to AC; draw LE par- allel to KA, and CG parallel to Bl : then DG is equal to BC, and the figure DHIG is equal to the square on BC, and EDGF is equal to BKLH. B K 108 GEOMETRT. If we add to the figure ABHE, the rectangle BKLH, we have the rectangle AKLE, which is equal to the rectangle of AB + BC and AB — BC. If to the same figure ABHE, we add the rectangle DGFE, equal to BKLH, we have the figure ABHDGF, which is equal to the differ- ence of the squares of AB and BC. But the sums of equals are equal (A. 2), hence, (AB + BC) (AB - BC) = AB' - BC' which was to be proved. C B K PROPOSITION XL THEOREM. The square described on the hypothenuse of a right-angled triangle, is equal to the sum of the squares described on the two other sides. Let ABC be a triangle, right- angled at A : then BC' = AB' -I- AC*. Construct the square BG on the side BC, the square AH on the side AB, and the square Al on the side AC ; from A draw AD perpendicular to BC, and prolong it to E : then DE is parallel to BF; draw AF and HC. In the triangles HBC and ABF, we have HB equal to AB, because they are sides of the same square ; BC equal BOOK IV. 109 to BF, for the same reason, and the included angles HBC and ABF equal, because each is equal to the angle ABC plus a right angle : hence, the triangles are equal in aU respects (B. I., P. V.). The 'triangle ABF, and the rectangle BE, have the same base BF, and because DE is the prolongation of DA, their altitudes are equal: hence, the triangle ABF is equal to half the rectangle BE (P. II.). The triangle HBC, and the square BL, have the same base BH, and because AC is the prolongation of LA (B. I, P. IV.), their altitudes are equal: hence, the triangle HBC is equal to half the square of AH. But, the triangles ABF and HBC are equal: hence, the rectangle BE is equal to the square AH. In the same manner, it may be shown that the rectangle DG is equal to the square Al : hence, the sum of the rectangles BE and DG, or the square BG, is equal to the sum of the squares AH and Al ; or, BC' = AB* + AC' ; which was to be proved. Cor. 1. The square of either side about the right angle is equal to the square of the hypothenuse dimin- ished by the square of the other side : thus, AB* = BC' - AC' ; or, AC= = BC' - AB'. Cor. 2. If from the vertex of the right angle, a per- pendicular be drawn to the hypothenuse, dividing it into two segments, BD and DC, the square of the hypothenuse is to the square of either of the other sides, as the hypoth- enuse is to the segment adjacent to that side. For, the square BG, is to the rectangle BE, as BC to BD (P. in.) ; but the rectangle BE is equal to the square AH : hence, BC* : AB' : : BC : BD. 110 GEOMETRY. In like manner, we have, BC* : AC" : : BC : DC. Cor. 3. The squares of the sides about the right angle are to each other as the adjacent segments of the hypothenuse. For, by combining " the propor- tions of the preceding corollary (B. n., P. IV., C), we have. AB' AC" BD DC. Cor. 4. The square described on the diagonal of a square is double the given square. . For, the square of the diagonal is equal to the sum of the squares of the two sides; but the square of each side is equal to the given square : hence. H 1 3 G A / \ \ / G E B F AC' = 2AB" ; or. AC = 2BC* Cor. 5. From the last corollary, we have, AC' : AB" : : 2 : 1 ; hence, by extracting the square root of each term, we have, AC : AB : : V2 : 1 ; that is, the diagonal of a square is to the side, as the square root of two is to one; consequently, the diagonal and the side of a square are incommensurable. BOOK IV. Ill PROPOSITION XII. THEOREM. In any triangle, the square of a side opposite an acute angle, is equal to the sum of the squares of the base and the other side, diminished by twice the rectangle of the base and the distance from, the vertex of the acute angle to the foot of the perpendicular drawn from the vertex of the opposite angle to the base, or to the base produced. Let ABC be a triangle, C one of its A acute angles, BC its base, and AD the per- pendicular drawn from A to BC, or BC produced ; tben AB' = BC' + AC' - 2BCxCD. B D~~C For, whether the perpendicular meets the base, or the base produced, we have BD equal to the k difference of BC and CD : hence (P. IX.), BD' = BC* + CD' - 2.BC X CD. Adding AD' to both members, we have, ^ E BD' + ad' = BC* + CD' + ad' - 2BC xCD. But, BD' + AD' = AB', and CD' + AD' = AC' : hence, fB* = BC' + AC' - 2BCxCD; which was to he proved. 112 GEOMETRY. PROPOSITION Xin. THEOREM. In any dbtiuse-anglecL triangle, the square of the side opposite the Muse angle is equal to the sum of the squares of the base and the other side, increased by twice the rect- angle of the base and the distance from the vertex of the obtuse angle to the foot of the perpendicular drawn from the vertex of the opposite angle to the base produced. Let ABC be an obtuse-angled triangle, B its obtuse angle, BC its base, and AD tbe ■ a perpendicular drawn from A to BC produced; then AC" = BC' + AB' + 2BC X BD. d"-| ^=-C For, CD is the sum of BC and BD : hence (P. VIII.), CD^ = BC" + BD" + 2BC X BD. Adding AD to both members, and reducing, we' have, AC" = BC* + AB* + 2BC x BD ; which was to he proved. Scholium. The right-angled triangle is the only one in which the sum of the squares described on two sides is equal to the square described on the third side. PROPOSITION XrV. THEOREM. In any triangle, the sum of the squares described on two sides is equal to twice the square of half the third side, increased by twice the square of the line draum from the middle point of that side to the vertex of the opposite angle. Let ABC be any triangle, and EA a line drawn from the middle of the base BC to the vertex A : then BOOK IV. lis AB" + AC** = 2BE' + 2EA» ' Draw AD perpendicular to BC ; then, from Proposition XIL, we have, AC" = EC* + EA* - 2EC X ED. * From Proposition XIII., we have, AB" = BE' + EA* + 2BE X ED. Adding these equations, member to member (A. 2), recol- lecting that BE is equal to EC, we have, AB' + AC* = 2BE* + 2EA' ; which was to be proved. t Cor. Let ABCD be a parallelogram, and BD, AC, its diagonals. Then, since the diagonals mutually bisect each other (B. L, P. g g XXXI.), we have, AB* -j-, BC* = 2AE* + 21? ; and, CD' + DA" = 2CE' + 2 DE* ; whence, by addition, recollecting that AE is equal to CE, and BE to DE, we have, AB" + BC' + CD* + DA' = 4CE' + 4DE' ; but, 4CF is equal to AC', and 4DE' to BD' (P. VIII., C.) : hence, AB' + BC' + CD' + DA' = AC' + BD'. That is, the sum of the squares of the sides of a parallelo- gram, is equal to the .sum of the squares of its diagonals. 114 GEOMETRY. \^ \ L.aMxi PROPOSITION XV. THEOREM. M,^,^".,^ In, any triangle, a line drawn parallel to the base divider the other sides proportionaUy. Let ABC be a triangle, and DE a line parallel to the base BC : then AD : DB : : AE : EC. Draw EB and DC. Then, because the triangles AED and DEB have their bases in the same line AB, and their vertices at the same point E, they have a common altitude: hence (P. VI., C), AED DEB AD DB. The triangles AED and EDC, have their bases in the same line AC, and their vertices at the same point D ; they have, therefore, a common altitude; hence. AED EDC AE EC. But the triangles DEB and EDC have a common base DE, and their vertices in the line BC, parallel to DE : they are, therefore, equal: hence, the two preceding proportions have a couplet in each equal; and consequently, the re- maining terms are proportional (B. IL, P. IV.), hence. AD : DB which was to be proved. AE EC; Cor. 1. We have, by composition (B. H, P. VI.), AD + DB : AD : : AE + EC : AE ; or, AB and, in like manner, AB BOOK IV. AD : : AC 116 DB AE AE; EC. Cor. 2. n any number of parallels be drawn cutting two lines, they divide the lines propor- tionally. •For, let be the point where AB and CD meet. In the triangle OEF, the line AC being parallel to the base EF, we have, OE : AE : : OF : CF. In the triangle OGH, we have, OE : EG : : OF : FH ; hence (B. II., P. IV., C), AE : EG : : CF : FH. In like manner, EG : GB :: FH : HD; and so on. PROPOSITION XVI. THEOREM. If a straight line divides two sides of a triangle propor- tionally, it is parallel to the third side. Let ABC be a triangle, and let DE di- vide AB and AC, so that AD : DB : : AE : EC ; then DE is parallel to BC. Draw DC and EB. Then the triangles B 116 GBOMETKY. ADE and DEB have a common altitude ; and consequently, we have, ADE : DEB : : AD : DB. The triangles ADE and EDC have also a common altitude; and consequently, we have, ADE : EDC : : AE : EC ; but, by hypothesis, AD : DB : : AE : EC ; hence (B. II., P. IV.), ADE : DEB : : ADE : EDC. The antecedents of this proportion being equal, the consequents are equal ; that is, the triangles DEB and EDC are' equal. But these triangles have ,a common base DE : hence, their altitudes are equal (P. VI., 0.) ; that is, the points B and C, of the line BC, are equally distant from DE, or DE prolonged: hence, BC and DE are parallel (B. I., P. XXX., C.) ; which was to he proved. PROPOSITION XVn. THEOREM. In any triangle, the straight line which bisects the angle at the vertex, divides the base into two segments propor- tional to the adjacent sides. Let AD bisect the vertical angle A of the triangle BAC : then the segments BD and DC are proportional to the adjacent sides BA and CA. From C, draw CE parallel to DA, and produce it until BOOK IV. 117 it meets BA prolonged, at E. Then, because CE and DA are parallel, the angles BAD and AEC are equal (B. I., P. XX., C. 3); the angles DAC and ACE are also equal (B. I., P. XX.," 0. 2). But, BAD and DAC are equal, by hypothesis ; consequently, AEC and ACE are equal : hence, the triangle ACE is isosceles, AE being equal to AC. In the triangle BEC, the line AD is parallel to the base EC : hence (P. XV.), BA : AE : : BD . : DC ; or, substituting AC for its equal AE, BA : AC : : BD : DC ; which was to he proved. PROPOSITION XVIII. THEOREM. Triangles which are mutually equiangular, are similar. Let the triangles ABC and DEF have the angle A equal to the angle D, the angle B to the angle E, and the angle C to the angle F: then they are similar. For, place the triangle DEF upon the triangle ABC, so that .the angle E shall coincide with the angle B ; then will the point F fall at some point H, of BC; the point D at some point G, of BA; 118 GEOMETRY. the side DF -will take the position GH, and BGH will be equal to EDF. Since the angle BHG is equal to BCA, GH will be paraUel to AC (B. L, P. XIX., C. 2) ; and consequently, we have (P. XV.), B iTt BA : EG : : BC : BH ; or, since BG is equal to ED, and BH to EF, BA : ED : : BC : EF. In like manner, it may be shown that BC : EF : : CA : FD ; and also, CA : FD : : AB : DE ; hence, the sides about the equal angles, taken in the same order, are proportional ; and consequently, the triangles are similar (D. 1); which was to be proved. Cor. If two triangles have two angles in one, equal to two angles in the other, each to each, they are similar (B. I, P. XXV., C. 2). PROPOSITION XIX. THEOREM. Triangles which have their corresponding sides proportional, are similar. In the triangles ABC and DEF, let the corresponding sides be proportional ; that is, let 119 CA FD; BOOK IV. BA : ED : : BC : EF then the triangles are similar. For, on BA lay off BG equal to ED ; on BC lay off BH equal to EF, and draw GH. Then, because BG is equal to ED, and BH to EF, we have, BA : BG : : hence, GH is parallel to AC (P. XVI.) ; and consequently, the triangles BAC and BGH are equiangular, and therefore similar :\/hence, BH CA BC But, by hypothesis, BC : EF : : CA hence (B. n., P. IV., C), we have, BH : EF : : HG HG. FD; FD. The. But, BH is equal to EF; hence, HG is equal to FD. xnejOvj, triangles BHG and EFD have, therefore, their sides equal,^^ . each to each, and consequently, they are equal in all re- ' spepts. Now, ,jit has just been shown that, BHG ajadjvBCA"' are^ similar I 'hence, EFD and BCA are also IfeimilarPttdiicA was to be proved. '#■■ '^■■ .W ■. A\ \xv SchoUum. In order that polygons may be similar, they must fulfill two conditions : they must be mutually equi- angular, and the corresponding sides must be proportional. In the case of triangles, either of these conditions involves the other, which is not true of any other species of polygons. 120 GEOMETRY. PEOPOSITION XX. THEOREM. Triangles which have an angle in each equal, and the in- cluding sides proportional, are similar. In the triangles ABC and DEF, let the angle B be equal to the angle E ; and suppose that BA : ED : : BC : EF ; then the triangles are similar. For, place the angle E upon its equal B ; F -will fall at some point of BC, as H ; D will fall at some point of BA, as G ; DF will take the position GH, and the triangle DEF will coincide with GBH, and consequently, is equal to it. But, from the assumed proportion, and because BG is equal to ED, and BH to EF, we have, BA : BG : : BC : BH ; hence, GH is parallel to AC; and consequently, BAG and BGH are mutually equiapgular, and therefore similar. But, EOF is equalXto BGH: hence, it is \ also similar to BAC ; which was to be proved. . , PROPOSITION XXI. THEOREM. L Triangles which have their sides parallel, each to 'each, or perpendicular, each to each, are similar. 1°. Let the triangles ABC and DEF have the side AB parallel to DE, BC to EF, and CA to FD ; then they are similar. BOOK IV. 121 For, since the side AB is parallel to DE, and BC to EF, the angle B is equal to the angle E (B. I., P. XXIV.) ; in like manner, the angle C is equal to the angle F, and the angle A to the angle D ; the triangles are, therefore, mutually equiangular, and con- sequently, are similar (P. XVIH.) ; which was to he proved. 2°. Let the triangles ABC and DEF have the side AB perpendicular to DE, BC to EF, and CA to FD : then they are similar. For, prolong the sides of the tri- angle DEF tiU they meet the sides of the triangle ABC. The sum of the interior angles of the quadrilateral BIEG is equal to four right angles (B. L, P. XXVI.); but, the angles EIB and EGB are each right angles, by hypothesis; hence, the simi of the angles I EG, IBG is equal to two right angles; the sum of the angles I EG and DEF is equal to two right angles, because they are adjacent; and since things which are equal to the same thing are equal to each other, the sum of the angles I EG and IBG is equal to the sum of the angles I EG and DEF; or, taking away the common part I EG, we have the angle IBG equal to the angle DEF. In like manner, the angle GCH may be proved equal to the angle EFD, and the angle HAI to the angle EDF; the triangles ABC and DEF are, therefore, mutually equiangular, and consequently similar ; which was to he proved. Cor. 1. In the first case, the parallel sides are homolo- 122 GEOMETBY. gOTis; in the second case, the perpendicular sides are homologous. Cor. 2. The homologous angles are those included by- sides respectively parallel or perpendicular to each other. Scholium. When two triangles have their sides perpen- dicular, each to each, they may have a different relative position from that shown in the figure. But we can always construct a triangle within the triangle ABC, whose sides shall be parallel to those of the other triangle, and then the demonstration will be the same as above. PROPOSITION XXn. THEOEEM. If a straight line is drawn parallel to the base of a tri- angle, and straight lines are drawn from the vertex of the triangle to points of the base, these lines divide the base and the parallel proportionally. Let ABC be a triangle, BC its base, A its vertex, DE parallel to BC, and AF, AG, AH, lines drawn from A to poiuts of the base : then Dl BF FG KL GH LE HC. For, the triangles AID and AFB, being similar (P. XXI.), we have, Al AF Dl BF; and, the triangles AIK and AFG, being similar, we have, Al : AF : : IK hence (B. 11., P. IV.), we have, FG BOOK IV. ' Dl : BF : : IK FG. In like manner, IK and, KL : FG :: GH :: KL LE GH, CH; hence (B. II., P. IV.), Dl : BF : : IK : which was to be proved. FG KL : GH LE HC; Cor. If BC is divided into equal parts at F, G, and H, then DE is divided into equal parts, at I, K, and L. PROPOSITION XXIII. THEOREM. If, in a right-angled triangle, a perpendicular is drawn from the vertex of the right angle to the hypothenuse : 1°. The triangles on each side of the perpendicular are similar to the given triangle, and to each other : 2°. Each side about the right angle is a mean propor- tional between the hypothenuse and the adjacent segment: 3°. The perpendicular is a mean proportional between the two segments of the hypothenuse. 1°. Let ABC be a right-angled triangle, A the vertex of the right angle, BC the hypothe- nuse, and AD perpendicular to BC : then ADB and ADC are similar to ABC, and consequently, similar to each other. The triangles ADB and ABC have the anerle B common, and the angles ADB and BAC equal, 124 GEOMETRY. because each is a right angle ; they are, therefore, simi- lar (P. XVIII., C). In like manner, it may be shown that the triangles ADC and ABC are similar ; and since ADB and ADC are each similar to ABC, they are similar to each other ; which was to be proved. 2°. AB is a mean proportional between BC and BD ; and AC is a mean proportional between CB and CD. For, the triangles ADB and BAC being similar, their homologous sides are proportional: hence, BC : AB : : AB : BD. In like manner, BC : AC : : AC : DC ; which was to be proved. 3°. AD is a mean proportional between BD and DC. iPor, the triangles ADB and ADC being similar, their homol- ogous sides are proportional ; hence, BD : AD : : AD : DC ; which was to be proved. Cor. 1. From the proportions, BC : AB : : AB : BD, and, BC : AC : : AC : DC, we have (B. n., P. I.), AB^ = BCxBD, and, AC* = BC X DC ; BOOK IV. whence, by addition, AB' + AC* = BC (BD + DC) ; or, AB' -h AC' = BC^ ; as was shown in Proposition XL 125 Cor. 2. If from any point A, in a semi-circumference BAC, chords are drawn to the extremi- ties B and C of the diameter BC, and a perpendicular AD is drawn to the diameter: then ABC is a right-angled triangle, right-angled at A ; and from what was proved above, each chord, is a mean proportional between the diameter and the adja- cent segment; and, the perpendicular is a mean propor- tional between the segments of the diameter. PROPOSITION XXIV. THEOREM. Triangles which have an angle in each equal, are to each other as the rectangles of the including sides. Let the triangles GHK and ABC have the angles G anji A equal : then are they to each other as the rectangles of the sides about these angles. ^ For, lay off AD equal to GH, AE to GK, and draw DE; then the triangles ADE and GHK are equal in all respects. Draw EB. H 126 GEOMETRY. The triangles ADE and ABE have their bases in the same hne AB, and a common vertex E ; therefore, they have the same altitude, and consequently, are to each other as their bases; that is, ADE : ABE : : The triangles ABE and ABC, have their bases in the same line AC, and a common vertex B : hence, ABE : ABC : : multiplying these proportions, .term by term, and omitting the common factor ABE (B. 11., P. VII.), we have, ADE : ABC :: ADxAE : ABxAC; substituting for ADE, its equal, GHK, and for ADxAE, its equal, GHxGK, we have, GHK : ABC :: GHxGK : ABxAC, which was to be proved. Cor. If ADE and ABC are similar, the angles D and B being homologous, DE is parallel to BC, and we have, AD : AB : : AE : AC ; hence (B. IL, P. IV.), we have, ADE : ABE : : ABE : ABC ; that is, ABE is a mean proportional between ADE and ABC. BOOK IV. 127 PROPOSITION XXV. THEOREM. Similar triangles are to each other as the squares of their homologous sides. Let the triaiigles ABC and DEF be similar, the angle A being equal to the angle D, B to E, and C to F: then the triangles are to each other as the squares of any two homologous sides. Because the angles A and D are equal, we have (P. XXIV.), ABxAC : DExDF; ABC DEF and, because the triangles are similar, we have. AB DE AC DF; multiplying the terms of this proportion by the correspond- ing terms of the proportion, AC : DF : : we have (B. II., P. Xn.), ABxAC : DExDF AC DF, AC* DT*: combining this with the first proportion (B. 11., P. IV,), we have, ABC : DEF : : AC' : DP. In like manner, it may be shown that the triangles are to each other as the squares of AB and DE, or of BC and EF ; which was to he proved. 128 GEOMETRY. PROPOSITION XXVI. THEOREM. Similar polygons may he divided into the same number of triangles, similar, each to each, and similarly -placed. Let ABCDE and FGHIK be two similar polygons, the angle A being equal to the angle F, B to G, C to H, and so on : then can they be divided into the same number of similar triangles, similarly placed. For, from A draw the C diagonals AC, AD, and from [ ,.■■''' \ H F, homologous with A, draw J .•• \ I / \ the diagonals FH, FI, to the Nv ^^ fir ^\ vertices H and I, homolo- ^;^ \/'^ gous with C and D. Because the polygons are similar, the triangles ABC and FGH have the angles B and G equal, and the sides about these angles proportional; they are, therefore, simi- lar (P. XX.). Since these triangles are similar, we have the angle ACB equal to FHG, and the sides AC and FH, proportional to BC and GH, or to CD and HI. The angle BCD being equal to the angle GHI, if we take from the first the angle ACB, and from the second the .equal angle FHG, we have the angle ACD equal to the angle FHl : hence, the triangles ACD and FHl have an angle in each equal, and the including sides proportional ; they are there- fore similar. In like manner, it may be shown that ADE and FIK are similar ; which was to he proved. Cor. 1. The corresponding triangles in the two poly- gons are homologous triangles, and the corresponding diagonals are homologous diagonals. BOOK IV. 129 Any two homologous triangles are like parts of the polygons to which they belong. For, the homologous triangles being similar, we have, and, whence, In like manner, - hence, ABC FGH ABC ACD ABC ACD ACD WhpTWA-Jhy r-.r>mpQ«i.tirm ("R. II., P. ^, ABC : FGH : : ACD + ABC + ADE : that is, ABC : FGH : : ABCDE : : FGH : : Atf : : FHl : : AC^ : : FGH : : ACD : : FHl : : ADE : : FHl, : : ADE : FH"; FfP; FHl. FIK; FIK. FHl -t- FGH + FIf FGHIK. Cor.' 2. If two polygons are made up of similar tri- angles, similarly placed, the polygons themselves are similar. PROPOSITION XXVII. THEOREM. The perimeters of similar polygons are to each other as any two hom,ologous sides; and the polygons are to each other as the squares of any two homologous sides. 1°. Let ABCDE and FGHIK be similar polygons: then their perimeters are to each other as any two homologous sides. For, any two homologous sides, as AB and FG, are like parts of the perimeters to which they belong : hence (B. 11., P. IX.), the perim- eters of the polygons are to each other as AB to FG, or as any other two homolo- gous sides ; which was to he proved. 130 GEOMETEY. 2°. The polygons are to eacli other as the squares of any two homologous sides. For, let the polygons be divided into homologous tri- angles (P. XXVI., G. 1) ; then, because the homologous tri- angles ABC and FGH are like parts of the polygons to which they belong, the polygons are to each other as these triangles ; but these triangles, being similar, are to each , other as the squares of AB and FG : hence, the polygons are to each other as the squares of AB and FG, or as the squares of any other two homologous sides; which was to he proved. Cor. 1. Perimeters of similar polygons are to each other as their homologous diagonals, or as any other homologous lines ; and the polygons are to each other as the squares of their homologous diagonals, or as the squares of any other homologous lines. Cor. 2. If the three sides of a right-angled triangle are made homologous sides of three similar polygons, these polygons are to each other as the squares of the sides of the triangle. But the square of the hjrpothenuse is equal to the sum of the squares of the other sides, and consequently, the polygon on the hypothenuse will be equal to the sum of the polygons on the other sides. PROPOSITION XXVm. THEOREM. If two chords intersect in a circle, their segments are reciprocally proportional. Let the chords AB and CD intersect at : then are BOOK IV. 131 their segments reciprocally proportional; that is, one seg- ment of the first will be to one segment of the second, as the remaining segment of the second is to the remain- ing segment of the first. For, draw CA and BD. Then the angles ODB and OAC are equal, because each is measured by half of the arc CB (B. m., P. XVIII.). The angles OBD and OCA are also equal, because each is naeasured by half of the arc AD : hence, the triangles OBD and OCA are similar (P. XVIII., C), and consequently, their homologous sides are proportional : hence, DO : AO : : OB : OC ; which was to be proved. Cor. From the above proportion, we have, DOxOC = AOxOB; that is, the rectangle of the segments of one chord is equal to the rectangle of the segments of the other. PROPOSITION XXIX. THEOREM. If from, a point without a circle, two secants are drawn terminating in the concave arc, they are reciprocally proportional to their external segments. Let OB and OC be two secants terminating in the concave arc of the circle BCD : then OB ; OC : : OD : OA. 132 GEOMETRY. For, draw AC and DB. The triangles ODB and OAC have the angle common, and the angles OBD and OCA equal, because each is measured by half of the arc AD : hence, they are similar, and consequently, their homologous sides are proportional ; "whence, OB : OC : : OD which was to be proved. .OA; Cor. From the above proportion, we have, OBxOA = OCxOD; that is, tfie rectangles of each secant and its external seg- ment are equal. PROPOSITION XXX. THEOREM. If from a point without a circle, a tangent and a secant are drawn, the secant terminating in the concave arc, the tangent is a mean proportional between the secant and its external segment. Let ADC be a circle, OC a secant, and OA a tangent: then OC : OA : : OA : OD. For, draw AD and AC. The triangles OAD and OAC have the angle common, and the angles OAD and ACD equal, be- cause each is measured by half of the arc AD (B. m., P. XVm., P. XXL); the tri- angles are therefore similar, and conse- quently, their homologous sides are propor- tional: hence, oc which was to be proved. BOOK IV. OA : : OA 138 OD; Cor. From the above proportion, we have, AO' = OCxOD; that is, the square of the tangent is equal to the rectangle of the secant and its external segment. PRACTICAL APPLICATIONS. PROBLEM I. To divide a given straight line into parts proportional to given straight lines: also into equal parts. 1°. Let AB be a given straight hne, and let it be re- quired to divide it into parts proportional to the lines P, Q, and R. From one extremity A, draw the indefinite line AG, making any angle with AB ; lay off AC equal to P, CD equal to Q, and DE equal to R ; draw EB, and from the points C and D, draw CI and DF parallel to EB : then Al, IF, and FB, are proportional to P, Q, and R (P. XV., C.-2). 134 QEOMETBT. 2°. Let AH be a given straight line, and let it be required to divide it into any number of equal parts, say five. From one extremity A, draw the indefinite line AG ; take Al equal to any con- venient line, and lay off IK, KL, LM, and MB, each equal to Al. Draw BH, and from I, K, L, and M, draw the lines IC, KD, LE, and MF, parallel to BH : then AH is divided into equal parts at C, D, E, and F (P. XV., C. 2). PROBLEM II. To construct a fourth proportional to three given straight lines. Let A, B, and C, be the ^ given lines. Draw DE and DF, making any convenient angle with each other. Lay off DA equal to A, DB equal to B, and DC equal to C ; draw AC, and from B draw BX parallel to AC : then DX is the fourth proportional required. For (P. XV., C), we have. or, DA A DB B DC C DX; DX. Cor. If DC is made equal to DB, DX is a third pro- portional to DA and DB, or to A and B. BOOK IV. 135 PROBLEM m. To construct a mean proportional between two given straight lines. Let A and B be the given lines. On an indefinite line, lay off DE equal to A, and EF equal to B ; on DF as a diameter describe the semicircle DGF, and draw EG perpendicular to DF: then EG is the mean proportional required. For (P. XXni, 0. 2), we have, DE : EG : : EG : EF ; A : EG : : EG : B. PROBLEM IV. To divide a given straight line into two such parts, that the greater part shall he a mean proportional between the whole line and the other part. Let AB be the given line. At the extremity B, draw BC perpendicular to AB, and make it equal to half of AB. With C as a centre, and CB as a radius, describe the arc DBE ; draw AC, and produce it till it terminates in the concave arc at E ; with A as centre and AD as radius, describe the arc DF: then AF is the greater part required. 136 GEOMETRY. For, AB being perpendicular to CB at B, is tangent to the arc DBE : hence (P. XXX.), AE : AB : : AB : AD ; and, by division (B. H., P. VL), AE — AB : AB : : AB — AD : AD. But, DE is equal to twice CB, or to AB : hence, AE — AB is equal to AD, or to AF ; and AB — AD is equal to AB — AF, or to FB : hence, by substitution, AF : AB : : FB : AF ; and, by inversion (B. 11., P. V.), AB : AF : : AF : FB. Scholium. When a straight line is divided so that the greater segment is a mean proportional between the whole line and the less segment, it is said to be divided in extreme and mean ratio. Since AB and DE are equal, the line AE is divided in extreme and mean ratio at D ; for we have, from the first of the above proportions, by substitution. AE DE DE AD. BOOK IV. 137 PEOBLEM V. Through a given point, in a given angle, to draw a straight line so that the segments between the point and the sides of the angle shall he equal. Let BCD be the given angle, and A the given point. Through A, draw AE parallel to DC; lay off EF equal to CE, and draw FAD : then AF and AD are the segments re- quired. For (P. XV.), we have, FA : AD : : FE : EC ; but, FE is equal to EC ; hence, FA is equal to AD, PROBLEM VI. To construct a triangle equal to a given polygon. Let ABCDE be the given polygon. Draw CA ; produce EA, and draw EG parallel to CA ; draw the line CG. Then the triangles BAC and GAC have the common base AC, and because their vertices B and G lie in the same line BG parallel to the base, their altitudes are equal, and consequently, the triangles are equal : hence, the polygon GCDE is equal to the polygon ABCDE. Again, draw CE ; produce AE and draw DF parallel to CE ; draw also CF; then will the triangles FCE and DCE be equal : hence, the triangle GCF is equal to the polygon GCDE, and consequently, to the given polygon. In like manner, a triangle may be constructed equal to any other given polygon. 138 GEOMETRY. PROBLEM Vn. To construct a square equal to a given triangle. Let ABC be the given triangle, AD its altitude, and BC its base. Construct a mean propor- tional between AD and half of BC (Prob. m.). Let XY be that mean proportional, and on it, as a side, con- struct a square : then this is the square required. For, from the construction, XT IBCxAD = area ABC. Scholium. By means of Problems VI. and VII., a square may be constructed equal to any given polygon. PROBLEM Vin. On a given straight line, to construct a polygon similar to a given polygon. Let FG be the given line, and ABCDE the given poly- gon. Draw AC and AD. At F, construct the angle GFH equal to BAC, and at G the angle FGH equal to ABC; then FGH is similar to ABC (P. XVin. C). In like manner, construct the triangle FHI similar to ACD, and FIK similar to ADE ; then the polygon FGH IK is similar to the polygon ABCDE (P. XXVI.. C. 2). BOOK IV. 139 PROBLEM IX. To construct a square equal to the sum of two given squares; also a square equal to the difference of two given squares. 1°. Let A and B be the sides of the given squares, and let A be the greater. Construct a right angle CDE ; make DE equal to A, and DC equal to B ; draw CE, and on it construct a square : , this square will be equal to the sum of the given squares (P. XI.). 2°. Construct a right angle CDE. Lay off DC equal to B ; with C as a centre, and CE, equal to A, as a radius, describe an arc cutting DE at E ; draw CE, and on DE construct a square : this square will be equal to the difference of the given squares (P. XL, C. 1). Scholium. A polygon may be constructed similar to either of two given polygons, and equal to their sum or difference. For, let A and B be homologous sides of the given polygons. Find a square equal to the sum or difference of the squares on A and B ; and let X be a side of that square. On X as a side, homologous to A or B, construct a polygon similar to the given polygons, and it will be equal to their sum or difference (P. XXVIL, C. 2). 140 GEOMETRY. EXERCISES. 1. The altitude of an isosceles triangle is 3 feet, each of the equal sides is 5 feet ; find the area. 2. The parallel sides of a trapezoid are 8 and 10 feet, and the altitude is 6 feet ; what is the area ? 3. The sides of a triangle are 60, 80, and 100 feet, the diameter of the inscribed circle is 40 feet; find the area. 4. Construct a square equal to the sum of the squares whose sides are respectively 16, 12, 8, 4, and 2 units in length. /v'guVShow that the sum of the three perpendiculars drawn from any point within an equilateral triangle to the three -sides is equal to the altitude of the triangle. 6. Show that the sum of the squares of two lines, dra^^^ ^°^^ ^^y point in the circumference - of a circle to tw(y!points on the diameter of the circle equidistant from the centre, will be always the same. 7. The distance of a chord, 8 feet long, from the centre of a circle is 3 feet ; what is the diameter of the circle,? •'^■fe; Construct a triangle, having given the vertical angle, the line bisecting the base, and the angle which the bisecting line makes with thp, base. "' 9. Show that if a line bisecting the exterior vertical ~"" angle of a triangle is not par- allel to the base, the distances of the point in which it meets the base produced, from the extremities of the base, are pro- portional to the other two sides of the triangle. BOOK IV. 141 10. The segments made by a perpendicular, drawn from a point on the circumference of a circle to a diam- eter, are 16 feet and 4 feet; find the length of the perpendicular. 11. Two similar triangles, ABC and DEF, have the homologous sides AC and DF equal respectively to 4 feet and 6 feet, and the area of DEF is 9 square feet; find the area of ABC. 12. Two chords of a circle intersect; the segments of one are respectively 6 feet and 8 feet, and one segment of the ^ther is 1 2 feet ; find the remaining segment. 13. Two circles, whose radii are 6 feet and 10 feet, intersect, and the line joining their points of ■ intersection is 8 feet ; find the distance between their centres. 14. Find the area of a triangle whose sides are re- spectively 31, 28, and 20 feet. 15. Show that the area of an equilateral triangle is equal to one fourth the square of one side, naultiplied by VS ; or to the square of one side multiplied by .433. 16. From a point, 0, in an eqtiilateral triangle, ABC, the dis- tances to the vertices were measured and found to be : OB = 20, OA = 28, OC = 31 ; find the area of the triangle and the length of each side. [AD is made equal to OA, CD to OB, CF to OC, BF to OA, BE to OB, AE to OC] BOOK V. REGULAR POLYGON S.— ARE A OF THE CIRCLE. DEFmiTION. 1. A Regular Polygon is a polygon ■which is both equilateral and equiangular. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar. Let ABCDEF and abcdef be regular polygons of the same number of sides: then they are similar. For, the corresponding angles in each are equal, because any angle in either polygon is equal to twice as many right angles as the polygon has sides, less four right angles, divided by the number of angles (B. L, P. XXVI., C. 4) ; and fur- ther, the corresponding sides are proportional, because all the sides of either polygon are equal (D. 1) : hence, the polygons are similar (B. IV., D. 1) ; which was to be proved. BOOK V. 148 PEOPOSITION n. THEOREM. The circumference of a circle may he circumscribed aibout any regular polygon; a circle may also be inscribed in it. 1°. Let ABCF be a regular polygon: then can the cir- cumference of a circle be circumscribed about it. For, through three consecutive ver- tices A, B, C, describe the circumfer- ence of a circle (B. III., Problem XIII., ^^ S.). Its centre lies on PO, drawn / , perpendicular to BC, at its middle h( point P ; draw OA and OD. \ " / Let the quadrilateral OPCD be (j^^^ ^^^£ ' turned about the line OP, until PC jp falls on PB ; then, because the angle C is equal to B, the side CD will take the direction BA; and because CD is equal to BA, the vertex D, wiU faU upon the vertex A; and consequently, the line OD will coincide with OA, and is, therefore, equal to it : hence, the circumference which passes through A, B, and C, passes through D. In like manner, it may be shown that it passes through each of the other vertices: hence, it is cir- cumscribed about the polygon ; which, was to he proved. 2°. A circle may be inscribed in the polygon. For, the sides AB, BC, &c., being equal chords of the circumscribed circle, are equidistant from the centre 0; hence, a circle described from as a centre, with OP as a radius, is tangent to each of the sides of the polygon, and consequently, is inscribed in it; which was to he vroved. 144 GEOMETRY. Scholium. If the circumference of a circle is divided into equal arcs, the chords of these arcs are sides of a regular inscribed polygon. For, the sides are equal, because they are chords of equal arcs, and the angles are equal, because they are measured by halves of equal arcs. If the vertices A, B, C, &c., of a regular inscribed polygon be joined with the centre 0, the triangles thus formed -will be equal, because their sides are equal, each to each : hence, all of the angles about the point are equal to each other. DEFmiTIONS. 1. The Centre of a Regular Polygon is the common centre of the circumscribed and inscribed circles. 2. The Angle at the Centre is the angle formed by drawing lines from the centre to the extremities of any side. The angle at the centre is equal to four right angles divided by the number of sides of the polygon. 3. The Apothem is the shortest distance from the cen- tre to any side. The apothem is equal to the radius of the inscribed circle. BOOK V . 146 PEOPOSITION III. PROBLEM. To inscribe a square in a given circle. Let A BCD be the given circle. Draw any two diameters AC and BD perpendicular to each other; they divide the circumference into four equal arcs (B. Ill, P. XVII., S.). Draw the chords AB, BC, CD, and DA: then the figure ABCD is the square required (P. n., S.). Scholium. The radius is to the side of the inscriljpd square as 1 is to a/2. ,• ' , ^ , — ^> i^ ^ ' J ' .Ji\ PROPOSITION IV. L THEOREM. 7'/\ If a regular hexagon is inscribed in a circle, any side is equal to the radius of the circle. Let ABD be a circle, and ABCDEH a regular inscribed hexagon: then any side, as AB, is equal to the radius of the circle. Draw the radii OA and OB. Then the angle AOB is equal to one sixth of four right angles, or to two thirds of one right angle, because it is an angle at the centre (P. n., D. 2). The sum of the two angles OAB and OBA is, consequently, equal to four thirds of a right angle (B. I., P. XXV., 0. 1) ; but, the angles OAB and OBA are equal, because the opposite sides OB and OA are equal : hence, each is equal to two thirds 146 GEOMETRY. of a right angle. The three angles of the triangle AOB are therefore equal, and consequently, the triangle is equilateral: hence, AB is equ^^l to OA; which was to be proved. 5 i M ' ,0-^ PROPOSITION^ V. PROBLEM. To inscribe a regular hexagon in a given circle. Let ABE be a circle, and its B centre. Beginning at any point of the circumference, as A, apply the ra- dius OA six times as a chord ; then ABCDEF is the hexagon re- quired (P. IV.). Cor. 1. If the alternate ver- tices of the regular hexagon are joined by the straight lines AC, CE, and EA, the inscribed triangle ACE is equilateral (P. IL, S.). Cor. 2. If we draw the radii OA and OC, the figure AOCB is a rhombus, because its sides are equal: hence (B. IV., P. XIV., 0.), we have, AB* + BC* + 6^ + OC' = AC* + OB' ; or, taking away from the first member the quantity OA*, and from the second its equal OB* and reducing, w6 have. whence (B. II., P. II.), AC* 30A' OA* AC*; 1; BOOK V or (B. n., P. Xn., 0. 2), AC : OA 147 a/S : 1; that is, the side of an inscribed equilateral triangle is to the radius, as the square root of 3 is to 1,; ^ PROPOSITION VI. THEOREM. If the radius of a circle is divided in extreme and mean ratio, the greater segment is equal to one side of a regu- lar inscribed decagon. Let ACG be a circle, OA its radius, and AB, equal to OM, the greater segment of OA when divided in extreme and mean ratio : then AB is equal to the side of a regu- lar inscribed decagon. Draw OB and BM. We have, by hypothesis, AO : OM : : OM : AM ; or, since AB is equal to OM, we have, AB : AM; AO AB hence, the triangleis OAB and BAM have the sides about their com- mon angle BAM, proportional ; they are, therefore, simUar (B. IV., P. XX.). But, the triangle OAB is isosceles ; hence, BAM is also isosceles, and conse- quently, the side BM is equal to AB. But, AB is equal to OM, by hypothesis: hence, BM is equal to OM, and conse- quently, the angles MOB and MBO are equal. The angle 148 GEOMETBT. AMB being an exterior angle of the triangle 0MB, is equal to the sum of the angles MOB and MBO, or to twice the angle MOB; and because AMB is equal to OAB, and also to OBA, the sum of the angles OAB and OBA is equal to four times the angle AOB: hence, AOB is equal to one fifth of two right angles, or to one tenth of four right angles; and consequently, the arc AB is equal to one tenth of the circumference : hence, the chord AB is equal to the side of a regular inscribed f^eoagon ; which was to he proved. "^1 ■ Kr - I ) Cor. 1. • If AB is '\ Applied tehytimes as a chord, the re- sulting polygon is a regular inscribed decagon. Cor. 2. If the vertices A, C, E, G, and I, of the alter- nate angles of the decagon are joined by straight lines, the resulting figure is a regular inscribed pentagon. Scholium 1. If the arcs subtended by the sides of any regular inscribed polygon are bisected, and chords of the semi-arcs drawn, the resulting figure is a regular inscribed polygon of double the number of sides. Scholium 2. The area of any regular inscribed polygon is less than that of a regular inscribed polygon of double the number of sides, because a part is less than the whole. BOOK V. 149 PEOPOSITION VII. PROBLEM. To oireumscribe, dbout a circle, a polygon which shall be similar to a given regular inscribed polygon. Let TNQ be a circle, its centre, and ABCDEF a regu- lar inscribed polygon. At tbe middle points T, N, P, &c., of the arcs sub- tended by the sides of the inscribed polygon, draw tan- gents to the circle, and pro- long them till they intersect ; then the resulting figure is the polygon required. 1°. The side HG being parallel to BA, and HI to BC, tJie angle H is equal /V^ ~- -- .A ^^ ..ii^; \ / ■■ ,•■■ \ ^ /O:. /i S\/D :^ ^ to the angle B. In like manner, it may be shown that any other angle of the circumscribed polygon is equal to the corresponding angle of the inscribed polygon : hence, the circumscribed polygon is equiangular. 2°. Draw the straight lines OG, OT, OH, ON, and 01. Then, because the lines HT and HN are tangent to the circle, OH bisects the angle NHT, and also the angle NOT (B. III., Prob. XTV., C.) ; consequently, it passes through the middle point B of the arc NBT. In like manner, it may be shown that the straight line drawn from the centre to the vertex of any other angle of the circum- scibed polygon, passes through the corresponding vertex of the inscribed polygon. The triangles OHG and OH I have the angles OHG and 150 GEOMETRY. OH I equal, from what has just been shown; the angles GOH and HOI equal, becaiise they are measured by the equal arcs AB and BC, and the side OH common ; they are, tiiere|ore, equal in aU respects Y lience, GH is equal to HI. 'In like manner, it may be shown that HI is equal to IK, IK to KL, and so on : hence, the circumscribed polygon is equilateral. The circumscribed poly- gon being both equiangular and equilateral, is regular; and since it has the same number of sides as the in- scribed polygon, it is similar to it. Cor. 1. If straight lines are drawn from the centre of a regular circumscribed polygon to its vertices, and the consecutive points in which they intersect the circumfer- ence joined by chords, the resulting figure is a regular inscribed polygon similar to the given polygon. Cor. 2. The sum. of the lines HT and HN is equal to the sum of HT and TG, or to HG ; that is, to one of the sides of the circumscribed polygon. Cor. 3. If at the vertices A, B, C, &c., of the inscribed polygon, tangents are drawn to the circle and prolonged till they meet the sides of the circumscribed polygon, the resulting figure is a circumscribed polygon of double the number of sides. Sch. 1. The area of any regular circumscribed polygon BOOK V. 151 is greater than that of a regular circumscribed polygon of double the number of sides, because the whole is greater than any of its parts. Sch. 2. By means of a circumscribed and inscribed square, we may construct, in succession, regular circum- scribed and inscribed polygons of 8, 16, 32, &c., sides. By means of the regular hexagon we may, in like man- ner, construct regular polygons of 12, 24, 48, &c., sides. By means of the decagon, we may construct regular poly- gons of 20, 40, 80, &c., sides. PROPOSITION Vin. THEOREM. The area of a regular polygon is equal to half the product of its perimeter and apothem. Let GHIK be a regular polygon, its centre, and OT its apothem, or the radius of the inscribed circle: then the area of the polygon is equal to half the product of the perimeter and the apothem. For, draw lines from the centre to the vertices of the polygon. These lines divide the polygon into triangles whose bases are the sides of the polygon, and whose altitudes are equal to the apothem. Now, the area of any triangle, as OHG, is equal to half the product of the side HG and the apothem: hence, the area of the poly- gon is equal to half the product of the perimeter and the apothem ; which was to he proved. 152 GEOMETKY. PROPOSITION IX. THEOREM. Tke perimeters of simUar regular polygons are to each other as the radii of their circumscribed or inscribed circles;' and their areas are to each other as the squares of those radii. 1°. Let ABC and KLM be similar regular polygons. Let OA and QK be the radii of their circumscribed, OD and QR be the radii of their inscribed circles: then the perim- eters of the polygons are to each other as OA is to QK, or as OD is to QR. For, the lines OA and QK are homolo- gous lines of the polygons to "which they belong, as are also the lines OD and QR : hence, the perimeter of ABC is to the perimeter of KLM, as OA is to QK, or as OD is to QR (B. IV., P. XXVII., C. 1) ; which was to be proved. 2°. The areas of the polygons are to each other as OA^ is to QK*, or as OD* is to QRI For, OA being homologous with QK, and OD with QR, we have, the area of ABC is to the area of KLM as OA' is to QK^ or as OD^ is to QR' (B. IV., P. XXVH., C. 1) ; which was to be proved. BOOK V. 153 PROFOSITION X. THEOREM. Two regular polygons of the same number of sides can he constructed, the one circumscribed about a circle and the other inscribed in it, which shall differ from each other by less than any given surface. Let ABCE be a circle, its centre, and Q the side of a square equal to or less than the given surface ; then can two similar regular polygons be constructed, the one circumscribed about, and the other inscribed in the given circle, which shall differ from each other by less than the square of Q, and consequently, by less than the given surface. Inscribe a square in the given circle (P. III.), and by means of it, inscribe, in succession, regular polygons of 8, 16, 32, &c., sides (P. Vn., S. 2), until one is found whose side is less than Q; let AB be the side of such a poly- gon. Construct a similar circum- scribed polygon abode : then these polygons differ from each other by less than the square of Q. For, from a and 6, draw the lines aO and 60 ; they pass through the points A and B. Draw also OK to the point of contact K; it bisects AB at I and is perpendicular to it. Prolong AO to E. Let P denote the- circumscribed, and p the inscribed polygon ; then, because they are regular and similar, we have (P. IX.), OK' or OA' "01' 154 GEOMETRY. hence, by division (B. n., P. VI.), we have, P : P -p :: M : OA'-OT; or, P-p OA* Al^ Multiplying the terms of the second couplet by 4 (B. 11., P. Vn.), we have P : P-p :: 40A" whence (B. IV., P. Vm., C), P : P -p U\': AE' AB^ But P is less than the square of AE (P. VII., S. 1) ; hence, P — p is less than the square of AB, and conse- quently, less than the square of Q, or than the given surface ; tvhich was to be proved. Definition. — The limit of a variable quantity is a quan- tity to which it may be made to approach nearer than any given quantity, and which it reaches under a partic- ular supposition. Lemma. — Two variable quantities which constantly approach to equality, and of which the difference becomes less than any finite magnitude, are ultimately equal. For if they are not ultimately equal, let D be their ultimate difference. Now, by hypothesis, the quantities have approached nearer to equality than any given quan- tity, as D ; hence D denotes their difference and a quantity greater than their difference, at the same time, which is impossible ; therefore, the two quantities are ultimately equal.* * Newton's Principia, Book I., Lemina I. BOOK V. 165 Cor. If we take any two similar regular polygons, the one circumscribed about, and the other inscribed in the circle, and bisect the arcs, and then circumscribe and inscribe two regular polygons having double the number of sides, it is plain that by continuing the operation, two new polygons may be found which shall differ from each other by less than any given surface ; hence, by the lemma, the two polygons wiU become ultimately equal. But this equality can not take place for any finite number of sides ; hence, the number of sides in each will be infinite, and each will coincide with the circle, which is their common limit. Under this hjrpothesis, the perimeter of each polygon will coincide with the circumference of the circle. Scholium. The circle may be regarded as a regular polygon having an infinite number of sides. The circumference may be regarded as the perimeter, and the radius as the apothem. PROPOSITION XI. PROBLEM. ITve area of a regular inscribed polygon, and that of a similar circumscribed polygon being given, to find the areas of the regular inscribed and circumscribed -polygons having double the number of sides. Let AB be the side of the given inscribed, and EF that of the given circumscribed polygon. Let C be their common centre, AMB a por- tion of the circumference of the circle, and M the middle point of the arc AMB. Draw the chord AM, and at A and B draw the tangents AP and BQ ; then AM is the side of the inscribed polygon, and PQ the side of the circum- scribed polygon of double the number of sides (P. VIL). Btaw CE CP_CM_andLCF. GEOMETRY. Denote the area of the given inscribed polygon by p, the area of the given circumscribed polygon by P, and the areas of the inscribed and circumscribed polygons having double the number of sides, respectively by p' and P'. 1°. The triangles CAD, CAM, and CEM, are like parts of the polygons to which they belong: hence, they are proportional to the polygons themselves. But CAM is a mean proportional be- tween CAD and CEM (B. IV., P. XXrV., 0.) ; consequently, p' is a mean proportional between p and P : hence. . p' = Vp X P. (1.) 2°. Because the triangles CPM and CPE have the com- mon altitude CM, they are to each other as their bases: hence, CPM : CPE :: PM : PE ; and because CP bisects the angle ACM, we have (B. IV., P. xvn.), PM : PE :: CM : CE :: CD : CA; hence (B. II., P. IV.), CPM : CPE :': CD : CA or CM. But, the triangles CAD and CAM have the common alti- tude AD ; they are, therefore, to each other as their bases : hence, CAD : CAM CD CM or, because CAD and CAM are to each other as the poly- gons to which they belong, "' ,'V-l n /■/,.,(?(/.'■ '"-'book v.'->T'^/-/5:-,-— — -. ^'rs? •^^ (p : p' :: CD : ' CM ; '^ ' '^^ Z'//'^//- hence (B. II., P. IV.), we have, CPM : CPE :: p : p' ; and, by .compositioii, CPM : .CPM + CPE or CME :: p -. p + p' ; hence (B. II., A Vn.), 2CPM or CMPA : CME : : 2p : p + p'. But, CMPA and CME are hke parts of P' and P; hence, P' : P : : 2p : p + p' \ or, ?' = ^P^, (2.) p + p' ^ ' Scholium. By means of Equation (1), we can find p and then, by means of Equation (2), we can find P'. PEOPOSITION XII. PROBLEM. To find the approximate area of a circle whose radius is 1. The area of an inscribed square is equal to twice the square described on the radius (P. III., S.) ; the area of a circumscribed square is equal to the square described on the diameter. If the radius be taken as the unit of linear measure, and the square described on it as the unit of area, the area of the inscribed square wiU be 2, and that of the circumscribed square will be 4. Making p equal to 2, and P equal to 4, we have, from Equations (1) and (2) of Proposition XI., p' = ^78= 2.8284271 . . inscribed octagon, 1 fi P' = — = 3.3137085 . . circumscribed octagon. 2 + V8 158 GEOMETRY. Making p equal to 2.8284271, and P equal to 3.3137085, we have, from the same equations, p' = 3.0614674 P' = 3.1825979 inscribed polygon of 16 sides, circumscribed polygon of 16 sides. By a continued application of these equations, "we find the areas radicated below: NniiBER OF Sides. ISSOBIBBD POLTOOKB. ClBOnllBOBIBED POLTOOKS 4 2.0000000 4.0000000 8 2.8284271 3.3137085 16 3.0614674 3.1825979 32 3.1214451 3.1517249 64 3.1365485 3.1441184 128 3.1408311 3.1422236 256 ' 3.1412772 3.1417504 512 3.1415138 3.1416321 1024 3.1415729 3.1416025 2048 3.1415877 3.1415951 4096 3.1415914 3.1415933 8192 3.1415923 8.1415928 16384 3.1415925 3.1415927 Now, the figures which express the areas of the last two polygons are the same for six decimal places ; hence, those areas differ from each other by less than one mill- ionth part of the measuring unit. But the circle differs from either of the polygons by less than they differ from each other. Hence, for all ordinary computation, it is sufficiently accurate to consider the area of a circle, whose radius is 1, equal to 3.141592; the unit of measure being, as shown above, the square described on the radius. This value, 3.141592, is represented by the Greek letter tt. Sch. For ordinary accuracy, it is taken equal to 3.1416. BOOK V. 159 ' ; PROPOSITION Xin. THEOREM. The circumferences of circles are to each other as their radii, and the areas are to each other as the squares ef. — their radii. Let C and be the centres of two circles whose radii are CA and OB : then the circumferences are to each other as their... radii, and the areas are to each other as the squares ^ their radii. I< — r — -f For, let similar regular polygons MNPST and EFGKL be inscribed in the circles: then the perimeters of these polygons are to each other as their apothems, and the areas are to each other as the squares of their apothems, whatever may be the number of their sides (P. IX.). If the number - of sides is made infinite (P. X., Sch.), the polygons coincide with the circles, the perimeters with the circumferences, and the apothems with the radii: hence, the circumferences of the circles are to each other as their radii, and the areas are to each other as the squares of the radii ; which was to be proved. Cor. 1. Diameters of circles are proportional to their radii : hence, the circumferences of circles are proportional to their diameters, and the areas are proportional to the squares of the diameters. 160 GEOMETRY. V Cor. 2. Similar arcs, as AB and DE, are like parts of the circumferences to which they belong, and similar sec- tors, as ACB and DOE, are like parts of the circles to which they belong : hence, similar arcs are to each other as their radii, and similar sectors are to each other as the squares of their radii. Scholium. The term infinite, used in the proposition, is to be understood in its technical sense. When it is pro- posed to make the number of sides of the polygons infinite, by the method indicated in the schoUum of Proposition X., it is simply meant to express the condition of things, when the inscribed polygons reach their hmits ; in which case, the dififerenoe between the area of either circle and its inscribed polygon, is less than any appreciable quantity. We have seen (P. XTT.), that when the number of sides is 16384,. the areas differ by less than the millionth part of the measuring unit. By increasing the number of sides, we approximate still nearer. PROPOSITION XIV. THEOREM. Ihe area of a cirde is equal to half the product of it^ circumference and radius. Let be the centre of a circle, OC its radius, and ACDE its circimaference : then the area of the circle is equal to half the product of the circumference and ra- dius. For, inscribe in it a regular poly- gon ACDE. Then the area of this polygon is equal to half the product BOOK V. 161 of its perimeter and apothem, -whatever may be the number of its sides (P. VIII.). If the number of sides is made infinite, the polygon coincides with the circle, the perimeter with the circum- ference, . and the apothem with the radius : hence, the area of the circle is equal to half the product of its cir- cumference and radius ; which was to he proved. _ Cor. 1. The area of a sector is equal to half the prod- uct of its arc and radius. - -'' f .'- / '.-) V, , •, ••'-.'/, '/ Cor. 2. The area of a sector is to the area of the circle, as the arc of the sector to the circumference. PROPOSITION XV. PROBLEM. Y - \ 1 To find an expression for the area of any circle in terms of its radius. Let C be the centre of a circle, and CA its radius. Denote its area by area CA, its radius by R, and the area of a circle whose radius is 1, by tt (P. XII., S.). Then, because the areas of circles are to each other as the squares of their radii (P. XTTT.), we have, area CA : jt : : R' : 1 ; whence, area CA = wR'. That is, the area, of any circle is 3.1416 times the square of its radius. PROPOSITION XVI. PROBLEM. To find an expression for the circumference of a circle, in terms of its radius, or diameter. Let C be the centre of a circle, and CA its radius. 162 AEOMEfBY. Denote its circumference by circ. CA, its radius by R, and its diameter by D. From the last Proposition, we have, area CA = wR^; and, from Proposition XIV., we have, area Ck = \ circ. CA x R ; hence, I circ. CA x R = tR* ; whence, by reduction, circ. CA = 27rR, or, circ. CA = ttD. That is, the circumference of any circle is equal to B.1416 times its diameter. Scholium 1. The abstract number tt, equal to 3.1416, denotes the number of times that the diameter of a circle is contained in the circumference, and also the number of times that the square constructed on the radius is con- tained in the area of the circle (P. XV.). Now, it has? been proved by the methods of higher mathematics, that the value of n- is incommensurable with 1 ; hence, it is impossible to express, by means of numbers, the exact length of a circumference in terms of the radius, or the exact area in terms of the square described on the radius. It is not possible, therefore, to square the circle — ^that is, to construct a square whose area shall be exactly equal to that of the circla Scholium 2. Besides the approximate value of tt, 3.1416, usually employed, the fractions -^ and'ffi are also sometimes used to express the ratio of the diameter to the circumference. BOOK V. 163 EXERCISES. 1. The side of an equilateral triangle inscribed in a circle is 6 feet; find the radius of the cirote;^ 2. The radius of a circle is 10 feet; find the apothem of a regular inscribed hexagon. < 'v 3. Find the side of a square inscribed in a circle whose radius is 5 feet. 4. Draw a line whose length shall be a/3. 5. The radius of a circle is 4 feet; find the area of an inscribed equilateral triangle. ,'|6. Show that the sums of the alternate angles of an octagon inscribed in a circle are equal to each other. 7. The area of a regular hexagon, whose side is 20 feet, is 1039.23 square feet; find the apothem. 8. One side of a regular decagon is 20 feet, and its apothem 15.4 feet; find the perimeter, and the area of a similar decagon whose apothem is 8 feet. 9. The area of a regular hexagon inscribed in a circle is 9 square feet, and the area of a similar circumscribed hexagon is 12 square feet; find the areas of regular in- scribed and circumscribed polygons of 12 sides. 10. Q-iven two diagonals of a regu- lar pentagon that intersect; show that the greater segments will be equal to each other and to a side of the penta- gon, and that the diagonals cut each other in extreme and mean ratio. 11. Show how to inscribe in a given circle a regular polygon of 15 sides. 12. Find the side and the altitude of an equilateral triangle in terms of the radius of the inscribed circle. 164 GEOMETRY. 13. Given an equilateral triangle inscribed in a circle, and a similar circumscribed triangle ; determine the ratio of the two triangles to each other. 14. The diameter of a circle is 20 feet; find the area of a sector whose arc is 120°. 15. The circumference of a circle is 200 feet; find its area. 16. The area of a circle is 78.54 square yards; find its diameter. 17. The radius of a circle is 10 feet, and the area of a circular sector 100 square feet; find the arc of the sector in degrees. A 18. Show that the area of an equilateral triangle cir- cumscribed about a circle is greater than that of a square circumscribed about the same circle. 19. Let AC and BD be diameters perpendicular to each other; from P, the middle point of the radius OA, as a centre, and a radius equal to PB, describe an arc cutting OC in Q ; show that the radius OC is divided in ex- treme and mean ratio at Q. 20. Show that the square of the side of a regular in- scribed pentagon is equal to the square of the side of a regular inscribed decagon increased by the square of the radius of the circumscribing circle. 21. Show how, from 19 and 20, to inscribe a regular pentagon in a given circle. 22. The side of a regular pentagon, inscribed in a circle, is 5 feet, and that of a regular inscribed decagon is 2.65 feet; find the side and the area of a regular hexagon inscribed in the same circle. BOOK VI. PLANES AND POLYEDIIAL ANGLES. DEFINITIONS. 1. A straight line is perpendicular to a plane, -when it is perpendicular to every straight line of the plane which passes through its foot; that is, through the point in which it meets the plane. In this ease, the plane is also perpendicular to the line. 2. A straight line is parallel to a plane, when it can not meet the plane, how far soever both may be pro- duced. In this case, the plane is also parallel to the line. 3. Two Planes are parallel, when they can not meet, how far soever both may be produced. 4. A Diedral angle is the amount of divergence of two planes. The line in which the planes meet is called the edge of the angUf and the planes themselves are called faces of the angle. The measure of a diedral angle is the same as that of a plane angle formed by two straight lines, one in each face, and both perpendicular to the edge at the same point. A diedral angle may be acute, obtuse, or a right angle. In the latter case, the faces are perpendicular to each other. 166 GEOMETRY. 5. A PoLTEDEAL ANGLE is the amount of divergence of several planes meeting at a common point. This point ■ is called the vertex of the angle; the lines in which the planes meet are called edges of the angle, and the portions of the planes lying between the edges are called faces of the angle. Thus, S is the vertex of the polyedral s angle, whose edges are SA, SB, SC, .a\\. SD, and whose faces are ASB, BSC, /' \ N. CSD, DSA. /j \ N. A polyedral angle which has but //^ \ ~^ three faces, is called a triedral £_ ^^ angle. * ^ POSTULATE. A straight line may be drawn perpendicular to a plane from any point of the plane, or from any point without the plane. PROPOSITION I. THEOREM. If a straight line has two of its points in a plane, it lies wholly in that plane. For, by definition, a plane is a surface such, that if any two of its points are joined by a straight line, that line lies wholly in the surface (B. I., D. 8). Cor. Through any point of a plane, an infinite num- ber of straight lines may be drawn which he in the plane. For, if a straight fine is drawn from the given point to any other point of the plane, that line lies wholly in the plane. Scholium. If any two points of a plane are joined by a straight line, the plane may be turned about that line as BOOK VI. 167 an axis, so as to take an infinite number of positions. Hence, we infer that an infinite number of planes may be passed through a given straight line. PROPOSITION n. THEOREM. Through three points, not in the same straight line, one plane can he passed, and only one. Let A, B, and C be the three points: then can one plane be passed through them, and only one. Join two of the points, as A and B, by the line AB. Through AB let a plane be passed, and let this plane be turned around AB until it contains the point C ; . in this position it will pass through the three points A, B, and C. If now, the plane be turned about AB, in either direction, it wUl no longer contain the point C: hence, one plane can always be passed through three points, and only one ; which was to he proved. Cor. 1. Three points, not in a straight line, determine the position of a plane, because only one plane can be passed through them. Cor, 2. A straight line and a point without that line determine the position of a plane, because only one plane can be passed through them. Cor. 3. Two straight lines which intersect determine the position of a plane. For, let AB and AC intersect at A: then either line, as AB, and one point of the other, as C, determine the position of a plane. Cor. 4. Two parallel straight lines determine the position 168 GEOMETRY. of a plane. For, let AB and CD be parallel. By definition (B. L, D. 16) two parallel lines always lie in the same plane. But either line, as * B AB, and any point of the other, as F, de- C * d termine the position of a plane : hence, two parallels determine the position of a plane. PROPOSITION m. THEOREM. The intersection of two planes is a straight line. Let AB and CD be two planes : then is their intersec- tion a straight line. For, let E and F be any two points com- D^ mon to the planes ; draw the straight line EF. This line having two points in the plane AB, lies wholly in that plane; and having two points in the plane CD, hes wholly in that plane : hence, every point of EF is common to both planes. Furthermore, the planes can have no common point lying without EF, otherwise there would be two planes passing through a straight line and a point lying without it, which is impos- sible (P. 11., C. 2) ; hence, the intersection of the two planes is a straight line ; which was to be proved. PROPOSITION IV. THEOREM. If a straight line is -perpendicular to two straight lines at their point of intersection, it is perpendicular to the plane of those lines. Let MN be the plane of the two lines BB, CC, and let AP be perpendiciilar to these lines at P : then is AP per- BOOK VI. 169 pendicular to every straight line of the plane which passes through P, and consequently, to the plane itseK. For, through P, draw in the plane MN, any line PQ ; through any point of this hne, as Q, draw the line BC, so that BQ shall be equal to QC (B. IV., Prob. V.) ; draw AB, AQ, and AC. The base BC, of the triangle BPC, being bisected at Q, we have (B. IV., P. XIV.), PC' + PB^ = 2PQ' + 2QCl In like manner, we have, from the triangle ABC, AC* + AB' = 2AQ' + 2QC'. Subtracting the first of these equations from the second, member from member, we have, AC" - PC* -r AB' - "PB' = 2AQ' — 2PQ'. But, from Proposition XI., 0. 1, Book IV., we have, AC' - PC' = AP^ and AB' - PB' = AP' ; hence, by substitution, 2AP' - 2AQ'— 2PQ'; whence. AP^ = AQ' - PQ' ; or, AP' + PQ" = AQ- The triangle APQ is, therefore, right-angled at P (B. IV., P. XIIL, S.), and consequently, AP is perpendicular to PQ: hence, AP is perpendicular to every line of the plane MN passing through P, and consequently, to the plane itself: which was to be proved. \ 170 GEOMETEY. Cor. 1. Only one perpendicular can be drawn to a plane from a point without the plane. For, suppose two perpendiculars, ii as AP and AQ, could be drawn from the point A to the plane MN. Draw PQ ; then the triangle APQ would have two right angles, APQ and AQP ; which is impossible (B. L, P. XXV., C. 3). Cor. 2. Only one perpendicular can be drawn to a plane from a point of that plane. For, suppose that two perpendiculars could be drawn to the plane MN, from the point P- Pass a plane through the perpendiculars, and let PQ be its intersection with MN ; then we should have two perpendiculars drawn to the same straight line from a point of that line ; which is impossible (B. I., P. XIV.). PROPOSITION V. THEOREM. If from a ■point without a jtlane, a perpendicular is drawn to the plane, and oblique lines drawn to different points of the plane: 1°. The perpendicular is shorter than any oblique line: 2°. Oblique lines which meet the plane at equal distances from the foot of the perpendicular, are equal .- 3°. Of two oblique lines which meet the plane at unequal distances from the foot of the perpendicular, the one which meets it at the greater distance is the longer. Let A be a point without the plane MN ; let AP be perpendicular to the plane ; let AC, AD, be any two oblique lines meeting the plane at equal distances from the foot of the perpendicular; and let AC and AE be any BOOK VI. 171 two oblique lines meeting the plane at unequal distances from the foot of the perpendicular: 1°. AP is shorter than any oblique -line AC. For, draw PC; then is AP less than AC (B. I, P. XV.) which was to he proved. 2°, AC and AD are equal. ' For, draw PD ; then the right-angled triangles APC, APD, have the side AP common, and the sides PC, PD, equal: hence, the triangles are equal ia all respects, i- and consequently, AC and AD are equal ; which was to be proved. 3°. AE is greater than AC. For, draw PE, and take PB equal to PC ; draw AB : then is AE greater than AB (B. I., P. XV.) ; but AB and AC are equal : hence, AE is greater than AC ; which was to be proved. Cor. The equal oblique lines AB, AC, AD, meet the plane MN in the circumference of a circle whose centre is P, and whose radius is PB : hence, to draw a perpendicular to a given plane MN, from a point A, without that plane, find three points B, C, D, of the plane equally distant from A, and then find the centre, P, of the circle whose circumference passes through these points: then AP is the perpendicular required. Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN. The equal oblique lines AB, AC, AD, are all equally inclined to the plane MN. The inclination of AE is less than the inclination of any shorter line AB. 172 GEOMETEY. PROPOSITION VI. THEOREM. If from tJve foot of a -perpendicular to a plane, a straight line is drawn at right angles to any straight line of that plane, and the point of intersection joined with any point of the perpendicular, the last line is perpendicular to the line of the plane. Let AP be perpendicular to the plane MN, P its foot, BC the given line, and A any point of the perpendicular; draw PD at right angles to BC, and join the point D with A : then is AD perpendicular to BC. For, lay off DB equal to DC, and draw PB, PC, AB," and AC. Because PD is perpendicular to BC, and DB equal to DC, we have, PB equal to PC (B. I., P. XV.) ; and because AP is perpendicular to the plane MN, and PB equal to PC, we have AB equal to AC (P. V.). The line AD has, therefore, two of its points A and D, each equally distant from B and C : hence, it is perpendicular to BC (B. L, P. XVI., C.) ; which was to be proved. Cor. 1. The line BC is perpendicular to the plane of the triangle APD ; because it is perpendicular to AD and PD, at D (P. IV.). Cor. 2. The shortest distance between AP and BC is measured on PD, perpendicular to both. For, draw BE between any other points of the lines: then BE is greater than PB, and PB greater than PD : hence, PD is less than BE. BOOK VI. 173 Scholium. The lines AP and BC, though not in the same plane, are considered perpendicular to each other. In general, any two straight lines not in the same plane are considered as making an angle with each other, which angle is equal to that formed by drawing, through a given point, two lines respectively parallel to the given lines. PROPOSITION VII. THEOREM. If one of two parallels is perpendicular to a plane, the other one is also perpendicular to the same plane. Let AP and ED be two parallels, and let AP be perpen- dicular to the plane MN: then is ED also perpendicular to the plane MN. For, pass a plane through the parallels; its intersection with MN is PD ; draw AD, and in the plane MN draw BC perpendicular to PD at D. Now, BD is perpendicular to the plane APDE (P. VI., C. 1) ; the angle BDE is consequently a right angle ; but the angle EDP is a right angle, because ED is parallel to AP (B. I, P. XX., C. 1): hence, ED is perpendicular to BD and PD, at their point of intersection, and consequently, to their plane MN (P. IV.) ; which was to be proved. Cor. 1. If the lines AP and ED are perpendicular to the plane MN, they are parallel to each other. For, if not, conceive a line drawn through D parallel to PA; it would be perpendicular to the plane MN, from what has just been proved; we would, therefore, have two perpen- diculars to the plane MN, at the same point; which is impossible (P. IV., 0. 2). 174 GEOMETRY. Cor. 2. If two straight lines, A and B, are parallel to a third line C, they are parallel to each other. For, pass a plane perpendicular to C ; it will be perpendicular to both A and B: hence, A and B are parallel. PROPOSITION" Vm. THEOREM. // a straight line is parallel to a line of a plane, it is parallel to that plane. Let the line AB be parallel to the line CD of the plane MN ; then is AB parallel to the plane MN. For, through AB and CD pass a plane (P. 11., 0. 4) ; CD is its inter- section with the plane MN. Now, since AB lies in this plane, if it can meet the plane MN, it wiU meet it at some point of CD ; but this is impossible, because AB and CD are parallel : hence, AB can not meet the plane MN, and consequently, it is par- allel to it ; which was to be proved. 7 PROPOSITION" IX. THEOREM. // two planes are perpendicular to the same straight line, they are parallel to each other. Let the planes MN and PQ be perpendicular to the line AB, at the points A and B : then are they parallel to each other. For, if they are not parallel, they will meet; and let be a / = / M / . U- / BOOK VI. 175 point common to both. From draw the lines OA and OB: then, since OA lies in the plane MN, it is perpendic- ular to BA at A (D. 1). For a like reason, OB is perpen- dicular to AB at B : hence, the triangle OAB has two right angles, -which is impossible; consequently, the planes can not "meet, and are therefore parallel; which was to be proved. PROPOSITION X. THEOREM. If a plane intersects two parallel planes, the lines of inter- section are parallel. Let the plane EH intersect the parallel planes MN and PQ, in the lines EF and GH : then are EF and GH par- aUel. For, if they are not parallel, they will meet if suflB.ciently prolonged, because they lie in the same plane ; but if the lines meet, the planes MN and PQ, in which they lie, also meet ; but this is impossible, because these planes are parallel: hence, the lines EF and GH can not meet; they are, therefore, parallel ; which was to he proved. PROPOSITION XI. THEOREM. j< If a straight line is perpendicular to one of two parallel planes, it is also perpendicular to the other. Let MN and PQ be two parallel planes, and let the line AB be perpendicular to PQ: then is it also perpendic- ular to MN. 176 6E0METEY. For, through AB pass any plane; its intersections with M N and PQ are parallel (P. X.) ; but, its intersection with PQ is perpendicular to AB at B (D. 1) ; hence, its inter- section with MN is also perpendicular n to AB at A (B. I., P. XX., C. 1): hence, AB is perpendicular to every line of the plane MN through A, and is, therefore, perpendicular to that plane : which was to he proved. / = ^y p N / • L/ f1 "4 PROPOSITION Xn. THEOREM. Parcdlel straight lines inoluded between parallel planes, are equal. Let EG and FH be any two parallel lines included be- tween the parallel planes M N and PQ : then are they equal. Through the parallels conceive a plane to be passed ; it will intersect the plane MN in the line EF, and PQ in the line GH ; and these lines are parallel (Prop. X.). The figure EFHG is, therefore, a parallelogram: hence, GE and HF are equal (B. I., P. XXVm.) ; which was to he proved. Got. 1. The distance between two parallel planes is measured on a perpendicular to both ; but any two per- pendiculars between the planes are equal : hence, parallel planes are every-where equally distant. Cor. 2. If a straight line GH is parallel to any plane MN, then can a plane be passed through GH parallel to MN : hence, if a straight line is parallel to a plane, all of its points are equally distant from that plane. BOOK VI. 177 PROPOSITION Xni. THEOREM. If two angles, not situated in the same -plane, have their Slides parallel, and lying in the same direction, the angles' are equal and their planes parallel. Let CAE and DBF be two angles lying in the planes MN and PQ, and let the sides AC and AE be respectively- parallel to BD and BF, and lying in the same direction : then are the angles CAE and DBF equal, and the planes MN and PQ parallel. Take any two points of AC and AE, as C and E, and make BD equal to AC, and BF to AE ; draw CE, DF, AB, CD, and EF. 1 °. The angles CAE and DBF are equal. For, AE and BF being parallel and equal, the figure ABFE is a parallelogram (B. L, P. XXX.) ; hence, EF is parallel and equal to AB. For a like reason, CD is paral- lel and equal to AB: hence, CD and EF are parallel and equal to each other, and consequently, CE and DF are also parallel and equal to each other. The triangles CAE and DBF have, therefore, their corresponding sides equal, and consequently, the corresponding angles CAE and DBF are equal ; which was to be proved. 2°. The planes of the angles, MN and PQ, are parallel. For, from A draw AG perpendicular to the plane PQ ; at the point G, where it meets the plane, draw in the plane PQ, GH and GK parallel, ' respectively, to BD and BF; then 178 6E0METEY. is AC paraUel to GH, and AE to GK (P. VH., 0. 2). AG, being perpendicular to GH and GK (D. 1), is perpendicular to their parallels, AC and AE (B. I., P. XX., 0. 1), and is, therefore, perpendicular to the plane MN (P. IV.). The planes MN and PQ, being perpendicular to the same straight hne, AG, are parallel to each other (P. IX.) ; which was to be proved. Cor. If two parallel planes, MN and PQ, are met by- two other planes, AD and AF, the angles CAE and DBF, formed by their intersections, are equal. PROPOSITION XIV. THEOREM. If three straight lines, not situated in the same plane, are equal and parallel, the triangles formed by Joining the extremities of these lines are equal, and their planes parallel. Let AB, CD, and EF be equal parallel lines not in the same plane : then are the triangles ACE and BDF equal, and their planes parallel. For, AB being equal and paral- /~i^ 7 lei to EF, the figure ABFE is a X/)^> / parallelogram, and consequently, ^—4 — \ — — — \ / AE is equal and parallel to BF. For a like reason, AC is equal and parallel to BD : hence, the included angles CAE and DBF are equal and their planes parallel (P. Xm.). Now, the triangles CAE and DBF have two sides and their included angles equal, each to each: hence, they are equal in aU respects. The triangles are, therefore, equal and their planes parallel; which was to he proved. BOOK VI. 179 PROPOSITION XV. THEOREM. // two straight lines are cut by three parallel planes, they are divided proportionally. Let the lines AB and CD be cut by the parallel planes MN, PQ, and RS, in the points A, E, B, and C, F, D; then AE : EB : : CF : FD. For, draw the line AD, and suppose it to pierce the plane PQ in G; draw AC, BD, EG, and GF. The plane ABD intersects the parallel planes RS and PQ in the lines BD and EG ; consequently, these lines are parallel (P. X.) : hence (B. IV., P. XV.), AE : EB : : AG : GD. M The plane ACD intersects the parallel planes MN and PQ, in the parallel lines AC and GF: hence, AG : GD : : CF : FD. Combining these proportions (B. n., P. IV.), we have, AE : EB : : CF : FD ; which was to he proved. Cor. 1. If two straight lines are cut by any number of parallel planes, they are divided proportionally. Cor. 2. If any number of straight lines are cut by three parallel planes, they are divided proportionally. Y ^1 s / ' '■ , \ Q L -V\ / ' \ I N / ^- \ \/ 180 GEOMETBf. PROPOSITION XVI. THEOREM. If a, straight line is perpendicular to a plane, every plane passed through the line is also perpendicular to that plane. Let AP be perpendicular to the plane MN, and let BF be a plane passed through AP : then is BF perpendicular to MN. In the plane MN, draw PD perpen- dicular to BC, the iatersection of BF and MN. Since AP is perpendicular to MN, it is perpendicular to BC and / DP (D. 1) ; and since AP and DP, in / the planes BF and MN, are perpendic- M ular to the intersection of these planes at the same point, the angle which they form is equal to the angle formed by the planes (D. 4) ; but this angle is a right angle : hence, BF is perpendicular to M N ; which was to be proved. Cor. If three lines AP, BP, and DP, are perpendicular to each other at a common point P, each line is perpen- dicular to the plane of the two others, and the three planes are perpendicular to each other. PROPOSITION XVII. THEOREM. If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their inter- section, is perpendicular to the other. Let the planes BF and MN be perpendicular to each other, and let the line AP, drawn in the plane BF, be per- pendicular to the intersection BC ; then is AP perpendicu- lar to the plane MN. to 181 BC at BOOK VI. For, in the plane MN, draw PD perpendicular P. Then because the planes BF and MN are perpendicular to each other, the angle APD is a right angle : hence, AP is perpendicular to the two lines PD and BC, at their intersection, and consequently, is perpendicular to their plane M N ; which was to he proved. Cor. If the plane BF is perpendicular to the plane MN, and if at a point P of their intersection, a perpendicular is erected to the plane MN, that perpendicular is in the plane BF. For, if not, draw in the plane BF, PA perpen- dicular to PC, the common intersection ; AP is perpendic- ular to the plane MN, by the theorem; therefore, at the same point P, there are two perpendiculars to the plane MN ; which is impossible (P. IV., C. 2). PROPOSITION XVIII. THEOREM. If two -planes cut each other, and are perpendicular to a third plane, their intersection is also perpendicular to that plane. Let the planes' BF, DH, be perpendicular to MN : then is their intersection AP perpendicular to MN. For, at the point P, erect a perpen- dicular to the plane MN ; that per- pendicular must be in the plane BF, and also in the plane DH (P. XVn., 0.) ; therefore, it is their common in- tersection AP ; which loas to he proved. 182 GEOMETRY. PROPOSITION XIX. THEOREM. The sum of any two of the plane angles formed by the edges of a triedral angle, is greater than the third. Let SA, SB, and SC, be the edges of a triedral angle: then is the sum of any two of the plane angles formed by them, as ASC and CSB, greater than the third ASB. If the plane angle ASB is equal to, or less than, either of the other two, the truth of the proposition is evident. Let us suppose, then, that ASB is greater than either. In the plane ASB, construct the angle BSD equal to BSC ; draw AB in that plane, at pleasure ; lay off SC equal to SD, and draw AC and CB. The triangles BSD and BSC have the side SC equal to SD, by construction, the side SB common, and the in- cluded angles BSD and BSC equal, by construction ; the triangles are therefore equal in all respects: hence, BD is equal to BC. But, from Proposition VII., Book I., we have, BC + CA > BD + DA. Taking away the equal parts BC and BD, we have, CA > DA; hence (B. I., P. IX.), we have, angle ASC > angle ASD ; and, adding the equal angles BSC and BSD, BOOK VI. 183 angle ASC + angle CSB > angle ASD + angle DSB ; or, angle ASC + angle CSB > angle ASB ; which was to be proved. PROPOSITION XX. THEOREM. The sum of the plane angles formed by the edges of any polyedral angle, is less than four right angles. Let S be the vertex of any polyedral angle whose edges are SA, SB, SC, SD, and SE ; then is the sum of the angles about S less than four right angles. For, pass a plane cutting the edges s in the points A, B, C, D, and E, and ysl the faces in the lines AB, BC, CD, DE, /// \\ and EA. From any point within the /'/ \\ polygon thus formed, as 0, draw the /y'^\j V i straight Unes OA, OB, OC, OD, and ^£;"' h\^^\\ "We then have two sets of triangles, b b one set having a common vertex S, the other having a common vertex 0, and both having com- mon bases AB, BC, CD, DE, EA. Now, in the set which has the common vertex S, the sum of all the angles is equal to the sum of all the plane angles formed by the edges of the polyedral angle whose vertex is S, together with the sum of a^l the angles at the bases: viz., SAB, SBA, SBC, &c. ; and the entire sum is equal to twice as many right angles as there are triangles. In the set whose common vertex is 0, the sum of all the angles is equal to the four right angles about 0, together with the interior angles of the polygon, and this sum is equal to twice as many right angles as there are triangles. Since 184 GEOMETRY. the number of triangles, in each set, is the same, it fol- lows that these sums are equal. But in the triedral angle whose vertex is B, we have (P. XIX.), s ABS 4- SBC > ABC; Ji and the hke may be shown at each of //I \\ the other vertices, C, D, E, A: hence, //I \\ the sum of the angles at the bases, in A-' v y^ the triangles whose common vertex is f^/;^_ h^/Q \ ' S, is greater than the sum of the \J/ \V angles at the bases, in the set whose B C common vertex is : therefore, the sum of the vertical angles about S, is less than the sum of the angles about : that is, less than four right angles ; which was to he proved. Scholium. The above demonstration is made on the supposition that the polyedral angle is convex, that is, that the diedral angles of the consecutive faces are each less than two right angles. PROPOSITION XXI. THEOREM. If the plane angles farmed by the edges of two triedral angles are equal, each to each, the planes of the equal angles are equally inclined to each other. Let S and T be the vertices of two triedral angles, and let the angle ASC be equal to DTF, ASB to DTE, and BSC to ETF: then the planes of the equal angles are' equally inclined to each other. For, take any point of SB, as B, and from it draw in the two faces ASB and CSB, the lines BA and BC, respect- ively perpendicular to SB ■ then the angle ABC measures the inclination of these faces. Lay off TE equal to SB. BOOK VI. 185 and from E draw in the faces DTE and FTE, the lines ED and EF, respectively perpendicular to TE : then the angle DEF measures the inclination of these faces. Draw AC and $ T DF. ■ /A /A The right-angled triangles // \ // \ SBA and TED, have the side V^.../..-...-.,J^ ^^1^--^^ SB equal to TE, and the angle / / ^ ' I ASB equal to DTE ; hence, kB ' ' is equal to DE, and AS to DT. In like manner, it may be shown that BC is equal to EF, and CS to FT. The triangles ASC and DTF, have the angle ASC equal to DTF, by hypothesis, the side AS equal to DT, and the side CS to FT, from what has just been shown ; hence, the triangles are equal in all respects, and consequently, AC is equal to DF. Now, the triangles ABC and DEF have their sides equal, each to each, and conse- quently, the corresponding angles are also equal ; that is, the angle ABC is equal to DEF: hence, the inclination of the planes ASB and CSB, is equal to the inclination of the planes DTE and FTE. In like manner, it may be shown that the planes of the other equal angles are equally in- clined ; which was to he proved. Cor. If the plane angles ASB and BSC are equal, respectively, to the plane angles DTE and ETF, and the inclination of the faces ASB and BSC is equal to that of the faces DTE and ETF, then are the remaining plane angles, ASC and DTF, equal to each other. Scholium 1. If the planfes of the equal plane angles are like placed, the triedral angles are equal in all respects, for they may be placed so as to coincide. If the planes of the equal angles are not similarly placed, the triedral angles are said to be angles equal by symmetry, or symmetrical 186 GEOMETRY. triedral angles. In this case, they may be placed so that two of the homologous faces shaU coincide, the triedral angles lying on opposite sides of the plane, which is then called a plane of symmetry. In this position, for every point on one side of the plane of symmetry, there is a corresponding point on the other side. Scholium 2. If the plane angles ASB and DTE are equal to each other, and the inclination of the face ASB to each of the faces BSC and ASC is equal, respectively, to the inclination of DTE to each of the faces ETF and DTF, then are the plane angles BSC and CSA equal, respective- ly, to the plane angles ETF and FTD. For, place the plane angle ASB upon its equal DTE, so that the point S shall coincide with T, the edge SA with TD, and the edge SB with TE, then will the face BSC take the direction of the face ETF, and the edge SC will lie somewhere in the plane ETF; the face ASC will take the direction of, the face DTF, and the edge SC will lie somewhere in the plane DTF. Since SC is at the same time in both the planes ETF and DTF, it must be on their intersection (P. ni.) : hence, the plane angles BSC and CSA coincide with and are equal, respectively, to ETF and FTD. If the triedral angle whose vertex is S can not be made to coincide with the triedral angle whose vertex is T, it may be made to coincide with its symmetrical tri- edral angle, and the corresponding plane angles would be equal, as before. Note 1. — The projection 'of a point on a plane is the foot of a perpendicular drawn from the point to the plane. Note 2. — The projection of a line on a plane is that line of the plane which joins the projection of the two extreme points of the given line on the plane. BOOK VI. 187 EXEEOISES. 1. Find a point in a plane eqioidistant from two given points without and on the same side of the plane. 2. From two given points on the same side of a given plane, draw two hnes that shall meet the plane in the same point and make equal angles with it. [The angle made hy a line with a plane is the angle which the line makes with its projection on the plane.] 3. What is the greatest number of equilateral triangles that can be grouped about a point so as to form a con- vex polyedral angle? /' f , ? j '^'-' ' ■'■,^, 4. Show that if from any two points in the edge of a diedral angle straight lines are drawn in each of its faces perpendicular to the edge, these lines contain equal angles. 5. From any point within a diedral angle, draw a per- pendicular to each of its two faces, and show that the angle contained by the perpendiculars is the supplement of the diedral angle. 6. Show that if a plane meets another plane, the sum of the adjacent diedral angles is equal to two right angles. 7. Show- that if two planes intersect each other, the opposite or vertical diedral angles are equal to each other. 8. Show that if a plane intersects two parallel planes, the sum of the interior diedral angles on the same side is equal to two right angles. 9. Show that if two diedral angles have their faces parallel and lying in the same or in opposite directions, they are equal. 10. Show that every point of a plane bisecting a diedral angle is equidistant from the faces of the angle. 188 GEOMETRY. 11. Show that the inclination of a line to a plane — that is, the angle which the hne makes with its own projection on the plane — ^is the least angle made by the line with any line of the plane. 12. Show that if three lines are perpendicular to a fourth at the same point, the first three are in the same plane. 13. Show that when a plane is perpendicular to a given line at its middle point, every point of the plane is equally distant from the extremities of the line, and that every point out of the plane is unequally distant from the extremities of the line. 14. Show that through a line parallel to a given plane, but one plane can be passed perpendicular to the given plane. 15. Show that if two planes which intersect contain two lines parallel to each other, the intersection of the planes is parallel to the lines. 16. Show that when a line is parallel to one plane and perpendicular to another, the two planes are perpen- dicular to each other. 17. Draw a perpendicular to two lines not in the same plane. 18. Show that the three planes which bisect the diedral angles formed by the consecutive faces of a tri- edral angle, meet in the same line. BOOK VII. POLYEDRONS. DEFmiTIONS. 1. A PoLYEDEON is & volume bounded by polygons. The bounding polygons are called faces of the poly- edron ; the lines in which the faces meet, are called edges of the polyedron ; the points in which the edges meet, are called vertices of the polyedron. 2. A Peism is a polyedron in which two of the faces are polygons equal in all re- yf y\ spects, and having their homologous sides //^..-fy parallel. The other faces are parallelograms III (B. L, P. XXX.). J44 The equal polygons are called bases of \A-^ the prism; one the upper, and the other the lower base; the paraIlelogram.s taken together make up the lateral or convex surface ai the prism ; the lines in which the lateral faces meet, are called lateral edges, and the lines in which the lateral faces meet either base are called hasal edges of the prism. 3. The Altitude of a prism is the perpendicular dis- tance between the planes of its bases. 4. A Right Peism is one whose lateral edges are perpendicular to the planes of the o In this case, any lateral edge is equal to the altitude. >i/ 190 GEOMETKY. 5. An Oblique Prism is one whose lateral edges are oblique to the planes of the bases. In this case, any lateral edge is greater than the altitude. 6. Prisms are named from the number of sides of their bases ; a triangular prism is one whose bases are triangles; a pentagonal prism is one whose bases are pen- tagons, &c. 7. A Paballelopipedon is a prism whose bases are parallelograms. A Right Parallelopipedon is one whose lateral edges are perpendicular to the planes of the bases. A Bectangular Parallelopipedon is one whose faces are all rectangles. A Cube is a rectangular parallelopipedon whose faces are squares. 8. A Pyramid is a polyedron bounded by a polygon called the base, and by tri- angles meeting at a common point, called the vertex of the pyramid. The triangles taken together make up the lateral or convex surface of the pyramid ; the lines in which the lateral faces meet, are called the lateral edges, and the lines in which the lateral faces meet the base are called basal edges of the pjn^amid. 9. Pyramids are named from the number of sides of their bases ; a triangular pyramid is one whose base is a triangle ; a quadrangular pyramid is one whose base is a quadrilateral, and so on. 10. The Altitude of a pyramid is the perpendicular distance from the vertex to the plane of its base. BOOK VII. 191 11. A Right PrBAMiD is one whose base is a regular polygon, and in -which the perpendictilar, drawn from the vertex to the plane of the base, passes through the cen- tre of the base. This perpendicular is called the axis of the p3rraniid. 12. The Slant Height of a right pyramid, is the per- pendicular .distance from the vertex to any side of the base. 13. A Truncated Pyramid is that portion of a pyramid included between the base and any plane which cuts the pyramid. When the cutting plane is parallel to the base, the truncated pyramid is called a FRUSTUM OF A PYRAMID, and the inter- section of the cutting plane with the pyramid, is called the upper base of the frustum ; the base of the pyramid is called the lower base of the frustum. 14. The Altitude of a frustum of a pyramid, is the perpendicular distance between the planes of its bases, 15. The Slant Height of a frustum of a right pyra- mid, is that portion of the slant height of the pyramid .which lies between the planes of its upper and lower bases. 16. Similar Polyedrons are those which are bounded by the same number of similar polygons, similarly placed. Parts which are similarly placed, whether faces, edges, or angles, are called homologous. 17. A Diagonal of a polyedron, is a straight line join- ing the vertices of two polyedral angles not in the same face. 192 GEOMETRY. 18. The Volume of a Poltedson is its numerical value expressed in terms of some other polyedron taken as a unit. The unit generally employed is a cube constructed on the linear unit as an edge. PROPOSITION I. THEOEEM. ■ The convex surface of a right prism is equal to the perim- eter of either base multiplied by the altitude. Let ABCDE-K be a right prism : then is its convex sur- face equal to, (AB + BC + CD + DE + EA) x AF. For, the convex surface is equal to the sum of all the rectangles AG, EH, CI, DK, EF, which compose it. Now, the altitude of each of the rectangles AF, BG, CH, &c., is equal to the altitude of the prism, and the area of each rec- tangle is equal to its base multiplied by its altitude (B. IV., P. V.) : hence, the sum of these rectangles, or the convex surface of the prism, is equal to, (AB + BC + CD + DE + EA) x AF ; that is, to the perimeter of the base multiplied by the altitude ; which was to he proved. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases. BOOK VII. 193 PROPOSITION II. THEOREM. In any prism, the sections made by paralieL planes are polygons eqwal in all respects. Let the prism AH be intersected by the parallel planes NP, SV: then are the sections NOPQR, STVXY, equal poly- gons. For, the sides NO, ST, are parallel, being the intersections of parallel planes with a third plane ABG F ; these sides, NO, ST, are included between the par- allels NS, OT: hence, NO is equal to ST (B. I., P. XXVIII., C. 2). For like reasons, the sides OP, PQ, QR, &c., of NOPQR, are equal to the sides TV, VX, &c., of STVXY, each to each ; and since the equal sides are parallel, each to each, it follows that the angles NOP, OPQ, &c., of the first section, are equal to the angles STV, TVX, &c., of the second section, each to each (B. VL, P. XIII.) : hence, the two sections NOPQR, STVXY, are equal in all respects ; which was to be proved. Cor. The bases of a prism and any section of a prism parallel to the bases, are equal in all respects. PROPOSITION m. THEOREM. If a pyramid is cut by a plane parallel to the base : 1°. The edges and the altitude are divided proportionally: 2°. The section is a polygon similar to the base. Let the pyramid S-ABCDE, whose altitude is SO, be cut by the plane abode, parallel to the base ABODE. 194 GEOMETBT. 1°. The edges and altitude are divided proportionally. For, let a plane be passed through the vertex S, parallel to the base AC; then the edges and the altitude are cut by three parallel planes, and are consequently divided proportion- ally (B. VL, P. XV., C. 2) ; which was to be proved. 2°. The section abcde is similar to the base ABCDE. For, each side of the section is paral- lel to the corresponding side of the base (B. VI., P. X.) ; hence, the corresponding angles of the section and of the base are equal (B. VI., P. Xm.) ; the two polygons are therefore mutually equi- angular. Again, because ab is parallel to AB, and be to BC, the triangle Sba is similar to SBA, and Sbc to SBC ; hence, S6 SB, aft : AB : : S6 : SB, and be : BC whence (B. II., P. IV.), ab : kB : : be : BC. In like manner, it may be shown that the remaining sides of dbede are proportional to the corresponding sides of ABCDE ; hence (B. IV., D. 1), the polygons are similar ; which was to be proved. Cor. 1. If two pyramids S-ABCD and S-XYZ, having a common vertex S and their bases in the same plane, are cut by a plane aoz parallel to the plane of their bases, the sections are to each other as the bases. BOOK VII. 195 so'; SS' SO'. S^' so^ For the polygons abed and ABCD, being similar, are to each other as the squares of any homologous sides (B. IV., P. XXVII.) ; but a6' : AB' : : Sa' : SA' : : So' : hence (B. II., P. IV.), we have, abed : ABCD In like manner, we have, xyz : XYZ hence, abed : ABCD : : xyz : XYZ. Cor. 2. If the bases are equal, any sections at equal distances from the vertex, or from the bases, are equal. Cor. 3. The area of any section parallel to the base is proportional to the square of its distance from the vertex. Cor. 4. If the two pyramids are cut by a plane KTR, so that ST is a mean proportional between So and SO, that is, so that ST* is a mean proportional between So' and SO', the section KLMN is a mean proportional between abed and ABCD, and also PQR is a mean proportional between xyz and XYZ. PROPOSITION IV. THEOREM. The convex su/rfaoe of a right pyramid is equal to the perimeter of its base multiplied by half the slant height. S Let S be the vertex, ABODE the base, and SF, perpendicular to EA, the slant height of a right pyramid: then is the convex surface equal to, (AB + BC + CD + DE + EA) x iSF. Draw SO perpendicular to the plane of the base. 196 GEOMETBY. From the definition of a right pyramid, the point is the centre of the base (D. 11) : hence, the lateral edges, SA, SB, &c., are all equal (B. VI., P. Y.) ; but the sides of the base are all equal, being sides of a regular polygon: hence, the lateral faces are aU equal, and consequently their altitudes are all equal, each being equal to the slant height of the pyramid. Now, the area of any lateral face, as SEA, is equal to its base EA, multiplied by half its altitude SF: hence, the sum of the areas of the lateral faces, or the convex sur- face of the pyramid, is equal to, (AB + BC + CD + DE + EA') x ^SF; which was to he proved. Scholium. The convex surface of a frustum of a right pyramid is equal to half the sum of the perimeters of its upper and lower bases, multiplied by the slant height. Let ABCDE-e be a frustum of a right pyramid, whose vertex is S : then the section ahcde is similar to the base ABCDE, and their homologous sides are parallel (P. III.). Any lateral face of the frustum, as AEea, is a trapezoid, whose altitude is equal to F/, the slant height of the frustum ; hence, its area is equal to i (EA + ea) x F/ (B. IV., P. VH.). But the area of the convex surface of the frustum is equal to the sum of the areas of its lateral faces; it is, therefore, equal to the half sum of the perimeters of its upper and lower bases, multiplied by the slant height. BOOK VII. 197 PROPOSITION" V. THEOREM. If the three faces which include a triedral an0e of a prism are equal in all respects to the three faces which include a triedral angle of a second prism, ^each to each, and are like placed, the two prisms are equal in all respects. Let B and 6 be the vertices of two triedral angles, included by faces respectively equal to each other, and similarly placed: then the prism ABCDE-K is equal to the prism ahcde-k in all respects. For, place the base abcde upon the equal base ABCDE, so that they shall coincide ; then because the triedral an- gles "whose vertices are 6 and B, are equal, the parallelogram bh will co- incide with BH, and the parallelogram bf with BF : hence, the two sides fg and gh, of one upper base, will coincide with the homologous sides FG and GH, of the other upper base; and because the upper bases are equal in all respects, and have been shown to coincide in part, they must coincide throughout; consequently, each of the lateral faces of one prism will coincide with the corresponding lateral face of the other prism ; the prisms, therefore, coincide throughout, and are therefore equal in all respects ; which was to be proved. Cor. If two right prisms have their bases equal in all respects, and have also equal altitudes, the prisms themselves are equal in all respects. For, the faces which include any triedral angle of the one, are equal in all respects to the faces which include the corresponding triedral angle of the other, each to each, and they are similarly placed. 198 GEOMETKY. PROPOSITION VL THEOREM. In any parallelopipedon, the opposite faces are equal in all respects, each to each, and their planes are parallel. Let ABCD-H be a parallelopipedon : then its opposite faces are equal and their planes are parallel. For, the bases, ABCD and EFGH are equal, and their planes parallel by definition (D. 7). The opposite faces AEHD and BFGC, have the sides AE and BF parallel, because they are opposite sides of the parallelogram. BE ; and the sides EH and FG parallel, because they are opposite sides of the parallelogram EG ; and conse- quently, the angles AEH and BFG are equal (B. VI. , P. XIII.). But the side AE is equal to BF, and the side EH to FG ; hence, the faces AEHD and BFGC are equal; and because AE is parallel to BF, and EH to FG, the planes of the faces are parallel (B. VI., P. XIII.). In like manner, it may be shown that the parallelograms ABFE and DCGH, are equal and their planes parallel: hence, the opposite faces are equal, each to each, and their planes are parallel; which was to he proved. Cor. 1. Any two opposite faces of a parallelopipedon may be taken as bases. Cor. 2. In a rectangular parallelo- pipedon, the square of any of the diagonals is equal to the sum of the squares of the three edges which meet at the same vertex. For, let FD be one of the diagonals, and draw FH, BOOK VII. Then, in the right-angled triangle FHD, we have, FD'' = DH' + FFP. But DH is equal to FB, and FH^ is equal to FA^ plus AH^ or FC'' : hence, 199 FD' FB' + FA^ + FC . Cor. 3. A parallelopipedon may be constructed on three straight lines AB, AD, and AE, intersecting in a common point A, and not lying in the same plane. For, pass through the extremity of each line, a plane parallel to the plane of the two others ; then will these planes, together with the planes of the given lines, be the faces of a par- allelopipedon. PROPOSITION Vn. THEOREM. If a plane is passed through the diagonally opposite edges of a parallelopipedon, it divides the parallelopipedon into two equal triangular prisms. Let ABCD-H be a parallelopipedon, and let a plane be passed through the edges BF and DH ; then are the prisms ABD-H and BCD-H equal in volume. For, through the vertices F and B let planes be passed perpendicular to FB, the former cutting the other lateral edges in the points e, h, g, and the latter cutting those edges produced, in the points a, d, and c. The sections Fehg and Bade are parallelograms, be- cause their opposite sides are parallel, 200 GEOMETRY. each to each (B. VI., P. X.) ; they are also equal (P. 11.) : hence, the polyedron Badc-g is a right prism (D. 2, 4), as are also the polyedrons Badr-h and Bcd-h. Place the triangle feh upon Bad, so that F shall co- incide "with B, e with a, and h with d ; then, because eE, 7iH, are perpendicular to the plane Feh, and aA, dD, to the plane Bod, the line eE takes the direction aA, and the line hH the direction dD. The lines AE and ae are equal, because each is equal to BF (B. I., P. XXVIIL). If we take away from the line aE the part ae, there remains the part eE ; and if from the same line, we take away the part AE, there remains the part Aa : hence, eE and aA are equal (A. 3) ; for a like reason hH is equal to dD : hence, the point E coincides with A, and the point H with D, and consequently, the polyedrons Feh-H and Bad-D co- incide throughout, and are therefore equal. If from the polyedron Bad-H, we take away the part Bad-D, there remains the prism BAD-H ; and if from the same polyedron we take away the part Feh-H, there re- mains the prism Bad-7i : hence, these prisms are equal in volume. In like manner, it may be shown that the prisms BCD-H and Bcd-h are equal in volume. The prisms Bad-h, and Bcd-h, have equal bases, because these bases are halves of equal parallelograms (B. I., P. XXVIIL, C. 1) ; they have also equal altitudes; they are therefore equal (P. V., C.) : hence, the prisms BAD-H and BCD-H are equal (A. 1) ; which was to be proved. Cor. Any triangular prism ABD-H, is equal to half of • the paraUelopipedon AG, which has the same triedral angle A, and the same edges AB, AD, and AE. BOOK VII. 201 PROPOSITION Vm. THEOREM. // two parcMelopipedons have a common lower base, and their, upper bases between the same parallels, they are equal in volume. Let the parallelopipedons AG and AL have the common lower base ABCD, and their upper bases EFGH and IKLM, between the same parallels EK and HL: then are they equal in volume. For, in the triangular prisms AEl-M and BFK-L, the faces AEI and BKF are equal, having their sides respect- ively equal; the faces AEHD and BFGC are equal (P. VI.) ; the faces EH Ml and FGLK are equal, as they consist, respectively, of the common part FGMI and the equal parts EHGF and IMLK: hence, the triangular prisms AEI-M and BFK-L are equal (P. V.). If from the polyedron ABKE-H, we take away the prism BFK-L, there remains the parallelopipedon AG ; and if from the same polyedron we take away the prism AEI-M, there remains the parallelopipedon AL: hence, these parallelo- pipedons are equal in volume (A. 3) ; which was to be proved. 202 GEOMBTBT. PROPOSITION IX. THEOREM. If two ■parallelopvpedons have a common lower base- and the same altitude, they are equal in volume. Let the parallelopipedons AG and AL have the common lower base A BCD and the same altitude : then are they equal in volume. Because they have the same altitude, their upper bases lie in the same plane. Let the sides IM and KL be prolonged, and also the sides FE and GH ; these prolonga- tions form a parallelogram OQ, which is equal to the common base of the given parallelopipedons, because its sides are respectively parallel and equal to the corresponding sides of that base. Now, if a third parallelopipedon be constructed, having for its lower base the parallelogram ABCD, and for its upper base NOPQ, this third parallelopipedon will be equal in volume to the parallelopipedon AG, since they will have the same lower base, and their upper bases between the same parallels, QG, NF (P. VITL). For a Kke reason, this third parallelopipedon will also be equal in volume to the parallelopipedon AL: hence, the two parallelopipedons AG, AL, are equal in volume ; which was to he proved. . Cor. Any oblique parallelopipedon is equal in volume to a right parallelopipedon having the same base and the same altitude. BOOK VII. 203 MQ LP PROPOSITION X. PROBLEM. To construct a rectangular parallelopipedon equal in volume to a. right parallelopipedon whose base is any parallelo- gram. Let ABCD-M be a right parallelopipedon, having for its base the parallelogram A BCD. Through the edges Al and BK pass the planes AQ and BP, respectively perpendic- ular to the plane AK, the former meeting the face DL in OQ, and the latter meet^ ing that face produced in N P : then the polyedron AP is a rectangular parallelopip- edon equal to the given parallelopipedon. It is a rectangular parallelopipedon, be- cause all of its faces are rectangles, and it is equal to the given parallelopipedon, because the two may be regarded as having the common base AK (P. VI., C. 1), and an equal altitude AO (P. IX.). s, ^ t K 1 c k H R Cor. 1. Since any oblique parallelopipedon is equal in volume to a right parallelopipedon, having the same base and altitude (P. IX., Cor.) ; and since any right parallelo- pipedon is equal in volume to a rectangular parallelopip- edon having an equal base and altitude ; it follows, that any oblique parallelopipedon is equal in volume to a rec- tangular parallelopipedon, having an equal base and an equal altitude. Cor. 2. Any two parallelopipedons are equal in volume when they have equal bases and equal altitudes. 204 GEOMETRY. PROPOSITION XI. THEOREM. Two rectangular parallelopipedons hairing a common lower base, are to each other as their altitudes. Let the parallelopipedons AG and AL have the common lower base ABCD : then are they to each other as then- altitudes AE and Al. 1°. Let the altitudes be commensurable, and suppose, for example, that AE is to Al, as 15 is to 8. Conceive AE to be divided into 15 equal parts, of which Al contains 8 ; through the points of division let planes be passed parallel to ABCD. These planes divide the parallelopipedon AG into 15 parallelopipedons, which have equal bases (P. II., C.) and equal altitudes ; hence, they are equal (P. X., Cor. j^. Now, AG contains 15, and AL 8 of these equal parallelopipedons; hence, AG is to AL, as 15 is to 8, or as AE is to AL In like manner, it may be shown that AG is to AL, as AE is to Al, when the altitudes are to each other as any other whole numbers. D\ Al 2°. Let the altitudes be incommen- B c surable. Now, if AG is not to AL, as AE is to Al, let us suppose that AG : AL : : AE : AO, in which AO is greater than Al. Divide AE into equal parts, such that each is less than 01 ; there is at least one point of division m, between BOOK VII. 205 and I. Let P denote the parallelopipedon, whose base is ABCD, and altitude km ; since the altitudes AE, Am, are to each other as two whole numbers, we have. AG AE But, by hypothesis, we have, AG : AL : : AE : therefore (B. II., P. IV., C), AL : P : : AO : Am. AO; Am. 0- y-- D\ But AO is greater than Am ; hence, if the proportion is true, AL must be greater than P. On the contrary, it is less ; consequently, the fourth term of the proportion can not be greater than Al. In like manner, it may be shown that the fourth term can not be less than Al ; it is, therefore, equal to Al. In this case, therefore, AG is to AL as AE is to Al. Hence, in all cases, the given parallelopipedons are to each other as their altitudes ; which was to he proved. Sch. Any two rectangular parallelopipedons whose bases are equal in all respects, are to each other as their alti- tudes. PROPOSITION XII. THEOREM. Two rectangular parallelopipedons having equal altitudes, are to each other as their bases. Let the rectangular parallelopipedons AG and AK have the same altitude AE : then are they to each other as their bases. 206 GEOMETRY. For, place them so that the plane angle EAO shall be common, and produce the plane of the face NL, until it intersects^ the plane of the face HC, in PQ ; we thus form a third rectangular par- aUelopipedon AQ. The parallelopipedons AG and AQ have a common base AH ; they are therefore to each other as their altitudes AB and AO (P. XL) : hence, we have the proportion, vol. AG vol. AQ : : AB AO. The parallelopipedons AQ and AK have the common base AL ; they are therefore to each other as their altitudes AD and AM : hence, vol. AQ : vol. AK AD AM. Multiplying these proportions, term by term (B. II., P. XII.), and omitting the common factor, vol. AQ, we have. vol. AG vol. AK : : AB x AD AO X AM. But ABxAD is equal to the area of the base ABCD, and AOxAM is equal to the area of the base AM NO: hence, two rectangular parallelopipedons having equal altitudes, are to each other as their bases ; which was to be proved. BOOK VII. 207 PROPOSITION XIII. THEOREM. Any two rectangular paraZlelopipedons are to each other as the products of their bases and altitudes; that is, as the products of their three dimensions. Let AZ and AG be any two rectangular paraUelopip- edons : then are they to each other as the products of their three dimensions. For, place them so that the plane angle EAO shall be common, and produce the faces necessary to complete the rectangular parallelopip- edon AK. The parallelopipe- dons AZ and AK have a com- mon base AN ; hence (P. XL), E \ K \K 1^ H V \ i X! ■■- - 1 \z l\ D A \ \ i c N' vol. AZ vol. AK AX AE. The parallelopipedons AK and AG have a common alti- tude AE ; hence (P. XIL), vol. AK vol. AG AMNO ABCD. Multiplying these proportions, term by term, and omitting the common factor, vol. AK, we have, vol. fJ. : vol. kG :: AMNOxAX : ABCDxAE; or, since AMNO is equal to AMxAO, and ABCD to ABxAD, vol. AZ : vol. AG : : AM x AO x AX : AS x AD x AE ; which was to be proved. 208 GEOMETRY. Coi: 1. If we make the three edges AM, AO, and AX, each equal to the linear unit, the parallelopipedon AZ be- comes a cube constructed on that unit, as an edge ; and consequently, it is the unit of volume. Under this sup- position, the last proportion becomes, 1 : vol. AG : : 1 : AB x AD x AE ; whence, vol. AG = AB x AD x AE. Hence, the volume of any rectangular parallelopipedon is equal to the product of its three dimensions; that is, the number of times which it contains the unit of volume, is equal to the continued product of the number of linear units in its length, the number of linear units in its breadth, and the number of linear units in its height. Cor. 2. The volume of a rectangular parallelopipedon is equal to the product of its base and altitude; that is, the number of times which it contains the unit of volume, is equal to the number of superficial units in its base, mul- tiplied by the number of linear units in its altitude. Cor. 8. The volume of any parallelopipedon is equal to the product of its base and altitude (P. X., C. 1). PROPOSITION" XIV. THEOREM. The volume of any prism is equal to the product of its base and altitude. Let ABCDE-K be any prism : then is its volume equal to the product of its base and altitude. For, through any lateral edge, as AF, and the other lateral edges not in the same faces, pass the planes AH, Al, dividing the prism into triangular prisms. These prisms all have a common altitude equal to that of the given prism. BOOK VII. 209 Now, the volume of any one of the triangular prisms, as ABC-H, is equal to half that of a parallelopipedon con- structed on the edges BA, BC, BG (P. VII., C.) ; but the volume of this paral- lelopipedon is equal to the product of its base and altitude (P. XIII., C. 3); and because the base of the prism is half that of the parallelopipedon, the volume of the prism is also equal to the product of its base and altitude : hence, the sum of the triangular prisms, "which make up the given prism, is equal to the sum of their bases, which make up the base of the given prism, into their common altitude ; which was to he proved. ^^ / ■\ v -^'"Gy F^ » r 'A .^'- '■/ ^-^ ^^ Cor. Any two prisms are to each other as the products of their bases and altitudes. Prisms having equal bases are to each other as their altitudes. Prisms having equal altitudes are to each other as their bases. PROPOSITION XV. THEOREM. Two triangular pyramids having equal bases and equal altitudes are equal in volume. Let S-ABC, and S-abc, be two pyramids having their equal bases ABC and ahc in the same plane, and let AT be their common altitude : then are they equal in vol- ume. For, if they are not equal in volume, suppose one of them, as S-ABC, to be the greater, and let their difference be equal to a prism whose base is ABC, and whose alti- tude is ka. ^10 OKOMETRY. Divide the altitude AT into equal parts, Ax, xy, &c., each of which is less than ka, and let k denote one of these parts ; through the points of division pass planes parallel to the plane of the bases ; the sections of the two pyramids, by each of these planes, are equal, namely, DEF to def, GHI to gU, &c. (P. in., 0. 2). On the triangles ABC, DEF, &c., as lower bases, con- struct exterior prisms whose lateral edges are parallel to AS, and whose altitudes are equal to k: and on the tri- angles def, ghi, &c., taken as upper bases, construct inte- rior prisms, whose lateral edges are parallel to aS, and whose altitudes are equal to k. It is evident that the sum of the exterior prisms is greater than the pyramid S-ABC, and also that the sum of the interior prisms- is less than the pyramid S-abc : hence, the difference be- tween the sum of the exterior and the sum of the interior prisms, is greater than the difference between the two pyramids. Now, beginning at the bases, the second exterior prism EFD-G, is equal to the first interior prism efd-a, because BOOK VII. 211 they have the same altitude k, and their bases EFD, efd, are equal: for a like reason, the third exterior prism HIG-K, and the second interior prism hig-d, are equal, and so on to the last in each set: hence, each of the exterior prisms, excepting the first BCA-D, has an equal correspond- ing interior prism ; the prism BCA-D, is, therefore, the difEerence between the sum of all the exterior prisms, and the sum of all the interior prisms. But the difference between these two sets of prisms is greater than that between the two pyramids, which latter difference was supposed to be equal to a prism whose base is BCA, and whose altitude is equal to Aa, greater than k; conse- quently, the prism BCA-D is greater than a prism having the same base and a greater altitude, which is impossible : hence, the supposed inequality between the two pyramids can not exist ; they are, therefore, equal in volume ; which was to be proved. PROPOSITION XVI. THEOREM. Any triangular prism may ie divided into three triangular pyramids, equal to each other in volume. Let ABC-D be a triangular prism : then can it be divided into three equal triangular pyramids. For, through the edge AC, pass the plane ACF, and through the edge EF pass the plane EFC. The pyramids ACE-F and ECD-F, have their bases ACE and ECD equal, because they are halves of the same parallelogram ACDE ; and they have a common altitude, because 212 GEOMETRY. their bases are in the same plane AD, and their vertices at the same point F; hence, they are equal in voliome (P. XV.). The pyramids ABC-F and DEF-C, have their bases ABC and DEF, equal, because they are the bases of the given prism, and their altitudes are equal because each is equal to the altitude of the prism ; they are, therefore, equal in volume : hence, the three pyramids into which the prism is divided, are all equal in volume ; which was to be proved. Cor. 1. A triangular pyramid is one third of a prism having an equal base and an equal altitude. Cor. 2. The volume of a triangular pyramid is equal *o one third of the product of its base and altitude. PROPOSITION XVn. THEOREM. TTie volume of any pip'amid is equal to one third of the product of its base and altitude. Let S-ABCDE, be any pyramid : then is its volume equal to one third of the product of its base and altitude. For, through any lateral edge, as SE, pass the planes SEB, SEC, dividing the pyramid into triangular pyramids. The altitudes of these pyramids are equal to each other, because each is equal to that of the given pyramid. Now, the volume of each triangular pyramid is equal to one third of the product of its base and alti- tude (P. XVI., C. 2); hence, the sum of the volumes of the triangular pyramids, is equal to one third of the product of the sum of their BOOK VII. 213 bases by their common altitude. But the sum of the triangular pyramids is equal to the given pyramid, and the sum of their bases is equal to the base of the given pyramid : hence, the volume of the given pyramid is equal to' one third of the product of its base and altitude ; which was to he proved. Cor. 1. The volume of a pyramid is equal to one third of the volume of a prism having an equal base and an equal altitude. Cor. 2. Any two .pyramids are to each other as the products of their bases and altitudes. Pyramids having equal bases are to each other as their altitudes. Pyramids having equal altitudes are to each other as their bases. Scholium. The volume of a polyedron may be found by dividing it into triangular pyramids, and computing their volumes separately. The sum of these volumes is equal to the volume of the polyedron. PEOPOSITION XVIII. THEOREM. Th& volume of a frustum of any triangular pyramid is equal to the sum of the volumes of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base of the frustum, and a mean proportional between the two bases. Let FG H-/i be a frustum of any triangular pyramid : then is its volume equal to that of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base FGH, the upper base fgh, and a mean proportional between these bases. For, through the edge FH, pass the plane FHgf, and 214 GEOMETBY. through the edge fg, pass the plane /grH, dividing the frustum into three pyramids. The pyra- mid g-FGH, has for its base the lower base FGH of the frustum, and its alti- tude is equal to that of the frustum, because its vertex g is in the plane of the upper base. The pyramid H-/grfe, has for its base the upper base fgh of the frustum, and its altitude is equal to that of the frustum, because its vertex Ues in the plane of the lower base. The remaining pyramid may be regarded as having the triangle F/H for its base, and the point g for its vertex. From g, draw gK parallel to /F, and draw also KH and K/. Then the pyramids K-F/H and g-FfH, are equal; for they have a common base, and their altitudes are equal, because their vertices K and g are in a line parallel to the base (B. VI., P. Xn., C. 2). Now, the pyramid K-F/H may be regarded as having FKH for its base and / for its vertex. From K, draw KL parallel to G H ; it is parallel to gh : then the triangle FKL is equal to fgh, for the side FK is equal to fg, the angle F to the angle /, and the angle K to the angle g. But, FKH is a mean proportional between FKL and FGH (B. IV., P. XXrV., C), or between fgh and FGH. The pyra- mid /-FKH, has, therefore, for its base a mean propor- tional between the upper and lower bases of the frustum, and its altitude is equal to that of the frustum; but the pyramid /-FKH is equal in volume to the pyramid g-F/H : hence, the volume of the given frustum is equal to that of three pyramids whose common altitude is equal to that of the frustum, and whose bases are the upper base, the lower base, and a mean proportional between them; which was to be proved. BOOK VII. 215 Cor. The volume of the frustum of any pyramid is equal to the sum of the volumes of three pyramids whose commjon altitude is that of the frustum^, and whose bases are the lower base of the frustum, the upper base of the frustum, and a mean proportional between them. For, let ABCDE-e be a frustum of a pyramid -whose vertex is S, and let PQ be a section parallel to the bases, such that ' distance from S is a mean proportional between the distances from S to the two bases of the frus- tum. Let planes be passed through SB, and SE, SD, dividing the frustum into triangular frustums ; the section of each of the triangular frustums is a mean proportional between its bases (P. III., C. 4). Now the sum of the tri- angular frustums is equal to the sum of three sets of pyramids, whose altitude is that of the given frustum. The sum of the bases of the first set is the lower base of the frustum, the sum of the bases of the second set is the upper base of the frustum, and the sum of the bases of the third set is a mean proportional between these bases. Hence, the sum of the partial frustums, that is, the given frustum, is equal to the sum of three pyramids having the same altitude as the given frustum, and whose bases are the two bases of the frustum and a mean proportional between them. PROPOSITION XIX. THEOREM. Similajr triangular prisms are to each other a.i the cubes of their homologous edges. Let CBD-P, chd-p, be two similar triangular prisms, and let BC, 6c, be any two homologous edges : then is the prism CBD-P to the prism cbdr-p, as BC' to 6c^ 216 GEOMETRY. For, the homologous angles B and b are equal, and the faces which bound them are similar (D. 16): hence, these triedral angles may be applied, one to the other, so that the angle cbd will coincide with CBD, the edge ba with BA. In this case, the prism cbd-p will take the position Bcd-p. From A draw AH perpendicular to the common base of the prisms : then the plane BAH is perpendicular to the plane of the common base (B. VI., P. XVI.). From a, in the plane BAH, draw ah perpendicular to BH : then ah is also per- pendicular to the base BDC (B. VI., P. XVII.) ; and AH, o.h, are the altitudes of the two prisms. Since the bases CBD, cbd, are similar, we have (B. IV., P. XXV.), base CBD base cbd CB' cb\ Now, because of the similar triangles ABH, aBh, and of the similar parallelograms AC, ac, we have, AH : ah : : CB : cb; hence, multiplying these proportions term by term, we have, base CBD X AH : base cbd x ah : CB^ : cb^. But, base CBD x AH is equal to the volume of the prism CDB-A, and base cbd x ah is equal to the volume of the prism cbd-p : hence, prism CDB-P : prism cbd-p : : CB' cb^ ; which was to be proved. BOOK VII. 217 Cor. 1. Any two similar prisms are to each other as the cubes of their homologous edges. For, since the prisms are similar, their bases are simi- lar polygons (D. 16); and these similar polygons may each be divided into the same number of similar triangles, similarly placed (B. IV., P. XXVI.) ; therefore, each prism may be divided into the same number of triangular prisms, having their faces similar and like placed ; conse- quently, the triangular prisms are similar (D. 16). But these triangular prisms are to each other as the cubes of their homologous edges, and being like parts of the polyg- onal prisms, the polygonal prisms themselves are to each other as the cubes of their homologous edges. Cor. 2. Similar prisms are to each other as the cubes of their altitudes, or as the cubes of any other homolo- gous lines. PROPOSITION XX. THEOREM. Similar pyramids are to each other as the cubes of their hom,ologaus edges. Let S-ABCDE, and S-ahcde, be two similar pyramids, so placed that their homologous angles at the vertex shall coincide, and let AB and ah be any two homologous edges : then are the pyra- mids to each other as the cubes of AB and ah. For, the face SAB, being similar to Sa6, the edge AB is parallel to the edge ah, and the face SBC being similar to She, the edge BC is parallel to hc\ hence, the planes of the bases are parallel (B. VI., P. XIIL). 218 GEOMETBY. Draw so perpendicular to the base ABCDE ; it will also be perpendicular to the base dbcde. Let it pierce that plane at the point o; then SO is to So, as SA is to Sa (P. IIL), or as A B is S to ab\ hence, i-SO : |So : : AB : ah. But the bases being similar polygons, we have (B. IV., P. XXVIL), A' hose ABCDE base abcde : : AB^ : oft*. Multiplying these proportions, term by term, we have, base ABCDE x|SO : base o6cdex|So : : AB* : W. But, hose ABCDE x^SO is equal to the volume of the pyra- mid S-ABCDE, and hase dbcde x^So is equal to the volume of the pyramid S-ahcde ; hence, pyramid S-ABCDE : pyramid S-ahcde : : AB* : a6* ; which was to he proved. Cor. Similar pyramids are to each other as the cubes of their altitudes, or as the cubes of any other homolo- gous lines. BOOK VII. 219 GENERAL FORMULAS. If we denote the volume of any prism by V, its base by B, 9,nd its altitude by H, -we shall have (P. XIV.), V=BxH (1.) If we denote the volume of any pyramid by V, its base by B, and its altitude by H, we have (P. XVII.), V = B X iH (2.) If we denote the volume of the frustum of any pjrra- mid by V, its lower base by B, its upper base by b, and its altitude by H, we shall have (P. XVIIL, C), V = (B + 6 + VB X 6) X iH • • • (3.) REGULAR POLYEDRONS. A Regular Polyedron is one whose faces are all equal regular polygons, and whose polyedral angles are equal, each to each. There are five regular polyedrons, namely: 1. The Tetraedron, or regular pyramid — a polyedron bounded by four equal* equilateral triangles. 2. The Hexaedron, or cube — a polyedron bounded by six equal squares. 3. The Octaedron — a polyedron bounded by eight equal equilateral triangles. 4. The DoDEGAEDRON— a polyedron bounded by twelve equal and regular pentagons. 220 GEOMETRY. 5. The IcosAEDBON — a polyedron bounded by twenty equal equilateral triangles. In the Tetraedron, the triangles are grouped about the polyedral angles in sets of three, in the Octaedron they are grouped in sets of four, and in the Icosaedron they are grouped in sets of five. Now, a greater number of equi- lateral triangles can not be grouped so as to form a salient polyedral angle ; for, if they could, the sum of the plane angles formed by the edges would be equal to, or greater than, four right angles, which is impossible (B. VI., P. XX.). In the Hexaedron, the squares are grouped about the polyedral angles in sets of three. Now, a greater number of squares can not be grouped so as to form a salient polyedral angle ; for the same reason as before. In the Dodecaedron, the regular pentagons are grouped about the polyedral angles in sets of three, and for the same reason as before, they can not be grouped in any greater number so as to form a salient polyedral angle. Furthermore,, no other regular polygons can be grouped so as to form a salient polyedral angle ; therefore. Only five regular polyedrons can be formed. TETRA£1>S0» OCTAEDBOM ICOSASDBON Hexabbbon SOSECAEDBON BOOK VII. 221 EXERCISES. 1. What is the convex surface of a right prism whose altitude is 20 feet and whose base is a pentagon each side of which is 15 feet? 2. The altitude of a pyramid is 10 feet and the area of its base 2 5 square feet ; find the area of a section made by a plane 6 feet from the vertex and parallel to the base. 3. Find the convex surface of a right triangular pyra- mid, each side of the base being 4 feet and the slant height 12 feet. 4. A right pyramid whose altitude is 8 feet and whose base is a square each side of which is 4 feet, is cut by a plane parallel to the base and 2 feet from the vertex ; required the convex surface of the frustum included be- tween the base and the cutting plane. 5. The three concurrent edges of a rectangular paral- lelopipedon are 4, 6, and 8 feet ; find the length of the diagonal. 6. Of two rectangular parallelopipedons having equal bases, the altitude of the first is 12 feet and its volume is 275 cubic feet; the altitude of the second is 8 feet — find its volume. 7. Two rectangular parallelopipedons having equal alti- tudes are respectively 80 and 45 cubic feet in volume, and the area of the base of the first is 12 square feet; find the base of the second and the altitude of both. 8. Find the volume of a triangular prism whose base is an equilateral triangle of which the altitude is 3 feet, the altitude of the prism being 8 feet. 9. The volumes of two pyramids having equal altitudes are respectively 60 and 115 cubic yards and the base of the smaller is 8 square yards ; find the base of the larger. 222 GEOMETRY. 10. Given a pyramid whose volume is 512 cubic feet and altitude 8 feet ; find the volume of a similar pyramid whose altitude is 12 feet, and find also the area of the base of each. 11. Find the volume of the frustum of a right trian- gular pyramid with each side of the lower base 6 feet and each side of the upper base 4 feet, the altitude being 5 feet. 12. Find the volume of the pyramid of which the frustum given in the last example is a frustum. [Find the radii of the inscribed circles of the upper and lower bases (B. lY., P. VL, C. 2) ; then the altitude of the pyramid, slant height, and the two radii form two similar triangles from which the altitude may be found.] 13. Given two similar prisms; , the base of the first contains 30 square yards and its altitude is 8 yards; the altitude of the second prism is 6 yards — find its volume and the area of its base. 14. A pyramid, whose base is a regular pentagon of which the apothem is 3.5 feet, contains 129 cubic feet; find the volume of a similar pyramid, the apothem of whose base is 4 feet. 15. Show that the four diagonals of a paraUelopipedon bisect each other in a common point. 16. Show that the two lines joining the points of the opposite faces of a paraUelopipedon, in which the diago- nals of those faces intersect, bisect each other at the point in which the diagonals of the paraUelopipedon intersect. 17. Show that two regular polyedrons of the same kind are similar. 18. Show that the surfaces of any two similar polye- drons are to each other as the squares of any two homologous edges BOOK VIII. THE CYLINDER, THE CONE, AND THE SPHERE. DEFEsriTioisrs. 1. A Cylinder is a volume which may be generated by a rectangle revolving about one of its sides as an axis. Thus, if the rectangle A BCD be turned about the side AB, as an axis, it will generate the cylinder FGCQ-P. The fixed line AB is called the axis of the cylinder; the curved surface generated by the side CD, opposite the axis, is called the convex surface of the cylinder ; the equal circles FGCQ, and EH DP, generated by the remaining sides BC and AD, are called bases of the cylinder ; and the perpendicular dis- tance between the planes of the bases is called the altitude of the cylinder. The line DC, which generates the convex surface, is, in any position, called an element of the surface; the ele- ments are all perpendicular to the planes of the bases, and any one of them is equal to the altitude of the cylinder. Any line of the generating rectangle ABCD, as IK, which is perpendicular to the axis, will generate a circle whose plane is perpendicular to the axis, and which is equal to either base : hence, any section of a cylinder by a plane perpendicular to the axis, is a circle equal to either base. Any section, FCDE, made by a plane through the axis, is a rectangle double the generating rectangle. 224 GEOMETRY. 2. Similar Cylinders are those which may be gener- ated by similar rectangles revolving about homologous sides. The axes of similar cylinders are proportional to the radii of their bases (B. IV., D. 1) ; they are also propor- tional to any other homologous lines of the cylinders. 3. A prism is said to be inscribed in a cylinder, when its bases are inscribed in the bases of the cylinder. In this case, the cylinder is said to be circumscribed about the prism. The lateral edges of the inscribed prism are elements of the surface of the circum- scribing cylinder. 4. A prism is said to be circumscribed about a cylinder, when its bases are circumscribed the bases of the cylinder. In this case, the cylinder to be inscribed in the prism. The straight lines which join the cor- responding points of contact in the upper and lower bases, are common to the sur- face of the cylinder and to the lateral faces of the prism, and they are the only lines which are common. The lateral faces of the prism are tangent to the cylinder along these lines, which are then called elements of contact. about is said 5. A Cone is a volume which may be generated by a right-angled triangle revolving about one of the sides ad- jacent to the right angle, as an axis. BOOK VIII. 225 Thus, if the triangle SAB, right.-aiigled at A, be turned about the side SA, as an axis, it will generate the cone S-CD.BE. The fixed line SA, is called the axis S of the cone J the curved surface gener- /f|y ated by the hypothenuse SB, is called the // n\ F/ * ■ 1 "\h convex surface of the cone ; the circle /c£lp\ generated by the side AB, is called the /?/ ! 1 \ base of the cone: and the point S, is / X"! \ \ called the vertex of the cone; the dis- v ^ "X r tance from the vertex to any point in the circumference of the base, is called the slant height of the cone; and the perpendicular dis- tance from the vertex to the plane of the base, is called the altitude of the cone. The line SB, which generates the convex surface, is, in any position, called an element of the surface; the ele- .ments are all equal, and any one is equal to the slant height ; the axis is equal to the altitude. Any line of the generating triangle SAB, as GH, which is perpendicular to the axis, generates a circle whose plane is perpendicular to the axis : hence, any section of a cone by a plane perpendicular to the axis, is a circle. Any section SBC, made by a plane through the axis, is an isosceles triangle, double the generating triangle. ft. A Truncated Cone is that portion of a cone in- cluded between the base and any plane which cuts the one. When the cutting plane is parallel to the plane of the base, the truncated cone is called a Frustum of a Cone, and the intersection of the cutting plane with the cone is called the upper base of the frustum ; the base of the cone is called the lower base of the frustum. 226 GKOMETK Y. K the trapezoid HGAB, right-angled at A and G, be revolved about AG, as an axis, it will generate a frustum of a cone, whose bases are ECDB and FKH, whose altitude is AG, and whose slant height is BH. 7. Similar Oones are those which may be generated by similar right-angled triangles revolving about homolo- gous sides. The axes of similar cones are proportional to the radii of their bases (B. lY., D. 1) ; they are also proportional to any other homologous lines of the cones. 8. A pyramid is said to be in- scribed in a cone, when its base is inscribed in the base of the cone, and when its vertex coincides with that of the cone. The lateral edges of the inscribed pyramid are elements of the surface of the circumscribing cone. 9. A pyramid is said to be circumscribed about a cone, when its base is circumscribed about the base of the cone, and when its vertex coincides with that of the cone. In this case, the cone is inscribed in the pyramid. The lateral faces of the circumscribing pyramid are tangent to the surface of the inscribed cone, along lines which are called elements of contact. 10. A frustum of a pyramid is inscribed in a frustum of a cone, when its bases are inscribed in the bases of the frustum of the cone. BOOK VIII. 227 The lateral edges of the inscribed frustum of a pyra- mid are elements of the surface of the circumscribing frustum of a cone. 11. A frustum of a pyramid is circumscribed about a frustum of a cone, when its bases are circumscribed about those of the frustum of the cone. Its lateral faces are tangent to the surface of the frustum of the cone, along lines which are called elements of contact. 12. A Sphere is a volume bounded by a surface, every point of which is equally distant from a point within called the centre. A sphere may be generated by a semi- circle revolving about its diameter as an axis. 13. A Radius of a sphere is a straight line drawn from the centre to any point of the surface. A Diameter is a straight line through the centre, limited by the surface. All the radii of a sphere are equal : the diameters are also equal, and each is double the radius. 14. A Spherical Sector is a volume gen- erated by a sector of the semicircle that generates the sphere. The surface generated by the arc of the circular sector is the base of the sector. The other bounding surfaces are either surfaces of cones or planes. The spherical sector generated by ACB is bounded by the surface generated by the arc AB and the conic surface gen- erated by BC ; the sector generated by BCD is bounded by the surface generated by BD and the conic surfaces generated by BC and DC, and so on. 15. A plane is Tangent to a Sphere when it touches it in a single point. 16. A Zone is a portion of the surface of a sphere included between two parallel planes. The bounding lines 228 GEOMETRY. of the sections are called bases of the zone, and the dis- tance between the planes is called the altitude of the zone. If one of the planes is tangent to the sphere, the zone has but one base. 17. A Spherical Segment is a portion of a sphere in- cluded between two parallel planes. The sections made by the planes are called bases of the segment, and the distance between them is called the altitude of the segment. If one of the planes is tangent to the sphere, the seg- ment has but one base. The Cylinder, the Cone, and the Sphere, are some- times called The Three Round Bodies. PROPOSITION I. THEOREM. 7%e convex surface of a cylinder is equal to the circum- ference of its base Tnultiplied by its altitude. Let ABD be the base of a cylinder whose altitude is H : then is its convex surface equal to the circumference of its base multiphed by the altitude. For, inscribe in the cylinder a prism whose base is a regular polygon. The convex surface of this prism is equal to the perimeter of its base multiphed by its altitude (B. VII., P. I.), what- ever may be the number of sides of its base. But, when the number of sides is infinite (B. V., P. X., Sch.), the convex surface of the prism coincides with that of the cyUnder, the perimeter BOOK VIII. 229 of the base of the prism coincides with the circumference of the base of the cylinder, and the altitude of the prism is the same as that of the cylinder : hence, the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude ; which was to be proved. Gar. The convex surfaces of cylinders having equal altitudes are to each other as the circumferences of their bases. PROPOSITION" II. THEOREM. The volume of a cylinder is equal to the product of its base and altitude. f M — c Let ABD be the base of a cylinder whose altitude is H ; then is its volume equal to the product of its base and altitude. For, inscribe in it a prism whose base is a regular polygon. The volume of this prism is equal to the product of its base and altitude (B. VII., P. XIV.), whatever may be the number of sides of its base. But, when the number of sides is infinite, the prism coincides with the cylinder, the base of the prism with the base of the cylinder, and the alti- tude of the prism is the same as that of the cylinder : hence, the volume of the cylinder is equal to the product of its base and altitude ; which was to be proved. Cor. 1. Cylinders are to each other as the products of their bases and altitudes ; cylinders having equal bases are to each other as their altitudes; cylinders having equal altitudes are to each other as their bases. ^ 230 GEOMETRY. Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. For, the bases are as the squares of their radii (B. V., P. XIIL), and the cylinders being similar, these radii are to each other as their altitudes (D. 2) : hence, the bases are as the squares of the altitudes; therefore, the bases multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes. PROPOSITION m. THEOREM. The convex surface of a cone is equal to the circumference of its base multiplied by half its slant height. Let S-ACD be a cone whose base is ACD, and whose slant height is SA: then is its convex surface equal to the circumference of its base multiplied by half its slant height. For, inscribe in it a right pyramid. The convex surface of this pyramid is equal to the perimeter of its base mul- tiplied by half its slant height (B. VII., P. rV.), whatever may be the number of sides of its base. But when the number of sides of the base is infi- nite, the convex surface coincides with that of the cone, the perimeter of the base of the pyramid coincides with the circumference of the base of the cone, and the slant height of the pyra- mid is equal to the slant height of the cone : hence, the convex surface of the cone is equal to the circumference of its base multiplied by half its slant height; which was to he proved. BOOK VIII. 231 PROPOSITION IV. THEOREM. The convex surface of a frustum of a cone is equal to half -the sum of the circumferences of its two bases multiplied by its slant height. Let BIA-D be a frustum of a cone, BIA and EGD its two bases, and EB its slant height : then is its convex surface equal to half the sum of the circumferences of its two bases multiplied by its slant height. For, inscribe in it the frustum of cX'''^^^^^^" a right pyramid. The convex sur- TW face of this frustum is equal to half /// the sum of the perimeters of its L'^"j~~ bases, multiplied by the slant height J/.L.L (B. VII., P. IV., C), whatever may be \|/ ""_ '^ ^^ the number of its lateral faces. But j when the number of these faces is infinite, the convex surface of the frustum of the pyra- mid coincides with that of the cone, the perimeters of its bases coincide with the circumferences of the bases of the frustum of the cone, and its slant height is equal to that of the cone : hence, the convex surface of the frustum of a cone is equal to half the sum of the circumferences of its bases multiplied by its slant height; which was to he proved. Scholium. From the extremities A and D, and from the middle point I, of a line AD, let the lines AO, DC, and IK. be drawn perpendicular to the axis OC : then will IK. be equal to half the sum of AO and DC. For, draw Dd and li, perpendicular to AO : then, because kl is equal to ZD, we shall have ki equal to id (B. IV., P. XV.), and conse- quently to Is ; that is, AO exceeds IK as much as IK 232 GEOMETBY. exceeds DC : hence, IK is equal to the half sum of AO and DC. Now, if the line AD be revolved about OC, as an axis, it will generate the surface of a frustum of a cone whose slant height is AD ; the point I will generate a circum- ference which is equal to half the sum of the circum- ferences generated by A and D : hence, if a straight line is revolved about another straight line, it generates a sur- face whose measure is equal to the -product of the gener- ating line and the circumference generated by its middle point. This proposition holds true when the line AD meets OC, and also when AD is parallel to OC. PROPOSITION V. THEOREM. I%e volume of a cone is equal to its base multiplied by one third of its altitude. Let ABDE be the base of a cone whose vertex is S, and whose altitude is So ; then is its volume equal to the base multiplied by one third of the altitude. For, inscribe in the cone a right pyrar- mid. The volume of this pyramid is equal to its base multiplied by one third / of its altitude (B. Vn., P. XVn.), what- / ever may be the number of its lateral Jy faces. But, when the number of lateral ^|l. — faces is infinite, the pyramid coincides \\. with the cone, the base of the pyramid B coincides with that of the cone, and their altitudes are equal : hence, the volume of a cone is equal to its base multiplied by one third of its altitude ; which was to be proved. BOOK VIII. 233 Cor. 1. A cone is equal to one third of a cylinder •having an equal base and an equal altitude. Cor. 2. Cones are to each other as the products of their bAses and altitudes. Cones having equal bases are to each other as their altitudes. Cones having equal altitudes are to each other as their bases. PROPOSITION VI. THEOREM. The volum>e of a frustum of a cone is equal to the sum, of the volumes of three cones, having for a common alti- tude the altitude of the frustum, and for bases the lower base of the frustum, the upper base of the frus-' turn, and a mean proportional between the bases. Let BIA be the lower base of a frustum of a cone, EGD its upper base, and OC its altitude: then is its vol- ume equal to the sum of three cones whose common altitude is OC, and whose bases are the lower base, the upper base, and a mean proportional between them. For, inscribe a frustum of a right pyramid in the given frustum. The cX°''^^TN,n volume of this frustum is equal to J\ the sum of the volumes of three / / pyramids whose common altitude is l/'i~ that of the frustum, and whose bases rkI /....... are the lower base, the upper base, nJ J/ and a mean proportional between the j two (B. VII., P. XVIII.), whatever may be the number of lateral faces. But when the number of faces is infinite, the frustum of the pyramid coincides with the frustum of the cone, its bases with the bases of the cone, the three pyramids become cones, and their 234 GEOMETRY. altitudes are equal to that of the frustum : hence, the volume of the frustum of a cone is equal to the sum of the volumes of three cones whose common altitude is that of the frustum, and -whose bases are the lower base of the frustum, the upper base of the frustum, and a mean proportional between them ; which was to be proved. PROPOSITION VII. THEOREM. Any section of a sphere made hy a plane is a circle. Let C be the centre of a sphere, CA one of its radii, and AM B any section made by a plane : then is this sec- tion a circle. For, draw a radius CO perpendicular D to the cutting plane, and let it pierce /„ — h::^'^~~i:^ the plane of the section at 0. Draw /v^,^c)h>^^^!X radii of the sphere to any two points I .]/-■ ^ \ M, M', of the curve which bounds the \ I section, and join these points with : \ / then, because the radii CM, CM' are ^- -^ equal, the points M, M', will be equally distant from (B. VI., P. V., 0.) ; hence, the section is a circle ; which was to be proved. Cor. 1, When the cutting plane passes through the centre of the sphere, the radius of the section is equal to that of the sphere ; when the cutting plane does not pass through the centre of the sphere, the radius of the section will be less than that of the sphere. A section whose plane passes through the centre of the sphere, is called a great circle of the sphere. A section whose plane does not pass through the centre of the sphere, BOOK VIII. 235 is called a small circle of the sphere. All great circles of the same, or of equal spheres, are equal. Cor. 2. Any great circle divides the sphere, and also the surface of the sphere, into equal parts. For, the parts may be so placed as to coincide, otherwise there would be some points of the surface unequally distant from the centre, which is impossible. Cor. 3. The centre of a sphere, and the centre of any small circle of that sphere, are in a straight line perpen- dicular to the plane of the circle. Cor. 4. The square of the radius of any small circle is equal to the square of the radius of the sphere diminished by the square of the distance from the centre of the sphere to the plane of the circle (B. IV., P. XL, C. 1 ) : hence, circles which are equally distant from the centre, are equal ; and of two circles which are unequally distant from the centre, that one is the less whose plane is at the greater distance from the centre. Cor. 5. The circumference of a great circle may always be made to pass through any two points on the surface of a sphere. For, a plane can always be passed through these points and the centre of the sphere (B. VI., P. II.), and its section will be a great circle. If the two points are the extremities of a diameter, an infinite number of planes can be passed through them and the centre of the sphere (B. VI., P. I., S.) ; in this case, an infinite number of great circles can be made to pass through the two points. Cor. 6. The bases of a zone are the circumferences of circles (D. 16), and the bases of a segment of a sphere are 286 GEOMETRY. PROPOSITION Vm. THEOREM. Any plane perpendicular to a radius of a sphere at its outer extremity, is tangent to the sphere at that point. Let C be the centre of a sphere, CA any radius, and FAG a plane perpendicular to CA at A : then is the plane FAG tangent to the sphere at A. For, from any other point of the plane, as M, draw the line MC : then because CA is a perpendicular to the plane, and CM an oblique line, CM is greater than CA (B. VI., P. V.) : hence, the point M lies without the sphere. The plane FAG, therefore, touches the sphere at A, and consequently is tan- gent to it at that point ; which was to he proved. Scholium. It may be shown, by a course of reasoning analogous to that employed in Book III., Propositions XI., Xn., Xin., and XIV., that two spheres may have any one of six positions with respect to each other, viz. : 1°. When the distance between their centres is greater than the sum of their radii, they are external one to the other : 2°. When the distance is equal to the sum of their radii, they are tangent externally: 3°. When this distance is less than the sum, and greater than the difference of their radii, they intersect each other: 4°. When this distance is equal to the difference of their radii, they are tangent internally: 5°. When this distance is less than the difference of their radii, one is wholly within the other: 6°. When this distance is equal to zei'o, they have a common centre, or are concentric. BOOK VIII. 237 DEFINITIONS. 1°. If a semi-circumference is divided into equal arcs, the chords of these arcs form half of the perimeter of a regular ■ inscribed polygon; this half perimeter is called a regular semi-perimeter. The figure bounded by the regu- lar semi-perimeter and the diameter of the semi-circum- ference is called a regular semi-polygon. The diameter itself is called the axis of the semi-polyfeon. 2°. If lines are drawn from the extrem- ities of any side perpendicular to the axis, the intercepted portion of the axis is called the projection of that side. The broken line ABCDGP is a regular semi-perimeter ; the figure bounded by it and the diameter AP, is a regular semi- polygon, AP is its axis, HK is the projec- tion of the side BC, and the axis, AP, is the projection of the entire semi-perimeter. PROPOSITION IX. LEMMA. !f a regidar semi-polygon is revolved about its axis, the surface generated by the semi-perimeter is equal to the axis multiplied by the circumference of the inscribed circle. Let ABCDEF be a regular semi-polygon, AF its axis, and ON its apothem : then is the surface generated by the regular semi-perimeter equal to AFxcirc. ON. From the extremities of any side, as DE, draw Dl and' EH perpendicular to AF ; draw also N M perpendicular to AF, and EK perpendicular to Dl. Now, the surface gener- ated by DE is equal to DExcirc. NM (P. rV., S.). But, 238 GEOMETRY. circ. NM because the triangles EDK and ONM are similar (B. IV., P. XXL), we have, DE : EK or IH : : ON : NM : : circ. ON whence, DExcirc. NM = \Hxcirc. ON; that is, the surface generated by any side is equal to the projection of that side multiplied by the circumference of the in- scribed circle : hence, the surface gener- ated by the entire semi-perimeter is equal to the sum of the projections of its sides, or the axis, multiplied by the circtunfer- ence of the inscribed circle ; which was to be proved. Cor. The surface generated by any portion of the pe- rimeter, as CDE, is equal to its projection PH, multiplied by the circumference of the inscribed circle. PROPOSITION X. THEOREM. The surface of a sphere is eqwal to its diameter multiplied by the circumference of a great circle. Let ABODE be a semi-circumference, its centre, and AE its diameter : then is the surface of the sphere generated by re- volving the semi-circumference about AE, equal to AE x circ. OE. For, the semi-circumference may be re- garded as a regular semi-perimeter with an infinite number of sides, whose axis is AE, and the radius of whose inscribed circle is OE : hence (P. IX.), the surface generated by it is equal to AE X circ. OE ; which was to be proved. BOOK VIII. 239 Cor. 1. The circumference of a great circle is equal to 2TrOE (B. v., P. XVI.) : hence, the area of the surface of the sphere is equal to 20Ex2wOE, or to 47rOE^, that is, the area of the surface of a sphere is equal to four great eirdles. Cor. 2. The surface generated by any arc of the semicircle, as BC, is a zone, whose altitude is equal to the projection of that arc on the diameter. But, the arc BC is a portion of a semi-perimeter having an infinite number of sides, and the radius of ■whose inscribed circle is equal to that of the sphere : hence (P. IX., C), the surface of a zone is equal to its altitude multiplied by the circumference of a great circle of the sphere. Cor. 3. Zones, on the same sphere, or on equal spheres, are to each other as their altitudes. PROPOSITION XI. LEMMA. If a triangle and a rectangle having the same base and equal altitudes, are revolved about the common base, the volume generated by the triangle is one third of that generated by the rectangle. Let ABC be a triangle, and EFBC a rectangle, having the same base BC, and an equal altitude AD, and let them both be revolved about BC : then is the volume generated by ABC one third of that generated by EFBC. For, the cone generated by the right-angled triangle ADB, is equal to one third of the cylinder generated by the rectangle ADBF (P. V., C. 1), and the cone generated 240 GEOMETRY. by the triangle ADC, is equal to one third of the cylinder generated by the rectangle ADCE. When AD falls within the triangle, the sum of the cones generated by ADB and ADC, is equal to the volume generated by the triangle ABC ; and the sum of the cylinders generated by ADBF and ADCE, is equal to the volume generated by the rectangle EFBC. When AD falls without the triangle, the difference of the cones generated by ADB and ADC, is equal to the volume generated by ABC ; and the dif- ference of the cylinders generated by ADBF and ADCE, is equal to the volume generated by EFBC : hence, in either case, the volume generated by the tri- angle ABC, is equal to one third of the volume generated by the rectangle EFBC ; which was to he proved. Cor. The volume of the cylinder generated by EFBC, is equal to the product of its base and altitude, or to ttAD'' X BC : hence, the volume generated by the triangle ABC, is equal to iirAD* x BC. PROPOSITION XII. LEMMA. If an isosceles triangle is revolved about a straight line passing through its vertex, the volume generated is equal to the surface generated by the base multiplied by one third of the altitude. Let CAB be an isosceles triangle, C its vertex, AB its base, CI its altitude, and let it be revolved about the line CD, as an axis : then is the volume generated equal to surf. ABx^CI. There may be three cases: BOOK VIII. 241 . Suppose the base, when produced, to meet the axis at D ; draw AM, IK, and BN, perpendic- ular to CD, and BO parallel to DC. Now, the volume generated by CAB is equal to the differ- ^ ,. ^ ence of the volumes generated by CAD and CBD ; hence CP- XL, C), vol. CAB = IttAIVPxCD — |rrBN'xCD = ^77 (AM' — BN^xCD. But, AM' - BN' is equal to (AM + BN) (AM - BN) (B. IV., P. X.) ; and because AM + BN is equal to 2IK (P. IV., S.), and AM — BN to AO, we have, vol. CAB = IttIKxAOxCD. But, the right-angled triangles AOB and CDl are similar (B. IV., P. XVIII.) ; hence, AO : AB :: CI : CD; or, AOxCD = ABxCI. Substituting, and changing the order of the factors, we have, vol. CAB = ABx27rlKxiCI. But, ABx2itIK = the surface generated by AB ; hence, vol. CAB = surf. ABxiCI. 2°. Suppose the axis to coincide with one of the equal sides. Draw CI perpendicular to AB, and AM and IK, perpendicular to CB. Then, vol. CAB — -iwAM^xCB = ^ttAMxAMxCB. But, since AMB and CIB are similar, AM : AB :: CI : CB; whence, AMxCB = ABxCI. Also, AM = 2 IK; hence, by substitution, we have, vol. CAB = ABx2TrlKx-lCI = surf. ABxiCI. a42 GEOMETRY. 3°. Suppose the base to be parallel to the axis. Draw AM and BN perpendicular to the axis. The volume generated by CAB, is equal to the cylinder generated by the rectangle ABNM, diminished by the sum of the cones generated by the triangles CAM and CBN ; hence, vol. CAB = ttCI^'xAB -inCPxM - iTrCl'xIB. But the sum of Al and IB is equal to AB : hence, we have, by reducing, and changing the order of the factors, vol. CAB = ABx27rCIxiCI. But ABx27tCI is equal to the surface generated by AB ; consequently, vol. CAB = surf. AB x ^Cl ; hence, in aU cases, the volume generated by CAB is equal to surf. ABx|CI; which was to be proved. PROPOSITION Xm. LEMMA. Jf a regular semi-polygon is revolved about its axis, the volume generated is equal to the surface generated by the semi-perimeter multiplied by one third of the apothem. Let FBDG be a regular semi-polygon, FG its axis, 01 its apothem, and let the semi-polygon be revolved about FG : then is the volume generated equal to surf. FBDGxiOI. For, draw lines from the vertices to the centre 0. These lines wiU divide the semi-polygon into isosceles triangles whose bases are sides of the semi-polygon, and whose altitudes are each equal to 01. BOOK VIII. •Z4:8 Now, the sum of the volumes generated by these tri- angles is equal to the volume generated by the semi- polygon. But, the volume generated by any triangle, as OAB, is equal to surf. AB x JOI (P. XII.) ; hence, the volume generated by the semi-polygon is equal to surf. FBDGx^OI; which was to he proved. Cor. The volume generated by a portion of the semi- polygon, OABC, limited by OC, OA, drawn to vertices is equal to surf. ABCx^OI. PROPOSITION XrV. THEOREM. The volume of a sphere is equal to its surface multiplied by one third of its radius. Let ACE be a semicircle, AE its diam- eter, its centre, and let the semicircle be revolved about AE : then is the volume generated equal to the surface generated by the semi-circumference multiplied by one third of the radius OA. For, the semicircle may be regarded as a regular semi-polygon having an infi- nite number of sides, whose semi-perim- eter coincides with the semi-circumference, and whose apothem is equal to the ra- dius : hence (P. XIII.), the volume generated by the semi- circle is equal to the surface generated by the semi- circumference multiplied by one third of the radius; which was to be proved. Cor. 1. Any portion of the semicircle, as OBC, bounded by two radii, will generate a volume equal to the surface generated by the arc BC multiplied by one third of the 244 GEOMETRY. radius (P. XIII., C). But this portion of the semicircle is a circular sector, the volume -which it generates is a spherical sector, and the surface generated by the arc is a zone : hence, the volume of a spherical sector is equal to the zone which forms its base multiplied by one third of the radius. Cor. 2. If we denote the volume of a sphere by V, and its radius by R, the area of the surface will be equal to 4TrR2 (P. X., C. 1), and the volume of the sphere will be equal to ^txR^x^R; consequently, we have, V- = -IttRs. Again, if we denote the diameter of the sphere by D, we shall have R equal to JD, and R' equal to |D', and con- sequently, V = irrDB; hence, the volumes of spheres are to each other as the ciubes of their radii, or as the cubes of their diameters. Scholium. If the figure EBDF, formed by drawing lines from the extremities of the arc BD perpendicular to CA, be re- volved about CA, as an axis, it will gen- erate a segment of a sphere whose volume may be found by adding to the spherical sector generated by CDS, the cone generated by CBE, and subtracting from their sum the cone generated by CDF. If the arc BD is so taken that the points E and F fall on opposite sides of the centre C, the latter cone must be added, instead of subtracted. The area of the zone BD is equal to 2rTCDxEF (P. X., C. 2); hence, segment EBDF = i7r(2CD^xEF + BE^xCE qp DF^xCF). BOOK VIII. 245 PROPOSITION XV. THEOREM. The surface nf a sphere is to the entire surface of the circuT)%SGribed cylinder, including its bases, as 2 is to 3; and the volumes are to each other in the same ratio. Let PMQ be a semicircle, and PADQ a rectangle, whose sides PA and QD are tangent to the semicircle at P and Q, and -whose side AD, is tangent to the semicircle at M. If the semicircle and the rectangle be revolved about PQ, as an axis, the former will generate a sphere, and the latter a circumscribed cylinder. 1°. The surface of the sphere is to the entire surface of the cylinder, as 2 is to 3. For, the surface of the sphere is equal to four great circles (P. X., C. 1), the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude (P. I.) ; that is, it is equal to the circumference of a great circle multiplied by its diam- eter, or to four great circles (B. V., P. XV.) ; adding to this the two bases, each of which is equal to a great circle, we have the en- tire surface of the cylinder equal to six great circles : hence, the surface of the sphere is to the entire surface of the circumscribed cylinder, as 4 is to 6, or as 2 is to 3 ; which was to he proved. 2°. The volume of the sphere is to the volume of the cylinder as 2 is to 3. For, the volume of the sphere is equal to I-ttR^ (P. XIV., C. 2) ; the volume of the cylinder is equal to its base multiplied by its altitude (P. II.) ; that is, it is equal to 246 GEOMETRY. n-R''x2R, or to |^R': hence, the volume of the sphere is to that of the cyhnder as 4 is to 6, or as 2 is to 3 ; which was to be proved. Cor. The surface of a sphere is to the entire surface of a circumscribed cylinder, as the volume of the sphere is to the volume of the cylinder. Scholium. Any polyedron which is circumscribed about a sphere, that is, whose faces are all tangent to the sphere, may be regarded as made up of pyramids, whose bases are the faces of the polyedron, whose common ver- tex is at the centre of the sphere, and each of whose altitudes is equal to the radius of the sphere. But, the volume of any one of these pyramids is equal to its base multiplied by one third of its altitude : hence, the volume of a circumscribed polyedron is equal to its surface nnil- tipUed by one third of the radius of the inscribed sphere. Now, because the volume of the sphere is also equal to its surface multiplied by one third of its radius, it follows that the volume of a sphere is to the volume of any circumscribed polyedron, as the surface of the sphere is to the surface of the polyedron. Polyedrons circumscribed about the same, or about equal spheres, are proportional to their surfaces. GEISTERAL FORMULAS. If we denote the convex surface of a cyhnder by S, its volume by V, the radius of its base by R, and its alti- tude by H, we have (P. I, II.), S = 2TrR X H (1.) V = ttRs X H (2.) BOOK VIII. 247 If we denote the convex surface of a cone by S, its volume by V, the radius of its base by R, its altitude by H, and its slant height by H', we have (P. III., V.), S = ttR X H' (3.) V = irRs X iH (4.) If we denote the convex surface of a frustum of a cone by S, its volume by V, the radius of its lower base by R, the radius of its upper base by R', its altitude by H, and its slant height by H', we have (P. IV., VI.), S = 7r(R + R') X H' (5.) V = ^7T (R2 + R'3 + R X R') X H . . . . (6.) If we denote the surface of a sphere by S, its volume by V, its radius by R, and its diaraeter by D, we have (P. X., 0. 1, XIV., 0. 2, XIV., 0. 1), S = 47tR' (7.) V = ^ttRS = |7rDS (8.) If we denote the radius of a sphere by R, the area of any zone of the sphere by S, its altitude by H, and the volume of the corresponding spherical sector by V, we shall have (P. X., C. 2, XIV, C. 1), S = 27rR X H (9.) V = I^Rs X H (10.) If we denote the volume of the corresponding spherical segment by V, its altitude by H, the radius of its upper base by R', the radius of its lower base by R", the distance of its upper base from the centre by H', and of its lower base from the centre by H", we shall have (P. XIV., S.) : V = i7r{2R'xH + R'SH' q: R"2xH") . .(11.) 248 GEOMETRY. EXERCISES. 1. The radius of the base of a cylinder is 2 feet, and its altitude 6 feet; find its entire surface, including the bases. 2. The volume of a cylinder, of which the radius of the base is 10 feet, is 6283.2 cubic feet; find the vol- ume of a similar cylinder of "which the diameter of the base is 16 feet, and find also the altitude of each cylinder. 3. Two similar cones have the radii of the bases equal, respectively, to 4J and 6 feet, and the convex surface of the first is 667.59 square feet; find the convex surface of the second and the volume of both. 4. A line 12 feet long is revolved about another line as an axis ; the distance of one extremity of the hne from the axis is 4 feet and of the other extremity 6 feet ; find the area of the surface generated. 5. Find the convex surface and the volume of the frustum of a cone the altitude of which is 6 feet, the radius of the lower base being 4 feet and that of the upper base 2 feet. 6. Find the surface and the volume of the cone of which the frustum in the preceding example is a frustum. 7. A small circle, the radius of which is 4 feet, is 3 feet from the centre of a sphere ; find the circumference of a great circle of the same sphere. 8. The radius of a sphere is 10 feet; find the area of a small circle distant from the centre 6 feet. 9. Find the area of the surface generated by the semi- perimeter of a regular semihexagon revolving about its axis, the radius of the inscribed circle being 5.2 feet and the axis 12 feet. 10. The area of the surface generated by the semi- BOOK VIII. 249 perimeter of a regular semioctagon revolved about an axis is 178.2426 square feet, and th^ -^adius of the inscribefl circle is 3.62 feet: find the axis. 11. An isosceles triangle, whose base is 8 feet and altitude' 9 feet, is revolved about a line passing through its vertex and parallel to its base ; how many cubic feet in the volume generated? 12. The altitude of a zone is 3 feet and the radius of the sphere is 5 feet ; find the area of the zone and the volume of the corresponding spherical sector. 13. Find the surface and the volume of a sphere whose radius is 4 feet. 14. The radius of a sphere is 5 feet; how many cubic feet in a spherical segment whose altitude is 7 feef and the distance of whose lower base from the centre of the sphere is 8 feet? 15. A cone such that the diameter of its base is equal to its slant height is circumscribed about a sphere ; show that the surface of the sphere is to the entire surface of the cone, including its base, as 4 is to 9, and that the volumes are in the same ratio. 16. The radius of a sphere is 6 feet; find the entire surface and the volume of the circumscribing cylinder. 17. A cone, with the diameter of the base and the V slant height equal, is circumscribed about a sphere whose radius is 5 feet ; find the entire surface and the volume of the cone. 18. A cone, with the diameter of the base and the slant height equal, and a cylinder, are circumscribed about a sphere ; what relation exists between the entire surfaces and the volumes of the cylinder, the cone and the sphere? 19. The edge- of a regular octaedron is 10 feet, and the radius of the inscribed sphere is 4.08 feet; find the volume of the octaedron. BOOK IX. SPHERICAL GEOMETRY. DEFINITIONS. 1. A Spherical Angle is the amount of divergence of the arcs of two great circles of a sphere meeting at a point. The arcs are called sides of the angle, and their point of intersection is called the vertex of the angle. The measure of a spherical angle is the same as that of the diedral angle included between the planes of its sides. Spherical angles may be acute, right, or obtuse. 2. A Spherical Polygon is a portion of the surface of a sphere bounded by arcs of three or more great circles.' The bounding arcs are called sides of the polygon, and the points in which the sides meet are called vertices of the polygon. Each side is taken less than a semi-circum- ference. Spherical polygons are classified in the same manner as plane polygons. 3. A Spherical Triangle is a spherical polygon of three sides. Spherical triangles are classified in the same manner as plane triangles. 4. A LuNE is a portion of the surface of a' sphere bounded by semi-circumferences of two great circles. 5. A Spherical Wedge is a portion of a sphere bound- ed by a lune and two semicircles which intersect in a diameter of the sphere. BOOK IX. 261 6. A Sphesical Pyramid is a portion of a sphere bounded by a spherical polygon and sectors of circles "whose common centre is the centre of the sphere. The spherical polygon is called the base of the pyra- mid, and the centre of the sphere is called the vertex of the pyramid. 7. A Pole of a Circle is a point, on the surface of the sphere, equally distant from all the points of the cir- cumference of the circle. 8. A Diagonal of a spherical polygon is an arc of a great circle joining the vertices of any two angles which are not consecutive. PROPOSITION I. THEOREM. J.ny side of a, spherical triangle is less than the sum of the two others. Let ABC be a spherical triangle situated on a sphere whose centre is : then is any side, as AB, less than the sum of the sides AC and BC. For, draw the radii OA, OB, and OC : these radii form the edges of a triedral angle whose vertex is 0, and the plane angles included be- tween them are measured by the arcs AB, AC, and BC (B. Ill, P. XVII., Sch.). But any plane angle, as AOB, is less than the sum of the plane angles AOC and BOC (B. VI., P. XIX.) : hence, the arc AB is less than the sum of the arcs AC and BC ; which was to be proved. 252 GEOMETRY. Cor. 1. Any side AB, of a spherical polygon ABCDE, is less than the sum of all the other sides. For, draw the diagonals AC and AD, dividing the polygon into trian- gles. The arc AB is less than the sum of AC and BC, the arc AC is less than the sum of AD and DC, and the arc AD is less than the sum of DE and EA ; hence, AB is less than the sum of BC, CD, DE, and EA. Cor. 2. The arc of a small circle, on the surface of a sphere, is greater than the arc of a great circle joining its two extremities. For, divide the arc of the small circle into equal parts, and through the two extremities of each part suppose the arc of a great circle to be drawn. The sum of these arcs, whatever may be their number, will be greater than the arc of the great circle joining the given points (C. 1). But when this number is infinite, each arc of the great circle wUl coincide with the corresponding arc of the small circle, and their stmi is equal to the entire arc of the small circle, which is, consequently, greater than the arc of the great circle. Cor. 3. The shortest distance from one point to another on the surface of a sphere, is measured on the arc of a great circle joining them. PROPOSITION n. THEOREM. The sum of the sides of a spherical polygon is less than the circumference of a great circle. Let ABCDE be a spherical polygon situated on a sphere whose centre is : then is the sum of its sides less than the circumference of a great circle. BOOK IX. 253 For, draw the radii OA, OB, OC, OD, and OE : these radii form the edges of a polyedral angle whose vertex is at 0, and the angles included be- tween them are measured by the arcs AB,. BC, CD, DE, and EA. But the sum of these angles is less than four right angles (B. VI., P. XX.) : hence, the sum of the arcs which measure them is less than the circumference of a great circle ; which was to he proved. PROPOSITION III. THEOREM. // a diameter of a sphere is drawn perpendicular to the plane of any circle of the sphere, its extremities are poles of tka,t circle. Let C be the centre of a sphere, FNG any circle of the sphere, and DE a diameter of the sphere perpendicular to the plane of FNG : then are its extremities, D and E, poles of the circle FNG. The diameter DE, being per- pendicular to the plane of FNG, must pass through the centre (B. VIIL, P. YIL, C. 3). If arcs of great circles DN, DF DG, &c., are drawn from D to different points of the circum- ference FNG, and chords of these arcs are drawn, these chords are equal (B. YI., P. V.), consequently, the arcs them- selves are equal. But these arcs are the shortest lines that can be drawn from the point D to the different 254 GEOMETRY. points of the circumference (P. I., C. 3) : hence, the point D is equally distant from all the points of the circum- ference, and consequently is a pole of the circle (D. 7). In like manner, it may be shown that the point E is also a pole of the circle : hence, both D and E are poles of the circle FNG ; which was to be proved. Cor. 1. Let AMB be a great circle perpendicular to DE : then are the angles DCM, ECM, &c., right angles ; and consequently, the arcs DM, EM, &c., are each equal to a quadrant (B. III., P. XVII., S.) : hence, the two poles of a great circle are at equal distances from the circumference. Cor. 2. The two poles of a small circle are at unequal distances from the circumference, the sum of the distances being equal to a semi-circumference. Cor. 3. If any point, as M, in the circumference of a great circle, is joined with either pole by the arc of a great circle, such arc is perpendicular to the circumfer- ence AMB, since .its plane passes through CD, which is perpendicular to AMB. Conversely: if MN is per'pendicu- lar to the arc AMB, it passes through the poles D and E: for, the plane of MN being perpendicular to AMB and passing through C, contains CD, which is perpendicular to the plane AMB (B. VI., P. XVH., C). Cor. 4. If the distance of a point D from each of the points A and M, in the circumference of a great circle, is equal to a quadrant, the point D is the pole of the arc AM (the arc AM is supposed to be either less or greater than a semi-circumference). For, let C be the centre of the sphere, and draw the radii CD, CA, CM. Since the angles ACD, MCD, are right angles, the line CD is perpendicular to the two straight lines CA, CM : it is, therefore, perpendicular to their plane BOOK IX. '^255 (B. VI., P. IV.) : hence, the point D is the pole of the arc AM. Scholium. The properties of these poles enable us to describe arcs of a circle on the surface of a sphere, with the same faoiUty as on a plane surface. For, by turning the arc DF about the point D, the extremity F will de- scribe the small circle FNG ; and by turning the quadrant DFA round the point D, its extremity A will describe an arc of a great circle. PEOPOSITION IV. THEOREM. The angle formed by arcs of two great circles, is equal to that formed by the tangents to these arcs at their point of intersection, and is measured by the arc of a great circle described from the vertex as a pole, and limited by the sides, produced if necessary. Let the angle BAC be formed by the two arcs AB, AC : then is it equal to the angle FAG formed by the tangents AF, AG, and is measured by the arc DE of a great circle, described about A as a pole. For, the tangent AF, drawn in the plane of the arc AB, is perpendicular to the ra- dius AO ; and the tangent AG, drawn in the plane of the arc AC, is perpendicular to the same radius AO : hence, the angle FAG is equal to the angle contained by the planes ABDH, ACEH (B. VI., D. 4); which is that of the arcs AB, AC. Now, if the arcs AD and AE are both quadrants, the 266 GEOMETET. lines OD, OE, are perpendicular to OA, and the angle DOE is equal to the angle of the planes ABDH, ACEH : hence, the arc DE is the measure of the angle contained by these planes, or of the angle CAB ; which was to be proved. Cor. 1. The angles of spherical triangles may be com- pared by means of the arcs of great circles described from their vertices as poles, and included between their sides. A spherical angle can always be constructed equal to a given spherical angle. Cor. 2. Vertical angles, such as ACP and BCN, are equal; for either of them is the angle formed by the two planes ACB, PCN. When two arcs ACB, PCN, intersect, the sum of two adjacent angles, as ACP, PCB, is equal to two right angleSi PROPOSITION V. THEOREM. Tf from, the vertices of the angles of a spherical triangle, as poles, arcs he described forming a second spherical triangle, the vertices of the angles of this second triangle are respectively poles of the sides of the first. From the vertices A, B, C, as poles, let the arcs EF, FD, DE, be described, forming the triangle DFE : then are the vertices D, E, and F, respects ively poles of the sides BC, AC, AB. For, the point A being the BOOK IX. 257 pole of the arc EF, the distance AE is a quadrant; the point C being the pole of the arc DE, the distance CE is likewise a quadrant : hence, the point E is at a quadrant's distance from the points A and C : hence, it is the pole of the 'arc AC (P. III., 0. 4). It may be shown, in like manner, that D is the pole of the arc BC, and F that of the arc AB ; which was to be proved. Cor. The triangle ABC, may be described by means of DEF, as DEF is described by means of ABC. Triangles so related that any vertex of either is the pole of the side opposite it in the other, are called polar triangles. PROPOSITION VI. THEOEEM. Any angle, in one of two polar triangles, is measured by a semi-circumference, minus the side lying opposite to it in the other triangle. Let ABC, and EFD, be any two polar triangles on a sphere whose centre is : then is any angle in either triangle measured by a semi-circumference, minus the side lying opposite to it in the other triangle. For, produce the sides AB, AC, if necessary, till they meet EF in G and H. The point A being the pole of the arc GH, the angle A is measured by that arc (P. IV.). But, since E is the pole of AH, the arc EH is a quadrant ; and since F is the pole of AG, FG is a quad- rant : hence, the sum of the arcs EH and GF is equal to a semi-circumference. But, the sum of the arcs EH and 258 GEOMETRY. GF is equal to the sum of the arcs EF and GH : hence, the arc GH, which measures the angle A, is equal to a semi-circumference minus the arc EF. In like manner, it may be shown, that any other angle, in either triangle, is measured by a semi-circumference minus the side lying opposite to it in the other triangle ; which was to he proved Cor. 1. Beside the triangle DEF, three other triangles, polar to ABC, may be formed by the intersec- tion of the arcs DE, EF, DF, pro- longed. But the proposition is applicable only to the central tri- angle, ABC, which is distinguished from the three others by the cir- cumstance, that the vertices A and D lie on the same side of BC; B and E, on the same side of AC; C and F, on the same side of AB. The polar triangles ABC and DEF are called supplemental tri- angles, any part of either being the supplement of the part opposite it in the other. Cor. 2. Arcs of great circles, drawn from corresponding vertices of two supplemental polar triangles perpendicular to the respective sides opposite, are supplements of each other. For, from A draw the arc of a great circle, AN, perpendicular to BC ; it must, when prolonged, pass through D, the pole of BC, and must also, when prolonged to P, be perpendicular to EF (P. m., C. 3) : DN and AP being quadrants (P. IIL 0. 1), DP and AN are supplements of each other. BOOK IX. 269 PROPOSITION VII. THEOREM. If from the vertices of any two angles of a spherical tri' angle) as poles, arcs of circles are described passing through the vertex of the third angle; and if from the second point in which these arcs intersect, arcs of great circles are drawn to the vertices, used as poles, the parts of the triangle thus formed are equal to those of the given tri- angle, each to each. Let ABC be a spherical triangle situated on a sphere whose centre is 0, CED and CFD arcs of circles described about B and A as poles, and let DA and DB be arcs of great circles : then are the parts of the triangle ABD equal to those of the given triangle ABC, each to each. For, by construction, the side AD is equal to AC, the side BD is equal to BC, and the side AB is common: hence, the sides are equal, each to each. Draw the radii OA, OB, OC, and OD. The radii OA, OB, and OC, form the edges of a triedral angle whose vertex is ; and the radii OA, OB, and OD, form the edges of a second triedral angle whose vertex is also at ; and the plane angles formed by these edges are equal, each to each : hence, the planes of the equal angles are equally inclined to each other (B. VI., P. XXI.). But, the angles made by these planes are equal to the corresponding spherical angles ; conse- quently, the angle BAD is equal to BAC, the angle ABD to ABC, and the angle ADB to ACB : hence, the parts of the triangle ABD are equal to the parts of the triangle ACB, each to each ; which was to he proved. 260 GEOMETRY. Scholium 1. The triangles ABC and ABD, are not, in general, capable of superposition, but their parts are sym- metrically disposed with respect to AB. Triangles which have all the parts of the one equal to all the parts of the other, each to each, hut are not capable of superposition, are called sjpnmetrical triangles. Scholium 2. If symmetrical triangles are isosceles, they can be so placed as to coincide throughout: hence, they are equal in area. PROPOSITION Vin. THEOREM. Jf two spherical triangles, on the same, or on equal spheres, have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the remaining parts are equal, each to each. Let the spherical triangles ABC and EFG, on the sphere whose centre is 0, have the side EF equal to AB, the side EG equal to AC, and the angle PEG equal to BAC : then is the side FG equal to BC, the angle EFG to ABC, and the angle EGF to ACB. For, draw the radii OE, OF, OG, OA, OB, and OC, forming the trie- dral angles 0-EFG and 0-ABC. Since the sides EF and EG are equal, respectively, to the sides AB and AC, the plane angles EOF and EOG are equal, respectively, to the plane angles ,AOB and AOC ; and as the spherical angles FEG and BAC are equal, the inclination of the faces EOF and EOG of the triedral angle O-EFG, is equal to the inclination of the faces AOB and AOC of the triedral angle 0-ABC ; therefore (B. VL, P. XXI., C), the angle FOG is equal to BOC, and the BOOK IX. 261 side FG equals the side BC : again, since the angle EOF is equal to AOB, FOG to BOC, and GOE to COA, the planes of the equal angles are equally inclined to each other (B. VI., P. XXI.), and, consequently (D. 1), the angle EFG is equal' to ABC, and EGF to ACB — ^henoe, the remaining parts of the triangles are equal, each to each ; which was to he proved. PROPOSITION IX. THEOREM. If two spherical triangles orb the same, or on equal spheres, have two angles and the included side of the one equal to two angles and the included side of the other, each to each, the remaining -parts are equal, each to each. Let the spherical triangles ABC and EFG, on the sphere whose centre is 0, have the angle FEG equal to BAC, the angle EFG equal to ABC, and the side EF equal to AB : then is the side EG equal to AC, the side FG to BC, and the angle FGE to BCA. For, draw radii, as before, form- ing the triedral angles 0-EFG and 0-ABC. Since the side EF is equal to AB, the plane angle EOF is equal to AOB ; as the angle FEG is equal to BAC, and EFG to ABC, the inclination of the face EOF, of the triedral angle O-EFG, to each of the faces EOG and FOG, is equal, respectively, to the inclina- tion of the face AOB, of the triedral angle 0-ABC, to each of the faces AOC and BOC, and hence (B. VI., P. XXI., S. 2), the plane angles EOG and GOF are equal, respect- ively, to AOC and COB ; therefore, the sides EG and GF are equal to the sides AC and CB, and the angle FGE to BCA ; which was to he proved. 262 GEOMETRY. PEOPOSITION X. THEOREM. If two spherical triangles on the same, or on equal spheres, have their sides equal, each to each, their angles are equal, each to each, the equal angles lying opposite the equal sides. Let the spherical triangles EFG and ABC, on the sphere whose centre is 0, have the side EF equal to AB, EG equal to AC, and FG equal to BC : then the angle FEG is equal to BAC, EFG to ABC, and EGF to ACB, and the equal angles lie opposite the equal sides. For, draw the radii, as before, forming the triedral angles 0-EFG and 0-ABC. Because the sides of the triangles are respect- ively equal, the plane angle EOF is equal to AOB, FOG to BOC, and GOE to COA. Hence (B. VI., P. XXL), the planes of the equal angles are equally inclined to each other, and, consequently, the spherical angle EFG is equal to spherical angle ABC, FEG to BAC, and EGF to ACB, the equal angles lying opposite the equal sides ; which was to he proved. Note.— The tanangle EFG is equal in all respects to either ABC or its symmet- Tical triangle. PROPOSITION XI. THEOREM. In any isosceles spherical triangle, the angles opposite the equal sides are equal; and conversely, if two angles of a spherical triangle are equal, the triangle is isosceles. 1°. Let ABC be a spherical triangle, on a sphere whose centre is 0, having the side AB equal to AC : then is the angle C equal to the angle B. BOOK IX. 2166 For, draw the arc of a great circle from the vertex A, to the middle point D, of the base BC : then in the two triangles ADB and ADC, we shall have the side AB equal to AC, .by hypothesis, the side BD equal to DC, by construction, and the side AD common ; consequently, the tri- angles have their angles equal, each to each (P. X.) : hence, the angle C is equal to the angle B ; which was to be proved. 2°. Let ABC be a spherical triangle having the angle C equal to the angle B : then is the side AB equal to the side AC, and consequently the triangle is isosceles. For, suppose that AB and AC are not equal, but that one of them, as AB, is the greater. On AB lay off the arc BE equal to AC, and draw the arc of a great circle from E to C : then in the triangles ACB and EBC, we shall have the side AC equal to EB, by construction, the side BC common, and the included angle ACB equal to the included angle EBC, by hypothesis; hence, the remaining parts of the triangles are equal, each to each, and conse- quently, the angle ECB is equal to the angle ABC. But, the angle ACB is equal to ABC, by hypothesis, and there- fore, the angle ECB is equal to ACB, or a part is equal to the whole, which is impossible : hence, the supposition that AB and AC are unequal, is absurd ; they are therefore equal, and consequently, the triangle ABC is isosceles ; which was to be proved. Cpr. The triangles ADB and ADC, having all of their parts equal, each to each, the angle ADB is equal to ADC, and the angle DAB is equal to DAC ; that is, if an are if a great circle is drawn from the vertex of an isosceles 264 GEOMETRY. spherical triangle to the -middle of its base, it is perpen- dicular to the base, and bisects the vertical angle of the tri- angle. PROPOSITION Xn. THEOREM. In any spherical triangle, the greater side is opposite the greater angle; and conversely, the greater angle is oppo- site the greater side. 1°. Let ABC be a spherical triangle, on a sphere whose centre is 0, in which the angle A is greater than the angle B : then is the side BC greater than the side AC. For, draw the arc AD, making the angle BAD equal to ABD ; then is AD equal to BD (P. XI.). But, the sum of AD and DC is greater than AC (P. I.) ; or, putting for AD its equal BD, we have the sum of BD and DC, or BC, greater than AC ; which was to be proved. 2°. In the triangle ABC, let the side BC be greater than AC : then is the . angle A greater than the angle B. For, if the angles A and B were equal, the sides BC and AC would be equal; or if the angle A were less than the angle B, the side BC would be less than AC, either of which conclusions contradicts the hypothesis, and is im- possible : hence, the angle A is greater than the angle B ; which was to be proved. BOOK IX. 265 PROPOSITION XIII. THEOREM. If two triangles on the same, or on equal spheres, are mu- tually equiangular, they are also mutually equilateral. Let the spherical triangles A and B be mutually equi- angular : then are they also mutually equilateral. For, let P be the sup- plemental polar triangle of A, and Q, the supple- mental polar triangle of B : then, because the tri- angles A and B are mu- tually equiangular, their supplemental triangles P and Q must be mutually equilateral (P. VI.), and conse- quently mutually equiangular (P. X.). But, the triangles P and Q being mutually equiangular, their supplemental triangles A and B are mutually equilateral (P. VI.) ; which was to he proved. Scholium. Two plane triangles that are mutually equi- angular are not necessarily mutually equilateral ; that is, they may be similar without being equal. Two spherical triangles on the same or on equal spheres can not be similar without being equal in all respects. PROPOSITION XIV. THEOREM. The sum/ of the angles of a spherical triangle is less than six right angles, and greater than two right angles. Let ABC be a spherical triangle, on a sphere whose centre is 0, and DEF its supplemental triangle : then is 266 GEOMETBT. the sum of the angles A, B, and C, less than six right angles and greater than two. For, any angle, as A, being measured by a semi-circum- ference, minus the side EF (P. VI.), is less than two right angles: hence, the sum of the three angles is less than six right angles. Again, because the measure of each angle is equal to a semi-circum- ference minus the side lying op- posite to it, in the supplemental triangle, the measure of the sum of the three angles is equal to three semi-circumferences, minus the sum of the sides of the sup- plemental triangle DEF. But the latter sjiam. is less than a circum- ference ;■ ^ (fonsequently, the meas- ure of the sum of the angles A, B, and C, is greater than a semi-circumference, and therefore the sum of the angles is greater than two right angles: hence, the sum of the angles A, B, and C, is less than six right angles and greater than two ; which was to be proved. Cor. 1. The sum of the three angles of a spherical triangle is not constant, like that of the angles of a rec- tilineal triangle, but varies between two right angles and six, without ever reaching either of these limits. Two angles, therefore, do not serve to determine the third. Cor. 2. A spherical triangle may have two, or even three of its angles right angles; also two, or even three of its angles obtuse. Cor. 3. K a triangle, ABC, is bi-redangular, that is, has two right angles B and C, the vertex A is the pole of the other side BC, and AB, AC, will be quadrants. BOOK IX. 267 For, since the arcs AB and AC are perpendicular to BC, each must pass through its pole (P. III., Cor. 3) : hence, their intersection A is that pole, and consequently, AB and AC are quadrants. If the angle A is also a right angle, the triangle ABC is tri-rectangular ; each angles is a right angle, and its sides are quadrants, tri-rectangular triangles make up the surface of a hemi- sphere, and eight the entire surface of a sphere. of its Four Scholium. The right angle is taken as the unit of measure of spherical angles, and is denoted by 1. The excess of the sum of the angles of a spherical triangle over two right angles, is called the spherical ex- cess. If we denote the spherical excess by E, and the three angles expressed in terms of the right angle, as a unit, by A, B, and C, we have, E = A + B + C-2. The spherical excess of any spherical polygon is equal to the excess of the sum of its angles over two right angles taken as many times, less two, as the polygon has sides. If we denote the spherical excess by E, the sum of the angles by S, and the number of sides by n, we have, E = S - 2 (m - 2) = S - 2tc + 4. 268 GEOMETRY. PROPOSITION XV. THEOREM. Any lune is to the surface of the sphere, as the arc which measures its angle is to the circumference of a great circle; or, as the angle of the lune is to four right angles. Let AMBN be a lune, and MON the angle of the lune; then is the area of the lune to the surface of the sphere, as the arc MN is to the circumference of a great circle MNPQ; or, as the angle MON is to four right angles (B. m., P. XVIL, 0. 2). In the first place, suppose the arc MN and the circumference MNPQ to be commensurable. For example, let them be to each other as 5 is to 48. Di- vide the circumference MNPQ into 48 equal parts, beginning at M ; MN -will contain five of these parts. Join each point of division with the points A and B, by a quadrant; there will be formed 96 equal isosceles spherical triangles (P. VII., S. 2) on the surface of the sphere, of which the lune will contain 1 ; hence, in this case, the area of the lune is to the surface of the sphere, as 10 is to 96, or as 5 is to 48; that is, as the arc MN is to the circumference MNPQ, or as the angle of the lune is to four right angles. In like manner, the same relation may be shown to exist when the arc MN, and the circumference MNPQ are to each other as any other whole numbers. If the arc MN, and the circumference MNPQ, are not commensurable, the same relation may be shown to exist BOOK IX. 26y by a course of reasoning entirely analogous to that em- ployed in Book IV., Proposition III. Hence, in all cases, the area , of a lune is to the surface of the sphere, as the arc measuring the angle is to the circumference of a great circle ; or, as the angle of the lune is to four right angles ; which was to be proved. Cor. 1. LuneSj on the same or on equal spheres, are to each other as their angles. Cor. 2. If we denote the area of a tri-rectangular tri- angle hy T, the area of a lune by L, and the angle of the lune by A, the right angle being denoted by 1, we have, L : 8T : : A : 4 ; whence, L = T X 2A; hence, the area of a lune is equal to the area of a tri- rectangular triangle multiplied by twice the angle of the lune. Scholium. The spherical wedge, whose angle is MON, is to the entire sphere, as the angle of the wedge is to four right angles, as may be shown by a course of reasoning entirely analogous to that just employed: hence, we infer that the volume of a spherical wedge is equal to the lune which forms its base, multiplied by one third of the radius. PROPOSITION XVI. THEOREM. Symmetrical triangles are equal in area. Let ABC and DEF be symmetrical triangles, on a sphere whose centre is 0, the side DE being equal to AB, the side DF to AC, and the side EF to BC : .then are the triangles equal in area. 270 GEOMETRY. For, conceive a small circle to be drawn through A, B, and C, and let P be its pole ; draw arcs of great circles from P to A, B, and C : these arcs wiU be equal (D. 7). Draw the arc of a great circle FQ, making the angle DFQ equal to ACP, and lay off on it FQ equal to CP ; draw arcs of great circles QD and QE. In the triangles PAC and FDQ, we have the side FD equal to AC, by hypothesis; the side FQ equal to PC, by construction, and the angle DFQ equal to ACP, by construction : hence (P. VIII.), the side DQ is equal to AP, the angle FDQ to PAC, and the angle FQD to APC. Now, because the triangles QFD and PAC are isosceles and equal in all their parts, they may be placed so as to co- incide throughout, the base FD falling on AC, DQ on CP, and FQ on AP : hence, they are equal in area. If we take from the angle DFE the angle DFQ, and from the angle ACB the angle ACP, the remaining angles QFE and PCB, will be equal. In the triangles FQE and PCB, we have the side QF equal to PC, by construction, the side FE equal to BC, by hypothesis, and the angle QFE equal to PCB, from what has just been shown : hence, the triangles are equal in all their parts, and being isosceles, they may be placed so as to coincide through- out, the side QE falling on PC, and the side QF on PB ; these triangles are, therefore, equal in area. In the triangles QDE and PAB, we have the sides QD, QE, PA, and PB, all equal, and the angle DQE equal to APB, because they are the sums of equal angles: hence, the triangles are equal in all their parts, and because they are isosceles, they may be so placed as to coincide BOOK IX. 271 throughout, the side QD falling on PB, and the side QE on PA; these triangles are, therefore, equal in area. Hence, the sum of the triangles QFD and QFE, is equal to the sum of the triangles PAC and PBC. If from the former sum we take away the triangle QDE, there will remain the triangle DFE ; and if from the latter sum we take away the triangle PAB, there will remain the tri- angle ABC : hence, the triangles ABC and DEF are equal in area. If the point P falls within the triangle ABC, the point Q will fall within the triangle DEF, and we shall have the triangle DEF equal to the sum of the triangles QFD, QFE, and QDE, and the triangle ABC equal to the sum of the equal triangles PAC, PBC, and PAB. Hence, in either case, the triangles ABC and DEF are equal in area; which was to be proved. PROPOSITION XVII. THEOREM. If the circumferences of two great circles intersect on the surface of a hemisphere, the sum of the opposite tri- angles thus formed is equal to a lune, whose angle is equal to that formed by the circles. Let the circumferences ACB, ,-- ^. PCN, intersect on the surface of y^'" ... \. a hemisphere whose centre is : /i - - \ then is the sum of the opposite iV \ q. ^^\ ... | triangles ACP, NCB, equal to the \ ^^llT \ % lune whose angle is NCB. \ \^~~~ \7 For, produce the arcs CB, CN, \. ^ — _/N on the other hemisphere till they ^ '^ meet at D. Now, since ACB and CBD are semi-circumferences, if we take away the common 272 GEOMETRY. part CB, we shall have BD equal to AC. For a like rear son, we have DN equal to CP, and BN equal to AP: hence, the two triangles ACP, BND, have their sides respectively equal : they are therefore symmetrical ; consequently, they are equal in area (P. XVI.). But the sum of the triangles BDN, BCN, is equal to the lune CBDNC, whose angle is NCB: hence, the sum of ACP and NCB is equal to the lune whose angle is NCB ; which was to be proved. Scholium. It is evident that the two spherical pyramids, which have the triangles ACP, NCB, for bases, are together equal to the spherical wedge whose angle is NCB. PROPOSITION XVIII. THEOREM. The area of a spherical triangle is equal to its spherical excess multiplied by a tri-reebangular triangle. Let ABC be a spherical triangle on a sphere whose cen- tre is O : then is its surface equal to (A + B + C - 2) X T. For, produce its sides till they meet the great circle DEFG, drawn at pleasure, without the triangle. By the last theorem, the two tri- angles ADE, AGH, are together equal to the lune whose angle is A ; but the area of this lune is equal to 2AxT (P. XV., C. 2): hence, the sum of the triangles ADE and AGH, BOOK IX. 273 is equal to 2AxT. In like manner, it may be shown that the sum of the triangles BFG and BID is equal to 28 xT, and that the sum of the triangles CIH and CFE is equal to 2CxT. But 'the sum of these six triangles exceeds the hemi- sphere, or four times T, by twice the triangle ABC. We therefore have, 2 xarea ABC = 2AxT + 2BxT + 2CxT — 4T; or, by reducing and factoring, area ABC = (A+B + C — 2) xT; which was to he proved. Scholium 1. The same relation which exists between the spherical triangle ABC, and the tri-rectangular triangle, exists also between the spherical pyramid which has ABC for its base, and the tri-rectangular pyramid. The triedral angle of the pyramid is to the triedral angle of the tri- rectangular pyramid, as the triangle ABC to the tri-rectan- gular triangle. From these relations, the following conse- quences are deduced : 1°. Triangular spherical pyramids are to each other as their bases ; and since a polygonal pyramid may always be divided into triangular pyramids, it follows that any two spherical pyramids are to each other as their bases. 2°. Polyedral angles at the centre of the same, or of equal spheres, are to each other as the spherical polygons intercepted by their faces. Scholium 2. A triedral angle whose faces are perpen- dicular to each other, is called a right triedral angle; and . if the vertex is at the centre of a sphere, its faces intercept a tri-rectangular triangle. The right triedral J74 GEOMETRY. angle is taken as the unit of polyedral angles, and the tri-rectangular spherical triangle is taken as its measure.. If the vertex of a polyedral angle is taken as the centre of a sphere, the portion of the surface intercepted by its faces is the measure of the polyedral angle, a tri-rectan- gular triangle of the same sphere being the unit. PROPOSITION XIX. THEOREM. The area of a spherical polygon is equal to its spherical excess multiplied by the tri-rectangular triangle. Let ABCDE be a spherical polygon on a sphere whose centre is 0, the sum of whose angles is S, and the num- ber of whose sides is n : then is its area equal to (S - 2» + 4) X T. For, draw the diagonals AC, AD, dividing the polygon into spherical triangles : there are n — 2 such tri- angles. Now, the area of each tri- angle is equal to its spherical" excess into the tri-rectangular triangle : hence, the sum of the areas of all the triangles, or the area of the polygon, is equal to the sum of all the an- gles of the triangles, or the sum of the angles of the polygon diminished by 2 (n — 2), into the tri-rectangular triangle ; or, area ABCDE = [S - 2 (n - 2)] x T ; whence, by reduction, area ABCDE = (S — 2» -f- 4) x T ; which was to be proved. BOOK IX. 276 GENERAL SCHOLIUM 1. From any point P on a hemisphere, two arcs of a great circle, PC and PD, can always be drawn, which shall be perpendicular to the circumfer- ence of the base of the hemi- E sphere, and they will in general y^'^ f \\ be unequal. Now, it may be / M\ \ \ proved, by a course of reasoning / 7 \''^\^''' \ analogous to that employed in aC / j ...X^.\. jB Book L, Proposition XV. : ^o^----L_l3 — ^ C R s 1°. That the shorter of the two arcs, PC, is the shortest arc that can be drawn from the given point to the circumference ; and, therefore, that the longer of the two, PED, is the longest arc that can be drawn from the given point to the circumference : 2°. That two oblique arcs, PQ and PR, drawn from the sanae point, to points of the circumference at equal distances from the foot of the perpendicular, are equal : 3°. That of two oblique arcs, PR and PS, drawn from the same point, that is the longer which meets the cir- cumference at the greater distance from the foot of the perpendicular. GENERAL SCHOLIUM 2. The arc of a great circle drawn perpendicular to an arc of a second great circle of a sphere, passes through the poles of the second arc (P. III., C. 3). The measure of a spherical angle is the arc of a great circle included between the sides of the angle and at the distance of a quadrant from its vertex (P. IV.). It is evident, therefore, 276 GEOMETEY. that the pole of either side of an acute spherical angle lies without the sides of the angle; and that the pole of either side of an obtuse spherical angle lies within the sides of the angle. Now, let A be an acute spher- ical angle, ST its measure, MN any arc of a great circle, other than ST, drawn perpendicular to the side AQ, and included between the two sides AQ and AR, and P the pole of the side AQ : and Let B be an obtuse spherical angle, CD its measure, EF any arc of a great circle, other than CD, drawn per- pendicular to the side BH, and in- cluded between the two sides BH --S — ^E and BG, and P' the pole of the side BH : then It may readily be shown (P. m., 0. 1, and Gen. S. I., 1°), 1°. That ST is longer than MN, and, hence, is the longest arc of a great circle that can be drawn perpendicular to the side AQ and included be- tween the two sides AQ and AR : and 2°. That CD is shorter than EF, and, hence, is the shortest arc of a great circle that can be drawn perpen- dicular to the side BH and included between the two sides BH and BG. BOOK IX. 277 EXERCISES. 1. The sides of a spherical triangle are 80°, 100", and 110°; find the angles of its supplemental triangle, and the angles of each of its polar triangles. 2. Find the area of a tri-rectangular triangle, on a sphere whose diameter is 8 feet. 3. Find the area of a tri-rectangular triangle, on a sphere whose surface and volume may be expressed by the same number. 4. The angle of a lune, on a sphere whose radius is 5 feet, is 50° ; find the area of the lune and the volume of the corresponding wedge. 5. The area of a lune is 33.5104 square feet and the angle of the lune is 60° ; find the surface and the vol- ume of the sphere. 6. Show that if two spherical triangles on unequal spheres are mutually equiangular, they are similar. 7. Show how to circumscribe a circle about a given spherical triangle. 8. Show how to inscribe a circle in a given spherical triangle. 9. Show that the intersection of the surfaces of two spheres is a circle, and that the line which joins the cen- tres of two intersecting spheres is perpendicular to the circle in which their surfaces intersect. 10. Show that two spherical pyramids of the same or equal spheres, which have symmetrical triangles for bases, are equal in volume. [Proof analogous to that in P. XVI.] 11. The circumferences of two great circles intersect on the surface of a hemisphere whose diameter is 10 feet, and the acute angle formed by them is 40° ; find the sum of the opposite triangles thus formed and the sum of the corresponding spherical pyramids. 278 GEOMETRY. 12. Show that the volume of a triangtilar spherical pyramid is equal to its base multiplied by one third the radius of the sphere. 13. Show that the volume of any spherical pyramid is equal to its base multiplied by one third the radius of the sphere. 14. Find the volume of a spherical pyramid whose base is a tri-rectangular triangle, the diameter of the sphere being 8 feet. 15. The angles of a triangle, on a sphere whose radius is 9 feet, are 100°, 115°, and 120°; find the area of the triangle and the volume of the corresponding spherical pyramid. 16. A spherical pyramid, of a sphere whose diameter is 10 feet, has for its base a triangle of which the angles are 60°, 80°, and 85°; what is its ratio to a pyramid whose base is a tri-rectangular triangle of. the same sphere? 17. The sum of the angles of a regular spherical octa- gon is 1140°, and the radius of the sphere is 12 feet; find the area of the octagon. 18. The volume of a spherical pyramid, whose base is an equiangular triangle, is 84.8232 cubic feet, and the radius of the sphere is 6 feet; find one of the angles of the base. 19. Given a spherical angle of 40°; what is the num- ber of degrees in the longest arc of a great circle that can be drawn perpendicular to either side of the angle and included between the two sides? 20. Given a spherical angle of 115°; what is the num- ber of degrees in the shortest arc of a great circle that can be drawn perpendicular to either side of the angle and included between the two sides? APPENDIX GRADED EXERCISES IN PLANE GEOMETRY. ADDITIONAL DEFINITIONS. L The Distance of a point from a line is measured oij a perpendicular to that line. 2. The Bisectrix of an angle is a line that divides the anjle into two equal parts. 8. A Median is a line drawn from any vertex of a triangle to the middle of the opposite side. 4. The Projection of a point, on a line, is the foot of a perpendicular drawn from the point to the line. 5. The Projection of one straight line on another, is that part of the second line which is contained between the projections of the two extreme points of the first line, upon the second. PROPOSITIONS. I. Theorem. — Show that the bisectrices of two adjacent angles are perpendicular to each other. II. Theorem. — Show that the perimeter of any triangle is greater than the sum of the distances from any point 280 GEOMETRY. within the triangle to its three vertices, and less than twice that sum. ni. Theoeem. — Show that the angle between the bisec- trices of two consecutive angles of any quadrilateral, is equal to one half the sum of the other two angles. IV. Theorem. — Show that any point in the bisectrix of an angle is equally distant from the sides of the angle. V. Theorem. — If two sides of a triangle are prolonged beyond the third side, show that the bisectrices of this included angle and of the exterior angles all meet in the same point. VI. Theorem. — Show that the projection of a line on a parallel line, is equal to the line itself; and that the pro- jection of a line on a line to which it is oblique, is less than the line itself. VTI. Theorem. — If a line is drawn through the point of intersection of the diagonals of a parallelogram and limited by the sides of the parallelogram, show that the line is bi- sected at the point. Vni. Theorem. — ^The bisectrices of the four angles of any parallelogram form, by their intersection, a rectangle whose diagonals are parallel to the sides of the given paral- lelogram. IX. Theorem. — Show that the sum of the distances from any point in the base of an isosceles triangle to the two other sides, is equal to the distance from the vertex of either angle at the base to the opposite side. X. Theorem. — Show that the middle point of the hypoth- APPENDIX. 281 enuse of any right-angled triangle is equally distant from the three vertices of the triangle. XI. Problem. — Draw two lines that shall divide a given right angle into three equal parts. XII. Theorem. — Draw a line AP through the vertex A of a triangle ABF and perpendicular to the bisectrix of the angle A; construct a triangle PBF, having its vertex P on AP, and its base coinciding with that of the given tri- angle: then show that the perimeter of PBF is greater than that of ABF. XIII. Theorem. — Let an altitude of the triangle ABC be drawn from the vertex A, and also the bisectrix of the angle A ; then show that their included angle is equal to half the difference of the angles B and C. XIV. Problem. — Given two lines that would meet, if sufficiently prolonged: then draw the bisectrix of their included angle, without finding its vertex. XV. Problem. — From two points on the same side of a given line, to draw two lines that shall meet each other at some point of the given line, and make equal angles with that line. XVI. Theorem. — Show that the sum of the lines drawn to a point of a given line, from two given points, is the least possible when these lines are equally inclined to the given line. XVII. Problem. — Prom two given points, on the same side of a given line, draw two lines meeting on the given line and equal to each other. 282 GEOMETRY. XVIII. Problem. — Through a given point A, draw a line that shall he equally distant from two given points, B and C. XIX. Problem. — Through a given point, draw a line cut- ting the sides of a given angle and making the interior angles equal to each other. XX. Problem. — Draw a line PQ parallel to the base BC of a triangle ABC, so that PQ shall be equal to the sum of BP and CQ. XXI. Problem. — In a given isosceles triangle, draw a line that shall cut off a trapezoid whose base is the base of the given triangle and whose three other sides shall be equal to each other. XXn. Theorem. — If two opposite sides of a parallelo- gram are bisected, and lines are drawn from the points of bisection to the vertices of the opposite angles, show that these lines divide the diagonal, wLich they intersect, into three equal parts. XXTTT. Problem. — Construct a triangle, having given the two angles at the base and the sum of the three sides. XXrV. Problem. — Construct a triangle, having given one angle, one of its including sides, and the sum of the two other sides. XXV. Problem. — Construct an equilateral triangle, hav- ing given one of its altitudes. XXVI. Theorem. — Show that the three altitudes of a triangle all intersect in a common point. APPENDIX. 283 XXVn. Theoeem. — ^If one of the acute angles of a right- angled triangle is double the other, show that the hypoth- enuse is double the smaller side about the right angle. XXVIII. Theorem. — Let a median be drawn from the vertex of any angle A of a triangle ABC: then show that the angle A is a right angle when the median is equal to half the side BC, an acute angle when the median is greater than half of BC, and an obtuse angle when the median is less than half of BC. XXIX. Theorem. — Let any quadrilateral be circumscribed about a circle : then show that the sum of two opposite sides is equal to the sum of the other two opposite sides. XXX. Problem. — Draw a straight line tangent to two given circles. XXXI. Problem. — Through a given point P, draw a cir- cle that shall be tangent to a given line CB, at a given point B. XXXII. Theorem. — Let two circles intersect each other, and through either point of intersection let diameters of the circles be drawn : then show that the other extremi- ties of these diameters and the other point of intersec- tion lio in the same straight line. XXXIII. Problem. — Through two given points A and B, draw .'J- circle that shall be tangent to a given line CP- XVf.XrV". Problem. — Draw a circle that shall be tangent to a given circle C, and also to a given line DP, at a given point P. 284 ^ GEOMETRY. XXXV. Problem. — Draw a circle that shall be tangent to a given line TP, and also to a given circle C, at a given point Q. XXXVI. Problem. — Draw a circle that shall pass through a given point Q, and be tangent to a given circle C, at a given point P. XXXVII. Problem. — Draw a circle, with a given radius, that shall be tangent to a given line DP, and to a given circle C. XXXVIII. Problem. — Find a point in the prolongation of any diameter of a given circle, such that a tangent from it to the circumference shall be equal to the diam- eter of the circle. XXXIX. Theorem. — Show that when two circles inter- sect each other, the longest common secant that can be drawn through either point of intersection, is parallel to the hne joining the centres of the circles. XL. Problem. — Construct the greatest possible equilat- eral triangle whose sides shall pass through three given points A, B, and C, not in the same straight line. XLI. Theorem. — Show that the bisectrices of the four angles of any quadrilateral intersect in four points, all of which lie on the circumference of the same circle. XLn. Theorem. — If two circles touch each other exter- nally, and if two common secants are drawn through the point of contact and terminating in the concave arcs, show that the lines joining the extremities of these se- cants, in the two circles, are parallel APPENDIX. 285 XLIII. Theoeem. — ^Let an equilateral triangle be inscribed in a circle, and let two of the subtended arcs be bisected by a chord of the circle : then show that the sides of the triangle divide the chord into three equal parts. XLIV. Problem. — Find a point, within a triangle, such that the angles formed by drawing lines from it to the three vertices of the triangle shall be equal to each other. XLV. Pboblem. — Inscribe a circle in a quadrant of a given circle. XL VI. Problem. — Through a given point P, within a given angle ABC, draw a circle that shall be tangent to both sides of that angle. XL VII. Theorem. — Show that the middle points of the sides of any quadrilateral are the vertices of an inscribed parallelogram. XLVni. Problem. — Inscribe in a given triangle, a tri- angle Avhose sides shall be parallel to the sides of a sec- ond given triangle. XLIX. Problem. — Through a point P, within a given angle, draw a line such that it and the parts of the sides that are intercepted shall contain a given area. L. Problem. — Construct a parallelogram whose area and perimeter are respectively equal to the area and perimeter of a given triangle. LI. Problem. — Inscribe a square in a semicircle ; that is, a square two of whose vertices are in the diameter, and the other two in the semi-circumference. 286 GEOMETET. LII. Problem.— Through a given point P draw a line cutting a triangle, so that the sum of the perpendiculars to it, from the two vertices on one side of the line, shall be equal to the perpendicular to it from the vertex, on the other side of the line. LITE. Theorem. — Show that the line which joins the middle points of two opposite sides of any quadrilateral, bisects the line joining the middle points of the two diagonals. LIV. Theorem. — If from the extremities of one of the oblique sides of a trapezoid, lines are drawn to the middle point of the opposite side, show that the triangle thus formed is equal to one half the given trapezoid. LY. Problem. — Find a point in the base of a triangle, such that the lines drawn from it, parallel to and limited by the other sides of the triangle, shall be equal to each other. LVI. Theorem. — Show that the line drawn from the middle of the base of any triangle to the middle of any line of the triangle parallel to the base, will pass through the opposite vertex, if sufficiently produced. LVII. Theorem. — Show that the three medians of any triangle meet in a common point. LVin. Theorem. — On the sides AB and AC of any tri- angle ABC, construct any two parallelograms ABPE and ACFG ; prolong the sides DE and FG till they meet in H ; draw HA, and on the third side BC of the triangle, con- struct a parallelogram two of whose sides are parallel and equal to HA: then show that the parallelogram on BC is equal to the sum of the parallelograms on AB and AC. APPENDIX. 287 LIX. Theoeem. — Assuming the principle demonstrated in the last proposition, deduce from it the truth that the square on the hypothenuse of a right-angled triangle is equal to the sum of the squares on the two other sides. LX. Theorem. — If from the middle of the base of a right-angled triangle, a line is drawn perpendicular to the hypothenuse dividing it into two segments, show that the difference of the squares of these segments is equal to the square of the other side about the right angle. LXI. Theorem. — If lines are drawn from any point P to the four vertices of a rectangle, show that the sum of the squares of the two lines drawn to the extremities of one diagonal, is equal to the sum of the squares of the two lines drawn to the extremities of the other diagonal. LXII. Theorem. — ^Let a line be drawn from the centre of a circle to any point of any chord ; then show that the square of this line, plus the rectangle of the segments of the chord, is equal to the square of the radius. LXIII. Problem. — Draw a line from the vertex of anf scalene triangle to a point in the base, such that thi^ line shall be a mean proportional between the segments into which it divides the base. LXIV. Theorem. — Show that the sum of the squares of the diagonals of any quadrilateral is equal to the sum of the squares of the four sides of the quadrilateral, dimin- ished by four times the square of the distance between the middle points of the diagonals. LXV. Problem. — Construct an equilateral triangle equal in area to any given isosceles triangle. 288 GEOMETRY.' LXVI. Theorem. — In a triangle ABC, let two lines be drawn from the extremities of the base BC, intersecting at any point P on the median through A, and meeting the opposite sides in the points E and D : show that DE is parallel to BC. APPLICATION OF ALGEBRA TO GEOMETRY. To solve a geometrical problem by means of algebra, draw a figure which shall contain all the given and re- quired parts and also such other lines as may be neces- sary to establish the relations between them ; then denote the given parts by leading letters, and the required parts by final letters of the alphabet : next consider the rela- tions between the given and required parts and express these relations by equations, taking care to have as many independent equations as there are parts to be determined (Bourdon, Art. 92). The solution of these equations will give the values of the required parts. To indicate the method of proceeding, the solution of the first problem is given. LXVII. Problem. — In a right-angled triangle ABC, given the base BA and the sum of the hjrpothe- nuse and the perpendicular, to find the hypothenuse and the perpendicular. Solution. Denote BA by c, BC by x, AC by y, and the sum of BC and AC by s. Then, x + y = s. ' • (1.) From B. IV., P. XI., a? ^ y^ + d'. (2.) From (1), we have, x = s — y. Squaring, x^ = ^ — 2sy + y^. • • • (3.) APPENDIX. 289 Subtracting (2) from (3), = s' — 2sy — A gi (J) Transposing and dividing, y = — ^ , ■whence, x = s t: — = — ^ ■ 2s 2s If c = 3 and s = 9, we have a; = 5 and ?/ = 4. LXVin. Problem. — ^In a right-angled triangle, given the hypothenuse and the sum of the sides about the right angle, to find these sides. LXIX. Problem. — ^In a rectangle, given the diagonal and the perpendicular, to find the sides. LXX. Problem. — Given the base and perpendicular of a triangle, to find the side of an inscribed square. LXXI. PbOBLEM. — In an equilateral triangle, given the distances from a point within the triangle to each of the three sides, to find one of the equal sides. LXXn. Problem. — In a right-angled triangle, given the base and the difference between the hypothenuse and the perpendicular, to find the sides. LXXni. Problem. — ^In a right-angled triangle, given the hypothenuse and the difference between the base and the perpendicular, to determine the triangle. LXXIV. Problem. — ^Having given the area of a rectan- gle inscribed in a given triangle, to determine the sides of the rectangle. LXXV. Problem. — In a triangle, having given the ratio of the two sides together with both segments ot the base made by a perpendicular from the vertex, to determine the triangle. 290 GEOMETRY. LXXVI. Pboblem. — ^In a triangle, having given the base, the sum of the two other sides, and the length of a line drawn from the vertex to the middle of the base ; to find the sides of the triangle. LXXVn. Problem. — In a triangle, having given the two sides about the vertical angle, together with the line bi- secting that angle and terminating in the base ; to find the base. LXXVin. Problem. — To determine a right-angled tri- angle, having given the lengths of two lines drawn from the vertices of the acute angles to the middle points of the opposite sides. LXXIX. Problem. — To determine a right-angled triangle, having given the perimeter and the radius of the in- scribed circle. LXXX. Problem. — To determine a triangle, having given the base, the perpendicular, and the ratio of the two sides. LXXXI. Problem. — To determine a right-angled trian- gle, having given the hypothenuse and the side of the inscribed square. LXXXn. Problem. — To determine the radii of three equal circles, described within and tangent to a given circle, and also tangent to each other. LXXXm. Problem. — In a right-angled triangle, having given the perimeter and the perpendicular let fall from the right angle on the hypothenuse, to determine the tri- angle. APPENDIX. 291 •LXXXIV. Problem. — To determine a right-angled tri- angle, having given the hypothenuse and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. LXXXV. Problem. — To determine a triangle, having given the base, the perpendicular, and the difference of the two other sides. LXXXVI. Problem. — To determine a triangle, having given the base, the perpendicular, and the rectangle of the two sides. LXXXVII. Problem. — To determine a triangle, having given the lengths of three lines drawn from the three angles to the middle of the opposite sides. LXXXVni. Problem. — In a triangle, having given the three sides, to find the radius of the inscribed circle. LXXXIX. Problem. — To determine a right-angled tri- angle, having given the side of the inscribed square and the radius of the inscribed circle. XC. Problem. — To determine a right-angled triangle, having given the hypothenuse and the radius of the in- scribed circle. TRIGONOMETRY AND MENSURATION INTRODUCTION TO TRIGONOMETRY. LOGARITHMS. 1. The Logarithm of a given number is the exponent of t'-i power to which it is necessary to raise a fixed numbe to produce the given number. The fixed number is called the Base of the System. Any positive number, except 1, may be taken as the base of a system. In the common system, to which alone reference is here made, the base is 10. Every number is, therefore, regarded as some power of 10, and the expo- nent of that power is the logarithm of the number. 2. If we denote any positive number by n, and the corresponding exponent of 10 by x, we shall have the exponential equation, 10* = 71. (1.) In this equation, x is, by definition, the logarithm of n, which may be expressed thus, X = log n. (2.) 3. If a number is an exact power of 10, its logarithm is a whole number. Thus, 100, being equal to 10', has for its logarithm 2. If a number is not an exact power of 10, its logarithm is composed of two parts, a whole number called the Characteristic, and a decimal part called the Mantissa. Thus, 225 being greater than 10' and less than 10', its logarithm is found to be 2.352183, 4 INTBODUCTION. of which 2 is the characteristic and .352183 is the man 4. If, in the equation, logiior = p, (3.) we make p successively equal to 0, 1, 2, 3, &c., and also equal to — 0, — 1, — 2, — 3, &c., we may form the fol- lowing TABLE, log 1 = log 10 = 1 log .1 = — 1 log 100 = 2 log .01 = - 2 log 1000 = 8 log .001 = -8 &c., &c. &c., &c. If a num.ber lies between 1 and 10, its logarithm lies between and 1, that is, it is equal to plus a deci- mal; i£ a number lies between 10 and 100, its logarithm is equal to 1 plus a decimal; if between 100 and 1000, its logarithm is equal to 2 plus a decimal ; and so on ; hence, we have the following Rule. — The characteristic of the logarithm of an entire nuinber is positive, and numerically 1 less than the num- ber of places of figures in the given number. If a decimal fraction lies between .1 and 1, its loga- rithm lies between — 1 and 0, that is, it is equal to — 1 plus a decimal; if a number lies between .01 and .1, its logarithm is equal to — 2 plus a decimal ; if between .001 and .01, its logarithm is equal to — 3 plus ' a deci- mal; and so on: hence, the following Rui.K. — The characteristic of the logarithm of a decimal fraction is negative, and num,erically 1 greater than the number of O's that imimicdiately follow the decimal point. TEIGONOMETET. 5 The characteristic alone is negative, the mantissa being always positive. This fact is indicated by writing the negative sign over the characteristic: thus, 2.371465, is equivalent to — 2 + .371465. Note. — ^It is to be observed, that the characteristic of the logarithm of a mixed number is the same as that of its entire part. Thus, the characteristic of the logarithm of 725.4275 is the same as the characteristic of the log- arithm of 726. QENEEAL PRINCIPLES. 5. Let m and n denote any two numbers, and x and y their logarithms. We shall have, from the definition of a logarithm, the following equations, 10"= = m. (4.) lOv = n. (5.) Multiplying (4) and (5), member by member, we have lO^+y = mn; whence, by the definition, X + y = log {ran). (6.) That is, the logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers. 6. Dividing (4) by (5), member by member, we have n ' whence, by the definition, x-y = \og Q (7.) That is, the logarithm, of a quotient is equal to the loga- rithm of the dividend diminished by that of the divisor. 6 INTRODUCTION. 7. Raising both members of (4) to the power denoted t>y P, we have, whence, bj'^ the definition, xp = logm^ (8.) That is, the logarithm of any power of a number is equal to the logarithm, of the nuiriber multiplied by the exponent of the power. 8. Extracting the root, indicated by r, of both mem- bers of (4), we have X 10'- = V^; whence, by the definition, ^ = log ^/m. (9.) That is, the logarithm of any root of a number is equal to the logarithm of the nuniber divided by the index of the root. The preceding principles enable us to abbreviate the operations of multiplication and division, by converting them into the simpler ones of addition and subtraction. TABLE OF LOGARITHMS. 9. A Table of Logarithms is a table containing a set of numbers and their logarithms, so arranged that,, having given any one of the numbers, we can find its logarithm ; or, having the logarithm, we can find the corresponding number. In the table appended, the complete logarithm is given for all numbers from 1 up to 100. For other numbers, TEIGONOMETKY. 7 the mantissas alone are given ; the characteristic may be found by one of the rules of Art. 4. Before explaining the use of the table, it is to be shown that the mantissa of the logarithm of any number is not changed by multiplying or dividing the number by any exact power of 10. Let n represent any number whatever, and 10^ any power of 10, j3 being any whole number, either positive or negative. Then, in accordance with the principles of Arts. 5 and 3, we shall have log (nxl 0") = log n + log 10" = p + log n ; but p is, by hypothesis, a whole number : hence, the deci- mal part of the log (n x 1 0^) is the same as that of log n ; which was to be proved. Hence, in finding the mantissa of the logarithm of a number, the position of the decimal point may be changed at pleasure. Thus, the mantissa of the logarithm of 456357, is the same as that of the number 4568.57; and the mantissa of the logarithm of 759 is the same as that of 7590, MANNER OF USING THE TABLE. 1°. To find the logarithm of a number less than 100. 10. Look on the first page, in the column headed "N," for the given number; the number opposite is the loga- rithm required. Thus, log 67 = 1.826075. 8 INTRODUCTION. 2°. To find the logarithm of a number between 100 and 10,000. 11. Find the characteristic by the first rule of Art. 4. To determine the mantissa, find in the column headed "N" the left-hand three figures of the given number; then pass along the horizontal line in -which these figures are found, to the column headed by the fourth figure of the given number, and take out the four figures found there ; pass back again to the column headed " 0," and there will be found in this column, either upon the hori- zontal line of the first three figures or a few lines above it, a number consisting of six figures, the left-hand two figures of which must be prefixed to the four already taken out. Thus, log 8979 = 3.953228. If, however, any dots are found at the place of the four figures first taken out, or if in returning to the " " column any dots are passed, the two figures to be pre- fixed are the left-hand two of the six figures of the " " column immediately below. Dots in the number taken out must be replaced by zeros. Thus, log 3098 = 3.491081, log 2188 = 3.340047. Note. — The above method of finding the mantissa as- sumes that the given number has four places of figures. If, therefore, the number lies between 100 and 1000, and has but three places of figures, find the characteristic by the first rule of Art. 4, and then, to find the mantissa, fill out the given number to four places of figures (or conceive it to be so filled out) by annexing (see Art. 9), and find the mantissa corresponding to the resulting number, as above. TRIGONOMETRY. 9 3°. To find the logarithm of a number greater than 10,000. 13. Find the characteristic by the first rule of Art. 4. To find the mantissa: set aside all of the given num- ber except the left-hand four figures, and find the man- tissa corresponding to these four, as in Art. 11 ; multiply the corresponding tabular difference, found in column " D," by the part of the number set aside, and discard as many of the right-hand figures of the product as there are fig- ures in the multiplier, and add the result thus obtained to the mantissa already found. If the left-hand figure of those discarded is 5 or more, increase the number added by 1. Note. — It is to be observed that the tabular difference, found in column "D," is millionths, and not a whole number ; and that, therefore, the result to be added " to the mantissa already found" is millionths. Example. — To find the logarithm of 672887: the char- acteristic is 5 ; set aside 87, and the mantissa correspond- ing to 6728 is .827886; the corresponding tabular differ- ence is 65, which multiplied by 87, the part of the number set aside, gives 5655 ; as there are two figures in the multiplier, discard the right-hand two figures of this product, leaving 56 ; but as the left-hand figure of those discarded is 5, call the result 57 (which is mill- ionths) ; adding this 57 to the mantissa already found, will give .827943 for the required mantissa; hence, log 672887 = 5.827943. The explanation of the method just given is briefly this : for the purpose of finding the mantissa, the given number is conceived to be a mixed one, thus, 6728.87, the mantissa not being affected by the position of the decimal point (see Art. 9). The numbers in the column 10 INTEODUCTION. "D" are the difEerences between the logarithms of two consecutive whole numbers. In the example just given, the mantissa of the logarithm of 6728 is .827886, and that of 6729 is .827951, and their difference is 65 mill- ionths; 87 hundredths of this difiference is 57 millionths ; hence, the mantissa of the logarithm of 6728.87 is found by adding 57 millionths to .827886. The principle em- ployed is, that the differences of numbers are proportional to the differences of their logarithms, when these differ- ences are small. 4°. To find the logarithm of a decimal. 13. Find the characteristic by the second rule of Art. 4. To find the mantissa, drop the decimal point, and con- sider the decimal a whole number. Find the mantissa of the logarithm of this number as in preceding articles, and it will be the mantissa required. Thus, log .0327 = 2.614548, log .378024 = 1.577520. Note. — To find the logarithm of a mixed number, find the characteristic by the Note, Art. 4 ; then drop the decimal point and proceed as above. 5°. To find the number corresponding to a given logarithm. 14. The rule is the reverse of those just given. Look in the table for the mantissa of the given logarithm. If it can not be found, take out the next less mantissa, and also the corresponding number, which set aside. Find the difference between the mantissa taken out and that of the given logarithm ; annex any number of O's, and divide this result by the corresponding number in the column "D." Annex the quotient to the number set aside, and TRIGONOMETRY. 11 then, if the characteristic is positive, point off, from the left hand, a number of places of figures equal to the characteristic plus 1 ; the result will be the number re- quired. If the characteristic is negative, prefix to the figures obtained a number of O's one less than the number of units in the negative characteristic and to the whole prefix a decimal point ; the result, a pure decimal, will be the number required. Examples. 1. Let it be required to find the number correspond- ing to the logarithm 5.233568. ' The next less mantissa in the table is 233504; the corresponding number is 1712, and the tabular difference is 253. Operation. Given mantissa, 233568 Next less mantissa, • • • • 233504 ■ • 1712 253 ) 6400000 ( 25296 .-. The required number is 171225.296. The number corresponding to the logarithm 2.233568 is .0171225. 2. What is the number corresponding to the logarithm 2.785407? Ans. .06101084. 3. What is the number corresponding to the logarithm 1.846741? Ans. .702653. MULTIPLICATION BY MEANS OF LOGARITHMS. 15. From the, principle proved in Art. 5, we deduce the following Rule. — Find the logarithms of the factors, and take their 12 INTRODUCTION. sum; then find the nwmber corresponding to the resulting logarithm, and it will be the product required. Examples. 1. Multiply 23.14 by 5.062. Operation. log 23.14 • • ■ 1.364363 log 5.062 • • • 0.704322 2.068685 .-. 117.1347, product. 2. ^ind the continued product of 3.902, 597.16, and 0.0314728. Operation. log 3.902 • • . 0.591287 log 597.16 . • . 2.776091 log 0.0314728 • • . 2.497936 1.865314 .-. 73.3354, product. Here, the 2 cancels the + 2, and the 1 carried from the decimal part is set down. 3. Find the continued product of 3.586, 2.1046, 0.8372, and 0.0294. Ans. 0.1857615. DIYISION- BY MEANS OF LOGARITHMS. 16. From the principle proved in Art. 6, we have the following Rule. — Find the logarithms of the dividend and divisor, and subtract the latter from, the former; then find the number corresponding to the resulting logarithm, and it will be the quotient required. TBIGONOMETRY. 13 Examples. Divide 24163 by 4567. log 24163 log 4567 Operation. 4.383151 3.659631 0.723520 .-. 5.29078, quotient. . Divide 0.7438 by 12.9476. Operation. log 0.7438 • . • 1.871456 log 12.9476 ■ • . 1.112189 2.759267 0.057447, quotient. Here, 1 taken from 1, gives 2 for a result. The sub- traction, as in this case, is always to be performed in the algebraic sense. 3. Divide 37.149 by 523.76. Ans. 0.0709274. The operation of division, particularly when combined with that of multiplication, can often be simplified by using the principle of THE ARITHMETICAL COMPLEMENT. 17. The Abithmetical Complement of a logarithm is the result obtained by subtracting it from 10. Thus, 8.130456 is the arithmetical complement of 1.869544. The arithmetical complement of a logarithm may be writ- ten out by commencing at the left hand and subtracting each figure from 9, until the last significant figure is reached, which must he taken from 10. The arithmetical complement is denoted by the symbol (a. c.) Let a and b represent any two logarithms whatever, and a — b their difference. Since we may add 10 to, 14 INTEODUCTION. and subtract it from, a — b, without altering its value, we have, a - & = o + (10 - 6) - 10. • • • (10.) But 10 — 6 is, by definition, the arithmetical complement of 6: hence, Equation (10) shows that the difference be- tween two logarithms is equal to the first, plus the arith- metical complement of the second, minus 10. Hence, to divide one number by another by means of the arithmetical complement, we have the following Rule. — Find the logarithm of the dividend, and the arithmetical complement of the logarithm of the divisor, add them together, and diminish the swm hy 10 ; the number corresponding to the resulting logarithm^ will be the quotient required. Examples. 1. Divide 327.5 by 22.07 Operation. log 327.5 • • • 2.515211 (a. c.) log 22.07 • • • 8.656198 1.171409 .-. 14.839, quotient. The operation of subtracting 10 is performed mentally. 2. Divide 37.149 by 523.76. Ans. 0.0709273. 3. Divide the. product of 358884 and 5672, by the product of 89721 and 42.056. log 358884 • • • 5.554954 log 5672 .... 3.753736 (a. c.) log 89721 .... 5.047106 (a. c.) log 42.056 • ■ ■ 8.376182 2.731978 .-. 539.48, result. 20 is here subtracted, as (a. c.) has been twice used. TRIGONOMBTEY. 15 4. Solve the proportion, 3976 : 7952 : : 5903 : x. Applying logarithms, the logarithm of the 4th term is equal to the sum of the logarithms of the 2d and 3d terms, minus the logarithm of the 1st: Or, the arithmet- ical complement of the logarithm/ of the 1st term, -plus the logarithm, of the 2d term,, plus the logarithm of the 3d term,, minus 10, is equal to the logarithm of the ith term,. Operation. (a. 0.) log 3976 • • • 6.400554 log 7952 ■ • . 3.900476 log 5903 • • • 3.771073 log a; • • • 4.072103 .-. x = 11806. RAISING TO POWERS BY MEANS OF LOGARITHMS. 18. From Article 7, we have the following Rule. — Find the logarithm of the num,ber, and multiply it by the exponent of the power; then find the number corresponding to the resulting logarithm, and it will be the power required. Examples. 1. Find the 5th power of 9. Operation. log 9 • • • 0.954243 5 4.771215 .-. 59049, power 2. Find the 7th power of 8. Ans. 2097154, nearly. 16 INTRODUCTION. EXTRACTINQ ROOTS BY MEANS OP LOGARITHMS. 19. From the principle proved in Art. 8, we have the following Rule. — Find the logarithm of the nivmber, and divide it by the index of the root; then find the number correspond- ing to the resulting logarithm, and it will be the root re- quired. Examples. 1. Find the cuhe root of 4096. The logarithm of 4096 is 3.612360, and one third of this is 1.204120. The corresponding number is 16, -which is the root sought. If the characteristic of the logarithm of the given num- ber is negative and not exactly divisible by the index of the root, add to it such negative quantity as shall make it exactly divisible, and add also to the mantissa a nu- merically equal positive quantity. 2. Find the 4th root of .00000081. The logarithm of .00000081 is 7.908485, which is equal to 8 4- 1.908485, and one fourth of this is 2.477121. The number corresponding to this logarithm is .03 ; hence, .03 is the root required. PLANE TRIGONOMETRY. 20. Plane Trigonometry is that branch of Mathematics which treats of the solution of plane triangles. In every plane triangle there are six parts : three sides and three angles. When three of these parts are given, one being a side, the remaining parts may be found by computation. The operation of finding the unknown parts is called the solution of the triangle. 31. A plane angle is measured by the arc of a circle included between its sides, the centre of the circle being at the vertex, and its radius being equal to 1. Thus, if the vertex A is taken as a centre, and the radius AB is equal to 1, the intercepted arc BC measures the angle A (B. III., P. XVIL, S.). Let ABCD represent a circle whose radius is equal to 1, and AC, BD, two diameters perpendicu- lar to each other. These diameters divide the circumference into four equal parts, called quadrants; and because each of the angles at the centre is a right angle, it follows that a right angle is measured by a quadrant. An acute angle is measured by an arc less than a quadrant, and an obtuse angle, by an arc greater than a quadrant. 18 PLANE TKIGONOMBTEY. 33. In Gteometry, the unit of angular measure is a right angle j so in Trigonometry, the primary unit is a quadrant, which is the measure of a right angle. For convenience, the quadrant is divided into 90 equal parts, each of which is called a degree ; each degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds. Degrees, minutes, and seconds, are denoted by the symbols °, ', ". Thus, the expression 7° 22' 33", is read, 7 degrees, 22 minutes, and 33 seconds. Fractional parts of a second are expressed decimally. A quadrant contains 324,000 seconds, and an arc of 7° 22' 33" contains 26553 seconds; hence, the angle meas- ured by the latter arc is the /g^y^fe part of a right angle. In like manner, any angle may be expressed in terms of a right angle. 33. The complement of an arc is the difference between that arc and 90°. The complement of an angle is the difference between that angle and a right angle. Thus, EB is the complement of AE, and FB is the complement of CF. In hke manner, the angle EOB is the complement of the angle AOE, and FOB is the complement of COF In a right-angled triangle, the acute angles are complements of each other. 34. The supplement of an arc is the difference between that arc and 180°. The supplement of an angle is the difference between that angle and two right angles. Thus, EC is the supplement of AE, and FC the supple- ment of AF. In like manner, the angle EOC is the supple- ment of the angle AOE, and FOC the supplement of AOF. PLANE TEIGONOMETEY. 19 In any plane triangle, any angle is the supplement of the sum of the two others. 35. Instead of the arcs themselves, certain functions of the arcs, as explained below, are used. A function of a quantity is something which depends upon that quan- tity for its value. The following functions are the only ones needed for solving triangles: 36. The sine of an arc is the distance of one extrem- ity of the arc from the diameter through the other ex- tremity. Thus, PM is the sine of AM, and P'M' is the sine of AM'. If AM is equal to M'C, AM and AM' are supplements of each other; and because MM' is parallel to AC, PM is equal to P'M' (B. I, P. XXni.) : hence, the sine of an arc is equal to the sine of its supplement. 37. The cosine of an arc is the sine of the comple- ment of the arc, " complement sine " being contracted into cosine. Thus, NM is the cosine of AM, and NM' is the cosine of AM'. These lines are respectively equal to OP and OP'. It is evident, from the equal triangles ONM and ONM', that NM is equal to NM'; hence, the cosine of an arc is equal to the cosine of its supplement. 38. The tangent of an arc is the perpendicular to the radius at one extremity of the arc, limited by the pro- longation of the diameter drawn to the other extremity. 20 PLANE TRIGONOMETRY. Thus, AT is the tangent of the arc AM, and AT'" is the tan- gent of the arc AM'. If AM is equal to M'C, AM and AM' are supplements of each other. But AM'" and AM' are also supplements of each other : hence, the arc AM is equal to the arc AM'", and the correspond- ing angles, AOM and AOM'", are also equal. The right-angled triangles AOT and AOT'" have a common base AO, and the angles at the base equal; consequently, the remaining parts are respectively equal : hence, AT is equal to AT'". But AT is the tangent of AM, and AT'" is the tangent of AM' : hence, the tangent of an arc is equal to the tangent of its supplement. 29. The cotangent of an arc is the tangent of its complement, "complement tangent" being contracted into cotangent. Thus, BT' is the cotangent of the arc AM, and BT" is the cotangent of the arc AM'. It is evident, from the equal triangles OBT' and OBT'', that BT' is equal to BT" ; hence, the cotangent of an arc- is equal to the cotangent of its supplement. When it is stated that the cotangent, tangent, &c., of an arc are equal respectively to the. cotangent, tangent, &c., of its supplement, the numerical values only, of the functions are referred to ; no account being taken of the algebraic signs ascribed to the several functions in the different quadrants, as will be explained hereafter. The sine, cosine, tangent, and cotangent of an arc, a, are, for convenience, written sin o, cos a, tan a, and cot a. PLANE TEIGONOMETEY. 21 These functions of an arc have been defined on the supposition that the radius of the arc is equal to 1 ; in this case, they may also be considered as functions of the angle which the arc measures. Thus,' PM, NM, AT, and BT', are respectively the sine, cosine, tangent, and cotangent of the angle AOM, as well as of the arc AM. 30. It is often convenient to use some other radius than 1 ; in such case, the functions of the arc to the radius 1, may be reduced to corresponding functions, to the radius R, R denoting any radius. Let AOM represent any angle, AM an arc described from as a centre with the radius 1, PM its sine ; A'M' an arc described from as a centre, with any radius R, and P'M' its sine. Then, because 0PM and OP'M' are similar triangles, we shall have, OM : PM : or, whence, PM PM = OM' : P'M', R : P'M'; P'M' and P'M' = PM x R; and similarly for each of the other functions : hence, Any function of an arc whose radius is 1, is equal to the corresponding function of an are whose radius is R divided by that radius. Also, any function of an arc whose radius is R, is equal to the corresponding function of an arc whose radius is 1 multiplied by the radius R. By means of this principle, formulas may be rendered homogeneous in terms of any radius, 22 PLANE TBIGONOMETBT. TABLE OF NATURAL SINES. 31. A Natubal Sine, Cosine, Tangent, or Cotangent, is the sine, cosine, tangent, or cotangent of an arc whose radius is 1. A Table of Natubal Sines, Cosines, &c., is a table by- means of which the natural sine, cosine, tangent, or co- tangent of any arc, or angle, may be found. Such a table might be used for all the purposes of trigonometrical computation, but it is usually found more convenient to employ a table of logarithmic sines, as ex- plained in the next article. TABLE OF LOGARITHMIC SINES. 33. A Logarithmic Sine, Cosine, Tangent, or Cotan- gent is the logarithm of the sine, cosine, tangent, or co- tangent of an arc whose radius is 10,000,000,000. This value of the radius is taken simply for convenience in making the table, its logarithm being 10. A Table of Logabithmic Sines is a table from which the logarithmic sine, cosine, tangent, or cotangent of any arc, or angle, may be found. Any logarithmic function of an arc, or angle, may be found by multiplying the corresponding natural function by 10,000,000,000 (Art. 30), and then taking the loga- rithm of the result; or more simply, by taking the loga- rithm of the corresponding natural function, and then adding 10 to the result (Art. 5). 33. In the table appended, the logarithmic functions are given for every minute from 0° up to 90°. In addi- tion, their rates of change for each second are given in the column headed "D."' PLANE TRIGONOMETRY. 28 The method of computing the numbers in the column .'leaded "D," will be understood from a single example. The logarithmic sines of 27° 34', and of 27° 35', are, respectively, 9.665375 and 9.665617. The difference be- tween their mantissas is 242 millionths; this, divided by 60, the number of seconds in one minute, gives 4.03 millionths, which is the change in the mantissa for 1"^ between the limits 27° 34' and 27° 35'. For the sine and cosine, there are separate columns of differences, which are written to the right of the respect- ive columns; but for the tangent and cotangent there is but a single column of differences, which is written be- tween them. The logarithm of the tangent increases just as fast as that of the cotangent decreases, and the re- verse, their sum being always equal to 20. The reason of this is, that the product of the tangent and cotangent is always equal to the square of the radius ; hence, the sum of their logarithms must always be equal to twice the logarithm of the radius, or 20. The arc, or angle, obtained by taking the degrees from the top of the page and the minutes from the Ze/f-hand column, is the complement of that obtained by taking the degrees from the bottom of the page, and the minutes from the right-hand column on the same horizontal line. But, by definition, the cosine and the cotangent of an arc, or angle, are, respectively, the sine and the tangent of the complement of that arc, or angle (Arts. 26 and 28): hence, the columns designated sine and tang at the top of the page, are designated cosine and cotang at the bottom. 24 PLANE TRIGOKOMETET. USE OF THE TABLE. To find the logarithmic functions of an arc, or angle, which is expressed in degrees and minutes. 34. If the arc, or angle, is less than 45°, look for the degrees at the top of the page, and for the minutes in the Ze/f-hand column ; then follow the corresponding hori- zontal line till you come to the column designated at the top by sine, cosine, tang, or cotang, as the case may be; the number there found is the logarithm required. Thus, logsinl9°55' • • ■ 9.532312 logtanl9°55' • • • 9.559097 If the arc, or angle, is 45° or more, look for the de- grees at the bottom of the page, and for the minutes in the right-hand column; then follow the correspondmg horizontal line backward till you come to the column designated at the bottom by sine, cosine, tang, or cotang, as the case may be ; the number there found is the loga- rithm required. Thus, log cos 52° 18' . . . 9.786416 logtan52°18' • • -10.111884 To find the logarithmic functions of an arc or angle which is expressed in degrees, minutes, and seconds. 35. Find the logarithm corresponding to the. degrees and minutes as before ; then multiply the corresponding number taken from the column headed "D," which is millionths, by the number of seconds, and add the prod- uct to the preceding result for the sine or tangent, and subtract it therefrom for the cosine or cotangent. PLANE TRIGONOMETRY. 25 Examples. 1. Find the logaritlimic sine of 40° 26' 28". Operation. log sin 40° 26' 9.811952 Tabular difference 2.47 No. of seconds 28 Product • • • 69.16 to be added • • 69 log sin 40° 26' 28" 9.812021 The same rule is followed for decimal parts, as in Art. 12. 2. Find the logarithmic cosine of 53° 40' 40". Operation. log cos 53° 40' 9.772675 Tabular difference 2.86 ISTo. of seconds 40 Product • • • 114.40 to be subtracted 114 log cos 53° 40' 40" 9.772561 If the arc or angle is greater than 90°, find the re- C[uired function of its supplement (Arts. 26 and 28). 3. Find the logarithmic tangent of 118° 18' 25". Operation. 180° Given arc 118° 18' 25" Supplement 61° 41' 35" log tan 61° 41' 10.268556 Tabular difference 5.04 No. of seconds 35 Product • • • 176.40 to be added 176 logtanll8° 18'25" 10.268732 26 PLANE TKIGONOMETRT. 4. Find the logarithmic sine of 32° 18' 35". Ans. 9.727945. 5. Find the logarithmic cosine of 95° 18' 24". Am. 8.966080. 6. Find the logarithmic cotangent of 125° 23' 50". Ans. 9.851619. To find the arc or angle corresponding to any logarithmie function. 36. This is done by reversing the preceding rule : Look in the proper column of the table for the given logarithm; if it is found there, the degrees are to be taken from the top or bottom, and the minutes from the left or right hand column, as the case may be. If the given logarithm is not found in the table, then find the next less logarithm, and take from the table the corre- sponding degrees and minutes, and set them aside. Sub- tract the logarithm found in the table from the given logarithm, and divide the remainder by the corresponding tabular difference. The quotient will be seconds, which must be added to the degrees and minutes set aside in the case of a sine or tangent, and subtracted in the case of a cosine or a cotangent. Examples. 1. Find the arc or angle corresponding to the loga- rithmic sine 9.422248. Operation. Given logarithm • • • 9.422248 Next less in table • • • 9.421857 • ■ • 15° 19' Tabular difference 7.68 ) 391.00 (51", to be added. Hence, the required arc is 15° 19' 51". PLANE TKIGONOMETR Y. 27 2. Find the arc or angle corresponding to the loga- rithmic cosine 9.427485. Operation. Given logarithm • • • 9.427485 Next less in table • • • 9.427354 • ■ • 74° 29' Tabular difference 7.58) 131.00 ( 17", to be subt. Hence, the required arc is 74° 28' 43". 8. Find the arc or angle corresponding to the logar rithmic sine 9.880054. Ans. 49° 20' 50". 4. Find the arc or angle corresponding to the loga- rithmic cotangent 10.008688. Ans. 44° 25' 37". 5. Find the arc or angle corresponding to the loga- rithmic cosine 9.944599. Ans. 28° 19' 45'. SOLUTION OF RIGHT-ANGLED TRDLNGLES. 37. In what follows, the three angles of every triangle are designated by the capital letters A, B, and C, A de- noting the right angle ; and the sides lying opposite the angles by the corresponding small letters a, b, and c. Since the order in which these letters are placed may be changed, without affecting the demonstration, it follows that whatever is proved with the letters placed in any given order, will be equally true when the letters are cor- respondingly placed in any other order. Let CAB represent any triangle, right- angled at A. With C as a centre, and a radius CD, equal to 1, describe the arc DG, and draw GF and DE perpen- C F D dicular to CA : then will FG be the sine of the angle C, CF will be its cosine, and DE its tangent. 28 PLANE TRIGONOMETBT. Since the three triangles CFG, CDE, and CAB are simi- lar (B. IV., P. XVni.), we may write the proportions, CB : AB : : CG : FG, or. a : c : : 1 : sin C, CB : CA : : CG : CF, or. a : b ■ : 1 : cos C, CA : lence. AB : : CD we have (B. : DE, or, II., P. I.), b : c : : 1 : tanC; c = a sin C • • • {l.)~ ' sinC c ~ a' • • • (4.) b = a cos C • • • (2.) • ,', •■ cos C b ~ a' • • • (5.) c — 6 tan C • • • (3.) tan C c ~ b' • • • (6.) Translating these formulas into ordinary language, we have the following PRINCIPLES. 1. The perpendicular of any right-angled triangle is equal to the hypothenuse multiplied by the sine of the angle at the base. 2. ^e base is equal to the hypothenuse multiplied by the cosine of the angle at the base. 3. 27fce perpendicular is equal to the base multiplied by the tangent of the angle at the base. 4. The sine of the angle at the base is equal to the perpendicular divided by the hypothenuse. 5. The cosine of the angle at the base is equal to the base divided by the hypothenuse. 6. The tangent of the angle at the base is equal to the perpendicular divided by the base. PLANE TRIGONOMETRY. 29 Either side about the right angle may be regarded as the base ; the other is then to be taken as th6 perpen- dicular. B may be substituted for C in the formulas, provided that, at the same time, b is substituted for c, and c for b: from (4), (5), (6), we may thus obtain, sin B = ^, (4'.) cosB = ^, (5'.) tan B = - (6'.) From the relations shown in (4), (5), (6), (4'), (5'), (6'), the natural functions of the acute angles of a right-angled triangle are sometimes defined as ratios: thus, of either of such angles, the sine is the ratio of the hypothenuse to the side opposite; the cosine is the ratio of the hypothenuse to the side adjacent; the tangent is the ratio of the side adjacent to the side opposite. Formulas (1) to (6) are sufficient for the solution of every case of right-angled triangles. They are in proper form for use with a table of natural functions: when a table of logarithmic functions is used, as is done in this book, they must be made homogeneous in terms of R, R being equal to 10,000,000,000, as stated in Art. 32. The formulas may be made homogeneous by the principle of Art. 30 ; thus, for example, the second member of (4), being the value of sin C when the radius is 1 , must be multiplied by R for the value of sin C when the radius is R, giving 'iO PLANE TE160N0METBY. . _ Rc sm C = — ; a whence, by solving witli reference to c, a sin C c = R In like manner, the remaining formulas may be made homogeneous, giving a sin C ,_ , . _ Re ,^ „ , c = — r> — • • • (7.) sm C = — • • • (10.) R a cos C R (7.) . - Re sm C =- (8.) R6 cos C = — a (9.) . - Re tan C = -^ (11.) c = ^-^- • • • (9.) tanC = ^" • • ■ (12.) In applying logarithms to these formulas, care must be taken to observe the principles of logarithms (Arts. 5 and 6), giving, for example (as logarithm of R is 10), log c = log a + log sin C — 10, log sin C = log c + 1 — log a = log c + (a. c.) log a (see Art. 11) ; &c. In solving right-angled triangles, four cases arise : CASE I. Given the hypothenuse and one of the acute angles, to find the remaining parts. 38. The other acute angle may be found by subtracting the given one from 90° (Art. 23). The sides about the right angle may be found by formulas (7) and (8). PLANE TRIGONOMETRY. 31 Examples. 1. Given a = 749, and C = 47° 03' 10"; required B, c, and h. Operation. B = 90° - 47° 03' 10" = 42° 56' 50". Applying logarithms to formula (7), we have, log c = log a + log sin C — 1 ; log a (749) .... 2.874482 log sin C (47° 03' 10") • 9.864501 logc 2.738983 .'. c = 548.255. [The 10 is subtracted mentally.] Applying logarithms to formula (8), we have, log h = log a + log cos C — 1 ; log a (749) . ■ ■ . 2.874482 log cos C (47° 03' 10") • 9.833354 log 6 2.707836 .-. 6 = 510.31. Ans. B = 42° 56' 50", b = 510.31, and c = 548.255. 2. Given a = 439, and B = 27° 38' 50", to find C, c, and b. Ans. C = 62° 21' 10", 6 = 203.708, and c = 388.875. 8. Given a = 125.7 yds., and B = 75° 12', to find the other parts. Ans. C = 14° 48', b ^ 121.53 yds., and c = 32.11 yds. 4. Given a = 7.521 ft., and C = 57° 34' 48", to find the other parts. Ans. B = 32° 25' 12", c = 6.348 ft., b = 4.082 ft. 82 PLANE TBIGONOMETBY. CASE n. Given one of the sides about the right angle and one of the acute angles, to find the remaining parts. 39. The other acute angle may be found by subtract- ing the given one from 90°. The hypothenuse may be found by formula (7), and the unknovm side about the right angle by formula (8). Examples. 1. Given c = 56.293, and C = 54° 27' 39", to find B, a, and h. Operation. B = 90° - 54° 27' 39" = 35° 32' 21". Applying logarithms to formula (7), we have log a = log c + 10 — log sin C ; but, 10 — log sin C = (a. c.) of log sin C ; whence, logc (56.293) . . • 1.750454 (a. c.) log sin C (54° 27' 39") • 0.089527 log a 1.839981 .-. a = 69.18 Applying logarithms to formula (8), we have log 6 = log o + log cos C — 10 ; logo (69.18) • • • 1.839981 log cos C (54° 27' 39") • 9.764370 log 6 • 1.604351 .-. 6 = 40.2114. Ans. B = 35° 32' 21", a = 69.18, and b = 40.2114. 2. Given c = 358, and B = 28° 47', to find C, a, and 6. Ans. C = 61° 13', a = 408.466, and b = 196.676. PLANE TRIGONOMETRY. 33 3. Given 6 = 152.67 yds., and C = 50° 18' 32", to find the other parts. Ans. B = 39° 41' 28", c = 183.95, and a = 239.05. 4. Given c = 379.628, and C = 39° 26' 16", to find B, a, and b. Ans. B = 50° 33' 44", a = 597.613, and b = 461.55, CASE m. Oiven the two sides about the right angle, to find the r«- maining parts. 40. The angle at the base may be found by formtda (12), and the solution may be completed as in Case II. Examples. 1. Given 6 = 26, and c = 15, to find C, B, and a. Operation. Applying logarithms to formula (12), we have log tan C = log c + 1 — log b ; logc (15) .... 1.176091 (a. c.)log& (26) .... 8.585027 log tan C . . 9.761118 .-. C = 29° 58' 54". [From Art. 28, it is evident that log tan C here found corresponds to two angles, viz., 29° 58' 54", and 180° — 29° 58' 54", or 150° 1' 6". As, however, the triangle is right-angled, the angle C is acute, and the smaller value must be taken.] B = 90° - C = 60° 01' 06". 34 PLANE TEIGONOMETKY. As in Case 11., log a = log c + 1 — log sin C ; logo • • ■ (15) • • 1.176091 (a. c.) log s.in C (29° 58' 54") 0.301271 logo 1.477362 .-. a = 30.017. Ans. C = 29° 58' 54", B = 60° 01' 06", and a = 30.017. 2. Given b = 1052 yds., and c = 347.21 yds., to find B, C, and a. B = 71° 44' 05", C = 18° 15' 55", and a = 1107.82 yds. 3. Given b - 122.416, and c = 118.297, to find B, C, and a. B =: 45° 58' 50", C = 44° 1' 10", and a = 170.235. 4. Given b = 103, and c = 101, to find B, C, and a. B = 45° 83' 42", C = 44° 26' 18", and a = 144.256. CASE IV. Given the hypothenuse and either side about the right angle, to find the remaining -parts. 41. The angle at the base may be found by one of formulas (10) and 11), and the remaining side may then be found by one of formulas (7) and (8). Examples. 1. Given a = 2391.76, and b = 385.7, to fiiid C, B, and c. Operation. Applying logarithms to formula (11), we have log cos C = log 6 + 10 — log a ; PLANE TEIGONOMETBY. 35 log 6 (385.7) . . • 2.586250 (a. c.) log a (2391.76) ■ • 6.621282 log cos C • • • 9.207582 .'. C = 80" 43' 11"; B = 90° — 80° 43' 11" = 9° 16' 49". From formula (7), we have log c = log a + log sin C — 10 ; log a (2391.76) • 3.378718 log sin C (80° 43' 1 1") 9.994278 logc 3.372996 .". c = 2360.45. Ans. B = 9° 16' 49", C = 80° 43' 11", and c = 2360.45. 2. Given a = 127.174 yds., and c = 125.7 yds., to find C, B, and 6. Operation, From formula (10), we have log sin C = log c + 1 — log a ; logc (125.7) • • • 2.099335 (a. c.) log a (127.174) • • 7.895602 log sin C • . • 9.994987 .-. 0=81° 16' 6"; B = 90° - 81° 16' 6" = 8° 43' 54". From formula (8), we have log b = log a + log cos C — 10 ; logo (127.174) • • 2.104398 log cos C (81° 16' 6") • 9.181292 log 6 1.285690 .-. b = 19.8. Ans. B = 8° 43' 54", C = 81° 16' 6", and & = 19.8 yds. 36 PLANE TBIGONOM.ETKY. 3. Given a = 100, and 6 = 60, to find B, C, and c. B = 36° 52' 11", C = 53° 7' 49", and c = 80. 4, Given a = 19.209, and c = 15, to find B, C, and b. Ans. B = 38° 39' 30", C = 51° 20' 30", b = 12. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 42. In tlie solution of oblique-angled triangles, four cases may arise. We shaU discuss these cases in order. CASE 1. Given one side and two angles, to determine the remaining parts. 43. Let ABC represent any oblique-angled triangle. Prom the vertex C, draw CD perpendicular to the base, forming two right- angled triangles ACD and BCD. Assume the notation of the figure. From formula (1), we have CD = 6 sin A, CD = a sin B. Equating these two values, we have, 6 sin A = a sin B ; whence (B. IL, P. IL), a -. 6 : : sin A : sin B. ... (13.) Since a and b are any two sides, and A and B the angles lying opposite to them, we have the following principle : PLANE TKIGONOMETBT. 37 The sides of a plane triangle are proportional to the tines of their opposite angles. It is to be observed that formula (13) is true for any value of the radius. Hence, to solve a triangle, when a side and two angles are given : First find the third angle, by subtracting the sum of the given angles from 180°; then find each of the re- quired sides by means of the principle just demonstrated. Examples. 1. Given B = 58° 07', C = 22° 37', and a = 408, to find A, b, and c. Operation. B 58° 07' C 22° 37' A . . . 180° - 80° 44' = 99° 16'. To find b, write the proportion, sin A : sin B : : a : 6 ; that is, the sine of the angle opposite the given side, is to the sine of the angle opposite the required side, as the given side is to the required side. Applying logarithms, we have (Ex. 4, P. 15) log 6 = (a. c.) log sin A + log sin B + log a — 10 ; (a c.) log sin A (99° 16') • • • 0.005705 log sin B (58° 07') • • • 9.928972 log a (408) .... 2.610660 log 6 2.545337 .'.6 = 351.024. In like manner, sin A : sin C : : a : c; 38 PLANE TRIQONOMETET. and log c = (a. c.) log sin A + log sin C + log a — 10 ; (a. c.) log sin A (99° 16') • • • 0.005705' log sin C (22" 37') • • • 9.584968 log a (408) .... 2.610660 logc 2.201333 .-.0=158.976. Ans. A = 99° 16', 6 = 851.024, and c = 158.976. 2. Given A = 38° 25', B = 57° 42', and c = 400, to find C, a, and b. Ans. C = 83° 53', a = 249.974, b = 340.04. 3. Given A = 15° 19' 51", C = 72° 44' 05", and c = 250.4 yds., to find B, a, and 6. Ans. B = 91° 56' 04", a = 69.328 yds., b = 262.066 yds. 4. Given B = 51° 15' 35", C = 37° 21' 25", and a = 305.296 ft., to find A, b, and c. Ans. A = 91° 23', 6 = 238.1978 ft., c = 185.3 ft. CASE n. Given two sides and an angle opposite one of them, to find the remaining -parts. 44. The solution, in this case, is commenced by find- ing a second angle by means of formula (13), after which we may proceed as in Case I. ; or, the solution may be completed by a continued application of formula (13). Examples. 1. Given A = 22° 37', b = 216, and a = 111, to find B, C, and c. PLANE TRIGONOMETRY. 39 From formula (13), we have a : b : : sin A : sin B ; that is, the side opposite the given angle, is to the side opposite the reqitired angle, as the sine of the given angle is to the sine of the required angle. Whence, by the apphcation of logarithms, log sin B = (a. c.) log a + log b + log sin A — 1 ; (a. 0.) log a • (117) • • 7.981814 log 6 • (216) ■ • 2.334454 log sin A (22° 37') • 9.584968 log sin B • • . 9.851236 .-. 8= 45° 13' 55", and B'= 134° 46' 05". Hence, we find two values of B, which are supplements of each other, because the sine of any angle is equal to the sine of its supplement. This would seem to indicate that the problem admits of two solutions. It now remains to determine under what conditions there will be two so- lutions, one solution, or no solution. There may be two cases: the given angle may be acute, or it may be obtuse. Eepresent the given parts of the triangle by A, a, b. The particular letters employed are of no consequence in the discussion, and, therefore, in the results, C or B may be substituted for A, provided that, at the same time, like changes are made in the corresponding small letters. 40 PLANE TBIGONOMETH Y. 1st Case: A < 90°. Let ABC represent the triangle, in which the angle A and the sides a and 6 are given. From C let faU a perpendicular upon AB, prolonged if necessary, and denote its length by jp. We shall have, from formula (1), Art. 37, b sin A P = -R-; from which the value of p may be computed. If a is greater than p and less than b, there will be two solutions. For, if with C as a centre, and a as a radius, an arc be described, it will cut the line AB in two points, B and B', each of which being joined with C, will give a triangle, and we shall thus have two triangles, ABC and AB'C, which will conform to the conditions of the problem. In this case, the angles B' and B, of the two triangles AB'C and ABC, will be supplements of each other. If a = p, there will be but one solution. For, in this case, the arc will be tangent to AB, the two points B and B' will unite, and there will be but one triangle formed. In this case, the angle ABC will be equal to 90°. If a is greater than both p and 6, there will also be but one solution. For, although the arc cuts AB in two points, and consequently gives two triangles, only one of them, ABC, con- forms to the conditions of the problem. PLANE TRIGONOMETRY. 41 In this cade, the angle ABC will be less than A and consequently acute. C If a 90°. When the given angle A is obtuse, the angle ABC will be acute ; the side a will be greater than 6, and there will be but one solution. (See B. III., Prob. XI., S.) In the example under considera- tion, there are two solutions, the first corresponding to B = 45° 13' 55", and the second to B' = 134° 46' 05". In the first case, we have A 22° 37' B 45° 13' 55" C . . . 180° - 67° 50' 55" = 112° 09' 05". To find c, we have sin B : sin C : : b : c ; and log c = (a. c.) log sin B + log sin C 4- log 6 — 10; (a. c.) log sin B (45° 13' 55") • 0.148764 log sin 0(112° 09' 05") • 9.966700 log 6 • (216) 2.334454 logc 2.449918 .-.0=281.785, Ans. B = 45° 13' 55", C = 112° 09' 05", and c = 281.785. 42 PLANE TRIGONOMETBT. In the second case, we have, A 22° 37' B' 134° 46' 05" C . . . 180° - 157° 23' 05" = 22° 36' 55"; and as before, (a.c.)logsm B' (134° 46' 05") • 0.148764 log sin C (22° 36' 55") • 9.584943 log 6 • • • (216) • • 2.334454 logc' 2.068161 .-. c' = 116.993, Ans. B' = 134° 46' 05", C = 22° 36' 55", and c' = 116.993. 2. Qiven A = 32°, a = 40, and b = 50, to find B, C, and c. B= 41° 28' 59", C = 106° 31' 01", c = 72.368. Ans. B' = 138° 31' 01", C = 9° 28' 59", c' = 12.436. 3. Given B = 18° 52' 13", 6 = 27.465 yds., and a 13.189 yds., to find A, C, and c. Ans. A = 8° 56' 05", C = 152° 11' 42", c = 39.611 yds. 4. Given C = 32° 15' 26", b = 176.21 ft., and c 94.047 ft., to find B, A, and a. Ans. B = 90°, A = 57° 44' 34", a = 149.014 ft PLANE TBIGONOMETKY, 43 CASE III. Given two sides and their included angle, to find the re- maining parts. 45. The solution, in this case, is begun by finding the half sum and the half difference of the two required angles. The half sum of these angles may be found by subtracting the given angle from 180°, and dividing the remainder by 2 ; the half difference may be found by means of the following principle, now to be demonstrated, viz. : In any plane triangle, the sum of the sides including any angle, is to their difference, as the tangent of half the sum of the two other angles, is to the tangent of half their difference. Let ABC represent any plane tri- angle, c and h any two sides, and A their included angle. Then we are to show that c + 6 : c — h : : tan i (C + B) With A as a centre, and &, the shorter of the two sides, as a radius, describe a semicircle meeting AB in I, and the prolongation of AB in E. Draw EC and CI, and through I draw IH parallel to EC. Since the angle EC I is inscribed in a semicircle, it is a right angle (B. III., P. XVIII., C. 2) ; hence, EC is perpendicular to CI, at the point C; and since IH is parallel to EC, it is also perpen- dicular to CI. The inscribed angle CIE is half the angle at the centre, CAE, intercepting the same arc CE. Since the 44 PLAKE TBIGONOMETBY. angle CAE is exterior to the triangle ABC, we have (B. I., P. XXV., C. 6), CAE = C + B ; ■ hence, CIE = * (C + B). AC and AF, being radii of the same circle, are equal to each other, and therefore (B. L, P. XL), the angle AFC is equal to the angle C ; but the angle AFC is ex- terior to the triangle FBA, and hence we have AFC or C = FAB + B ; hence, FAB = C — B. But the inscribed angle, ICH, is half the angle at the centre, FAB, intercepting the same arc Fl ; hence, ICH = iiC- B). From the two right-angled triangles ICE and ICH, we have (formula 3, Art. 37), EC = IC tan CIE = IC tan i (C + B> and IH = IC tan ICH = IC tan i (C — B) ; hence, we have, after omitting the equal factor IC (B. IL, p. vn.), EC : IH : : tan i (C + B) : tan i (C - B), The triangles ECB and IHB being similar (B, IV., P. XXI.), PLANE TEiaONOMETRT. 46 EC : IH : : EB : IB, or, since EB = c + b, and IB = c — 6, EC : IH : : c + b : g - b. Combining the preceding proportions, we have c + b : c — b : : tan |(C + B) : tan i (C - B) ; • (14.) which was to be proved. By means of (14), the half difference of the two re- quired angles may be found. Knowing the half sum and the half difference, the greater angle is found by adding the half difference to the half sum, and the less angle is found by subtracting the half difference from the half sum. Then the solution is completed as in Case L Examples. 1. Qiven c = 540, b = 450, and A = 80°, to find B C, and a. Operation. e + b = 990; c — b = 90; 1(0 + B) = i(180°- 80«) = 50°. Applying logarithms to formula (14), we have 46 PLANE TRIGONOMETRY. log tan i (C — B) = (a. c.) log (c + 6) + log (c — 6) + logtaiii(C + B) - 10, (a. c.) log (c + 6) • . (990) 7.004365 log (c - 6) . . (90) 1.954243 log tan i (C + B) (50°) 10.076187 log tan i (C - B) 9.034795 .-. i(C - B) = 6" 11'. C = 50° + 6° 11' = 56° 11'; B = 50° - 6^ 11' = 43° 49'. From formula (13), we have sin C : sin A : : e : a ; whence, (a. c.) log sin C (56° 11') • 0.080492 log sin A (80°) • • 9.993351 logc • (540) • • 2.732394 logo 2.806237 .-. a = (540.082. Ans. B = 43° 49', C = 56° 11', a = 640.082. 2. Given c= 1686 yds., 6 = 960 yds., and A =128° 04', to find B, C, and a. Ans. B = 18° 21' 21", C = 33° 34' 39", a = 2400 yds. 3. Given a = 18.739 yds., c = 7.642 yds., and B = 45° 18' 28", to find A, 6, and C. Ans. A = 112° 34' 13", C = 22° 07' 19", & = 14.426 yds. PLANE TBIGONOMETBT. 47 4. aiven a = 464.7 yds., b = 289.3 yds., and C = 87° 03' 48", to find A, B, and c. Ans. A = 60° 13' 39", B = 32° 42' 33", c = 684.66 yds. 5. Given a = 16.9584 ft, 6 = 11.9613 ft., and C = 60° 43' 36", to find A, B, and c. Ans. A = 76° 04' 12", B = 43° 12' 12", c = 15.22 ft. 6. Given a = 3754, 6=3277.628, and C = 57°53'17", to find A, B, and c. Ans. A = 68° 02' 25", B = 54° 04' 18", c = 3428.512. CASE rv. Given the three sides of a triangle, to find, the reiwdining parts.* 46. Let ABC represent any A plane triangle, of -which BC is the y longest side. Draw AD perpendic- t/ ular to the base, dividing it into y/^ two segments CD and BD. ^ ^ [The longest side is taken as the base, to make it certain that the perpendicular from the vertex shall fall on the base, and not on the base produced.'\ From the right-angled triangles CAD and BAD, we have AD* = AC* - DC', and AD^ = AB' - BD'. * The angles may be found by formula (A) or (B), Iiemma, Art. 97, Mensu- ration. 48 PLANE TEIGONOMETRT. Equating these values of AD^, we have, y X)/ AC' — DC" = AB' - BD* ; X / 3 whence, by transposition, c D B AC' - AB' = DC' - BD'. Hence (B. IV., P. X), we have (AC + AB) (AC - AB) = (DC + BD) (DC - BD). Converting this equation into a proportion (B. II., P. II.), we have DC + BD : AC + AB : : AC - AB : DC - BD ; or, denoting the greater segment by s and the less seg- ment by s', and the sides of the triangle by a, 6, and c, s + s' : 6 + c : : 6 — c : s — s' ; • (15.) that is, if in any plane triangle, a line be drawn from the vertex perpendicular to the base, dividing it into two segments ; then, Th& sum of the two segments, or the whole base, is to the sum of the two other sides, as the difference of these sides is to the difference of the segments. The half difference of the segments added to the half sum gives the greater segment, and the half difference subtracted from the half sum gives the less segment. [The greater segment is, of course, adjacent to the greater side.] We shall then have two right-angled triangles, in each of which we know the hypothenuse and the base ; PLANE TEIGONOMETBY. 49 hence, the angles of these triangles may be found, and consequently, those of the given triangle. Examples. 1. Q-iven a = 40, b = 34, and c = 25, to find A, B, and C. Operation. Applying logarithms to formula (15), we have log (s - s') = (a. c.) log {s + s') + log (6+c) + log (6-c)-10 ; (a. c.) log (s + s') • • (40) • • 8.397940 log(6 + c) • ■ (59) • • 1.770852 log(6-c) • • (9) • • 0.954243 log(s-s') .... 1.123035 ■•. s-g' = 13.275. s = i (s -r s') + i (s - s') = 26.6375. s' z= i{s + s') -i{s- s') = 13.3625. From, formula (11), we find log cos C = logs + (a. c.) log 6 .-. C = 38° 25' 20", and log cos B = log s' + (a. c.) log c .-. B = 57° 41' 25" 96° 06' 45" A = 180° - 96° 06' 45" = 83° 53' 15". 2. Given a = 6, & = 5, and c = 4, to find A, B and C. Ans. A = 82° 49' 09", B = 55° 46' 16", C = 41° 24' 85". 3. Given a = 71.2 yds., & = 64.8 yds., and c = 37 ydfl., to find A, B, and C. Ans. A = 84° 01' 53", B = 64° 60' 51", C = 31° 07' 16". 50 PLANE TRIGONOMETBY. PROBLEMS. 1. Knowing the distance AB, equal to 600 yards, and the angles BAG = 57° 35', ABC = 64° 51', find the two distances AC and BC. Ans. AC BC 643.49 yds., 600.11 yds. 2. At what horizontal distance from a column, 200 feet high, will it subtend an angle of 31° 17' 12"? Ans. 329.114 ft. 3. Required the height of a hill D above a hori- zontal plane AB, the dis- tance between A and B being equal to 975 yards, and the angles of elevation at A and B being respectively 15° 36' and 27° 29'. Ans. DC = 587.61 yds. 4. The distances AC and BC are found by measurement to be respects ively, 588 feet and 672 feet, and their included angle 55° 40'. Required the distance AB. Ans. 592.967 ft. 5. Being on a horizontal plane, and wanting to ascer- tain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevar tion of the top of the hill 40°, and of the top of the tower 51°; then measuring in a direct line 180 feet PLANE TEIGONOMETBT. 61 farther from the hill, the angle of elevation of the top of the tower was 33° 45'; required the height of the tower. Ans. 83.998 ft. 6. Wanting to know the horizontal distance between two inaccessible objects E and W, the following measurements were made : VIZ. AB BAW WAE ABE EBW 536 yards 40° 16' 57° 40' 42° 22' 71° 07' Required the distance EW. Ans. 989.617 yds. 7. Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points C and D were chosen at a distance from each other equal to 200 yards; from the former of these points, A could be seen, and from the latter, B ; and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D, a distance DE equal to 200 yards, and the following angles taken: AFC = 83° 00', BDE = 54° 30', ACD = 53° 30', BDC = 156° 25', ACF = 54° 31', BED = 88° 30'. Required the distance AB. Ans. 845.459 yds. 62 PLANE TRIGONOMETRY. 8. The distances AB, AC, and BC, between the points A, B, and C, are known ; viz. : AB = 800 yds., AC = 600 yds., and BC = 400 yds. From a fourth point P, the angles A PC and BPC are measured ; viz. : APC = 33° 45', and BPC = 22° 30'. Required the distances AP, BP, and CP. C AP Ans. } BP ( CP 710.198 yds. 934.289 yds. 1042.524 yds. This problem is used in locating the position of buoys in maritime surveying, as follows. Three points. A, B, and C, on shore are known in position. The surveyor sta- tioned at a buoy P, measures the angles APC and BPC. The distances AP, BP, and CP, are then found as follows : Suppose the circumference of a circle to be described through the points A, B, and P. Draw CP, cutting the circumference in D, and draw the lines DB and DA. The angles CPB and DAB, being inscribed in the same segment, are equal (B. III., P. X'Viil., C. 1) ; for a like reason, the angles CPA and DBA are equal: hence, in the triangle ADB, we know two angles and one side ; we may, therefore, find the side DB. In the triangle ACB, we know the three sides, and we may compute the angle B. Sub- tracting from this the angle DBA, we have the angle DBC. Now, in the triangle DBC, we have two sides and their included angle, and we can find the angle DCB. Finally, in the triangle CPB, we have two angles and one side, from which data we can find CP and BP. In like man- ner, we can find A P. ANALYTICAL TRIGONOMETRY. 47. Ajstalytioal Teigonometby is that branch of Mathe- matics which treats of the general properties and relations of trigonometrical functions. DEFINITIONS AND GENERAL PRINCIPLES. 48. Let ABCD represent a circle whose radius is 1, and suppose its circumference to be divided into four equal parts, by the diameters AC and BD drawn perpendicular to each other. The horizontal diameter AC is called the initial diameter ; the vertical diameter BD is called the. secondary diameter; the point A, from which arcs are usually reckoned, is called the origin of arcs, and the point B, 90° distant, is called the secondary origin. Arcs estimated from A, around toward B, that is, in a direc- tion contrary to that of the motion of the hands of a watch, are considered positive; consequently, those reck- oned in a contrary direction must be regarded as negative. The arc AB, is called the first quadrant j the arc BC, the second quadrant; the arc CD, the third quadrant; and the arc DA, the fourth quadrant. The point at which 54 ANALYTICAL an arc terminates, is called its extremity, and an arc is said to be in that quadrant in which its extremity is situated. Thus, the arc AM is in the first quadrant, the arc AM' in the sec- ond, the arc AM" in the third, and the arc AM'" in the fourth. 49. The complement of an arc has been defined to be the difference be- tween that arc and 90° (Art. 23); geometrically considered, the comple- ment of an arc is the arc included between the extremity of the arc and the secondary origin. Thus, MB is the complement of AM ; M'B, the complement of AM' ; M"B, the complement of AM", and so on. When the arc is greater than a quadrant, the complement is negative, according to the conventional principle agreed upon (Art. 48). The supplement of an arc has been defined to be the difference between that arc and 180° (Art. 24); geomet- rically considered, it is the arc included between the ex- tremity of the arc and the left-hand extremity of the initial diameter. Thus, MC is the supplement of AM, and M"C the supplement of AM". The supplement is negative, when the arc is greater than two quadrants. 50. The sine of an arc is the distance from the initial diameter to the extremity of the arc. Thus, PM is the sine of AM, and P"M " is the sine of the arc AM". The term distance is used in the sense of shortest or perpendicular distance. TEIGONOMETR Y. 66 51. The cosine of an arc is the distance from the sec- ondary diameter to the extremity of the arc: thus, NM is the cosine of AM, and N'M' is the cosine of AM'. The cosine may be measured on the initial diameter: thus, OP is equal to the cosine of AM, and OP' to the cosine of AM' ; that is, the cosine of an arc is equal to the distance, measured on the initial diameter, from the cen- tre of the arc to the foot of the sine. 53. The versed-sine of an arc is the distance from the sine to the origin of arcs: thus, PA is the versed-sine of AM, and P'A is the versed-sine of AM'. 53. The co-versed-sine of an arc is the distance from the cosine to the secondary origin: thus, NB is the co- versed-sine of AM, and N"B is the co-versed-sine of AM". 54. The tangent of an arc is that part of a perpen- dicular to the initial diameter, at the origin of arcs, in- cluded between the origin and the prolongation of the diam- eter drawn to the extremity of the arc: thus, AT is the tangent of AM, or of AM", and AT'" is the tangent of AM', or of AM'". 55. The cotangent of an arc is that part of a perpen- dicular to the secondary diameter, at the secondary origin, included between the secondary origin and the prolongation of the diameter drawn to the extremity of the arc: thus, BT' is the cotangent of AM, or of AM", and BT" is the co- tangent of AM , or of AM'". 56. The secant of an arc is the distance from the cen- tre of the arc to the extremity of the tangent: thus, OT is the secant of AM, or of AM", and OT'" is the secant of AM', or of AM'". 56 ANALYTICAL 57. The cosecant of an arc is the distance from the centre of the arc to the extremity of the cotangent: thus, OT' is the cosecant of AM, or of AM", and OT" is the co- secant of AM', or of AM'". The prefix co, as used here, is equivalent to comple- ment; thus, the cosine of an arc is the "complement sine," that is, the sine of the complement, of that arc, and so on, as explained in Art. 27. The eight trigonometrical functions above defined are also called circular functions. RULES FOR DETERMINING THE ALGEBRAIC SIGNS OF CIROULAR FUNCTIONS. 58. All distances estimated upward are regarded as positive; consequently, all distances estimated downward must be considered negative. Thus, AT, PM, NB, P'M', are positive, and AT'", P"'M"', P"M", &c., are negative. AH distances estimated toward the right are regarded as positive ; consequently, aU distances esti- mated toward the left must be considered negative. Thus, NM, BT', PA, &c., are positive, and N'M', BT", &c., are negative. These two rules are sufficient for determining the alge- braic signs of all the circular functions, except the secant and cosecant. For the secant and cosecant, the following is the rule : All distances estimated from the centre in a direction toward the extremity of the arc are regarded as positive; TEIQONOMETRT. 57 consequently, all distances estimated in a direction away from the extremity of the arc must be considered negative. Thus, OT, regarded as the secant of AM, is estimated in a direction toward M, and is positive; but OT, regard- ed as the secant of AM", is estimated in a direction away from M", and is negative. These conventional rules enable us to give at once the proper sign to any function of an arc in any quadrant. 59. In accordance with the above rules, and the defi- nitions of the circular functions, we have the following principles : The sine is positive in the first and second quadrants, and negative in the third and fourth. The cosine is positive in the first and fourth quadrants, and negative in the second and third. The versed-sine and the co-versed-sine are always positive. The tangent and cotangent are positive in the first and third quadrants, and negative in the second and fourth. The sebant is positive in the first and fourth quadrants, and negative in the second and third. The cosecant is positive in the first and second quad- rants, and negative in the third and fourth. LIMITING VALUES OF THE CIEOULAR FUNCTIONS. 60. The limiting values of the circular functions are those values which they have at the beginning and the end of the different quadrants. Their numerical values are discovered by following them as the arc increases from 0° around to 360°, and so on around through 450°, 58 ANALYTICAL 540°, &c. The signs of these values are determined by the principle, that the sign of a varying magnitude up to the limit, is the sign at the liinit. For illustration, let us examine the limiting values of the sine and the tangent. If we suppose the arc to be 0, the sine will be ; as the arc increases, the sine increases until the arc becomes equal to 90°, when the sine becomes equal to +1, which is its greatest possible value ; as the arc increases from 90°, the sine diminishes until the arc becomes equal to 180°, when the sine becomes equal to + ; as the arc in- creases from 180°, the sine becomes negative, and increases numerically, but decreases algebraically, until the arc be- comes equal to 270°, when the sine becomes equal to — 1, which is its least algebraical value ; as the arc increases from 270°, the sine decreases numerically, but increases algebraically, until the arc becomes 360°, when the sine becomes equal to — 0. It is — 0, for this value of the arc, in accordance with the principle of limits. The tangent is when the arc is 0, and increases till the arc becomes 90°, when the tangent is +oo; in pass- ing through 90°, the tangent changes from -|- oo to — oo, and as the arc increases the tangent decreases numeric- ally, but increases algebraically, till the arc becomes equal to 180°, when the tangent becomes equal to — ; from 180° to 270° the tangent is again positive, and at 270° it becomes equal to +oo; from 270° to 360°, the tangent is again negative, and at 360° it becomes equal to -0. If we still suppose the arc to increase after reaching 360°, the functions will again go through the same changes, that is, the functions of an arc are the same as the functions of that arc increased by_ 360°, 720°, &c. By discussing the limiting values of all the circular functions we may form the following table : TRIGONOMETRY. 59 TABLE I. Aro = 0°. Arc = 90° Arc = 180°. Arc = 270°. Arc = 360°. sin = sin = 1 sin = Bin = -1 sin = -0 cos = 1 cos = cos = -1 COB = -0 cos = 1 T-sin = v-sin = 1 v-sin = 2 V-sin = 1 v-ain = co-v-sin = 1 oo-T-sin = co-T-aln = 1 oo-v-sin = 2 co-v-sin = 1 tan = tan = 00 tan = -0 tan = 00 tan = -0 cot = 00 cot = cot = — 00 cot = cot = —00 sec = 1 sec = 00 sec = -1 sec = -00 sec = 1 cosec = oo cosec = 1 cosec = 00 ooseo = -1 cosec 00 RELATIONS BETWEEN THE CIRCULAR FUNCTIONS OF ANY ARC. 61. Let AM, denoted by a, represent any arc whose radius is 1. Draw the lines as represented in the figure. Then we shall have, OM = OA = 1 ; PM = ON = sin a; NM = OP = cos a; PA = ver-sin a ; N B = co-ver-sin a ; AT = tan a ; BT' = cot a ; OT = sec a ; and OT' = cosec a. From the right-angled triangle OPM, we have, PM* + op' = OM', or, sin' « -f cos' a = : (1.) The symbols sin" a, cos' a, &c., denote the square of the sine of a, the square of the cosine of a, &c. From formula (1) we have, by transposition. sin' a = 1 — cos' a ; cos' a = 1 — sin' a. (2.) (3.) 60 ANALYTICAL We have, from the figure, PA = OA - OP, or, vernsin a = 1 — cos a ; • - . . . (4.) and, NB = OB - ON, or, co-ver-sin o = 1 — sin a. (5.) From the similar triangles OAT and 0PM, we have, OP : PM : : OA : AT, or, cos o : sin a : : 1 : tan a; whence, tan a = (6.) ' cos a ' From the similar triangles ONM and OBT', we have, ON : NM :: OB : BT', or, sin a : cos a : : 1 : cot a; , , cos a ,_ . whence, cot a = — (7.) ' sm a ' Multiplying (6) and (7), member by member, we have, tan a cot o = 1 ; (8.) whence, by division, tan a — — j— ; (9.) COu Of *^^ ^°*« = t^ ^1^-) From the similar triangles 0PM and OAT, we have, OP : OM : : OA : OT, or, cos a : 1 : : I : sec o ; whence, sec a = (11,) ' cos a ^ ' TBIGONOMETB Y. 61 From the similar triangles ONM and OBT', we have, ON : OM : : OB : OT', or, sin a : 1 : : 1 : cosec a ; 1 whence, ■ cosec a = sm a From the right-angled triangle OAT, we have. Of' = OA' + AT^ ; or, sec^ a = 1 + tan" a. ■ (12.) (13.) From the right-angled triangle OBT', we have. OT" 08' + BT ; or, cosec" a = 1 + cot« a. (14.) It is to be observed that formulas (5), (7), (12), and (14), may be deduced from formulas (4), (6), (11), and (13), by substituting 90° — a, for a, and then making the proper reductions. Collecting the preceding formulas, we have the follow- ing table : TABLE II. (1.) Bin' a + COB' a = 1. (9.) tana = 1 cot a (2.) sin' a = 1 — cos' a. 1 tana (3.) 008' a = 1 - sin' a. (10.) cot a — (4.) ver-sin a = 1 — cos a. (11.) sec a = 1 cos a (5.) co-ver-ain a = 1 - sin o. 1 sin a (6.) tana = sin a COB a (12.) cosec a ^ (7.) oot = cos a sin a (13.) sec' a = 1 + tan' a. (8.) tan a cot a = 1. (14.) cosec' a = 1 + oot* a. 62 ANALYTICAL FUNCTIONS OF NEGATIVE ARCS. 63. Let AM'", estimated from A toward D, be numeric- ally equal to AM ; then, if we denote the arc AM by a, the arc AM'" wUl be denoted by — a (Art. 48). A being the middle point of the arc M"'AM, the radius OA bisects the chord M"'M at right angles (B. III., P. VI.) ; therefore, PM'" is numeric- ally equal to PM, but PM'" being measured downward from the initial diameter is negative, while PM being measured upward is positive, and, therefore, PM'" = — PM ; OP is equal to the cosine of both AM"' and AM (Art. 61); hence, we have, sin (— a) = — sin a, (1.) cos (— a) = cos a. (2.) Dividing (1) by (2), member by member, and then divide ing (2) by (1), member by member, we have (formulas 6 and 7, Art. 61), tan (— a) = — tan (a) ; cot (— a) = — cot a. Taking the reciprocals of the members of (2), and then the reciprocals of the members of (1), we have (formulas 11 and 12, Art. 61), sec (— o) = sec a; cosec (—a) = — cosec a. TBIGONOMETE Y, 68 FUNCTIONS OF ARCS FORMED BY ADDING AN ARC TO, OE SUBTRACTING IT FROM, ANY NUMBER OF QUADRANTS. 63. Let a denote any arc less than 90°. By definition, we have, cos (90° — a) = sin a. cot (90° — a) = tan a. cosec (90° — a) = sec a. then B M>^ N~~^\ /!\ U / :\ .'i V : ■■■\ ..■- \ A---+ ; 1 p' " ■ 7 sin (90° — a) = cos a; tan (90° — a) = cot a ; sec (90° — o) = cosec a; Let the arc BM' = AM = a ; AM' = 90° + a. Draw hnes, as in the figure. Then PM = sin a ; OP = cos a ; ON = P'M' = sin (90° + a) ; NM' = cos (90° + a). The right-angled triangles ONM' and 0PM have the angles NOM' and POM equal (B. IE., P. XV.), the an- gles ONM' and 0PM equal, hoth being right angles, and therefore (B. L, P. XXV., C. 2), the angles OM'N and OMP equal; they have, also, the sides OM' and OM equal, and are, consequently (B. L, P. VI.), equal in all respects: hence, ON — OP, and NM' = PM. These are nu- merical relations; by the rules for signs. Art. 58, ON and OP are both positive, NM' is negative, and PM positive; and hence, algebraically, ON = OP, and NM' = — PM ; therefore, we have, sin (90° + a) = cos a; (1.) cos (90° + a) = — sina. (2.) Dividing (1) by (2), member by member, we have, sin (90° + g ) _ cos a cos (90° + a) ~ — sin a ' or (formulas 6 and 7, Art. 61). tan (90° + a) = — cot a. 64 ANALYTICAL In like manner, dividing (2) by (1), member by mem- ber, we bave, cot (90° + a) = — tan a. Taking the reciprocals of botb members of (2), we bave (formulas 11 and 12, Art. 61), sec (90° + o) = — cosec a. In like manner, taking the reciprocals of both members of (1), we have, cosec (90° + a) = sec a. Again, let M"C = AM = a; then AM" = 180° — a. As before, the right-angled triangles OP"M" and 0PM may be proved equal in all respects, giving the numerical velar tions, P"M" z= PM, and OP" = OP, and, by the application of the rules for signs. Art. 58, may be obtained, P"M" = PM, and OP" = - OP ; hence, sin (180° — a) = sin a ; ■ cos (180° — a) = — cos a. (2.) From these equations (1) and (2), and formulas (6), (7), (11), and (12), Art. 61, may be obtained, as before, tan (180° — a) = — tan a ; cot (180° — a) = — cot a; sec (180° — a) = — sec a ; cosec (180° — a) = cosec a. In like manner, the values of the several functions of the remaining arcs in question may be obtained in terms of functions of the arc a. Tabulating the results, we have the following TBIGONOMETRY. 65 TABIiE m. Aro = 90° + ffl. Arc = 270° - ffl. sin = COB a, cos = — Bin ffl. Bin = — cos ffl, cos = — Bin ffl. tan = — cot o, cot = — tan ffl. tan = cot a, cot = tana, sec = — oosec ffl, coseo = sec a. seo = — coseo a. cosec = — Bee ffl. Arc = 180° - a. Arc = 270° + a. sin = sin a, cos = - cos a, Bin = — cos a, cos = sin a, tan = — tana, cot = - cot a. tan = — cot a. cot = ~ tan o. sec = — sec 0, coseo = cosec a. sec = cosec a, cosec = — seo ffl. Arc = 180° + a. Arc = 360° — a. sin = — sin ffl, cos = — cos ffl, sin = — sin ffl, cos = OOBO, tan = tana. cot = cot ffl. tan = — tana. cot = — cot 0, sec = — seo 0, cosec = — cosec a. J seo = sec a. coseo = — oosec a. It will be observed that, -when the arc is added to, or subtracted from, an even number of quadrants, the name of the function is the same in both columns ; and when the arc is added to, or subtracted from, an odd number of quadrants, the names of the functions in the two col- umns are contrary: in all cases, the algebraic sign is determined by the rules already given (Art. 58). By means of this table, we may find the functions of any arc in terms of the functions of an arc less than 90°. Thus, sin 115° = sin (90° + 25°) = cos 25°, sin 284° = sin (270° + 14°) = -cos 14°, sin 400° = sin (360° + 40°) = sin 40°, tan 210° = tan (180° + 30°) = tan 30°. &c. &c. &c. 66 ANALYTICAL PARTICULAR VALUES OF CERTAIN FUNCTIONS. 64. Let MAM' be any arc, denoted by M "a, M'M its chord, and OA a radius drawn perpendicular to M'M: then will PM=|M'M, imd AM = iM'AM (B. HI., P. VL). But PM is the sine of AM, or, PM = sin a : hence, sin a = |M'M ; that is, the sine of an are is equal to one half the chord of tivice the arc. Let M'AM^eO"; then will AM = 30°, and M'M wiU equal the radius, or 1 (B. V., P. IV.) : hence, we have sin 30° = i; that is, the sine of 30° is equal to half the radius. Also, cos 30° = Vl - sinS 30° = 1^/3 ; , , „„o sin 80° /T hence, tan 30 = —^-^ = ^ -. Again, let MAM = 90°: then will AM = 45°, and M'M V2 (B. v., P. IIL) : hence, we have sin 45° = iA/2; Also, , cos 45° = VI — sin' 45° = |a/2 ; , 4. ^ CO sin 45° ^ hence, tan 45 = j-_-6 = 1. ' cos 45 Many other numerical values might be deduced. TBIGONOMETB Y. 67 / C P P' A CN = COS h. FORMULAS EXPRESSING RELATIONS BETWEEN THE CIRCULAR FUNCTIONS OF DIFFERENT ARCS. 65. Let AB and BM represent two arcs, having the common radius 1 ; denote the first by a, and the second by &; then, AM = a -\-h. From M draw PM perpendicular to CA, and NM perpendicular to CB; from N draw NP' perpendicular, and NL parallel, to CA. Then, by definition, we have PM = sin (a + h), NM = sin 6, and From the figure, we have PM = PL + LM. (1.) From the right-angled triangle CP'N (Art. 37), we have P'N = CN sin a ; or, since P'N = PL, PL = cos h sin a = sin a cos h. Since the triangle MLN is similar to CP'N (B. lY., P. XXL), the angle LMN is equal to the angle P'CN ; hence, from the right-angled triangle MLN, we have LM = N M cos a = sin & cos a = cos a sin 6. Substituting the values of PM, PL, and LM, in equation (1), we have sin (a + 6) = sin a cos 6 + cos a sin 6 ; • (A.) that is, the sine of the sum of two arcs is equal to the sine of the first into the cosine of the second, plus the co- sine of the first into the sine of the second. 68 ANALYTICAL Since the above formula is true for any values of a and 6, we may substitute — b for b ; -whence, sin (a — b)^ = sin a cos (— 6) + cos a sin (— b) ; but (Art. 62), cos (— b) = cos b, and sin (— 6) = — sin b ; hence, sin (a — b) = sin a cos 6 — cos a sin b ; • (B.) that is, the sine of the difference of two arcs is equal to the sine of the first into the cosine of the second, minus the cosine of the first into the sine of the second. If, in formula (B), we substitute (90° — a), for a, we have sin (90°— a— &) = sin (90°— a) cos 6— cos (90°— a) sin b ; (2.) but (Art. 63), sin (90° - a — b) = sin [90° - (a + 6)] = cos (a + &), and, sin (90° — a) = cos a, cos (90° — a) = sin a ; hence, by substitution in equation (2), we have cos (a + &) = cos a cos b — sin a sin b ; • (C.) that is, the cosine of the sum of two arcs is equal to the rectangle of their cosines, minus the rectangle of their sines. If, in formula (C), we substitute — b, for 6, we find cos (a — 6) = cos a cos (— b) — sin a sin (— 6), or, cos {a — b) = cos a cos 6 + sin a sin 6 ; • • (D.) TKIGONOMETK Y. 69 that is, the cosine of the difference of two arcs is equal to the rectangle of their cosines, plus the rectangle of their sines. If we divide formula (A) by formula (C), member by member, we have sin (g + h) _ sin a cos h + cos a sin h cos (a + h) ~ cos a cos b — sin a sin 6 Dividing both terms of the second member by cos a cos 6, recollecting that the sine divided by the cosine is equal to the tangent, we find , / , IS tan a + tan h ,'. tan (a + 6) = ^—^—-^—^, . . . (E.) that is, the tangent of the sum of two arcs, is equal to the sum of their tangents, divided by 1 mimus the rectangle of their tangents. K, in formula (E), we substitute — h for h, recollect- ing that tan (—6).= — tan 6, we have , , , , tan a — tan 6 ,_ , tan (a -6) =3-^^^--^^; • • • (F.) that is, the tangent of the difference of two arcs is equal to the difference of their tangents, divided by 1 plus the rectangle of their tangents. In like manner, dividing formula (C) by formula (A), member by member, and reducing, we have . , , , , cot a cot 6 — 1 ,_ , cot (a + 6) = — ^,— r-r- i • ■ • (G.) ^ ' cot a + cot 6 ' ^ ' 70 ANALYTICAL and thence, by the substitution of — 6 for 6, , / , , cot a cot 6+1 /„ » cot (a — 6) = :rT 1. — • • • ("•) ^ ' cot — cot a ^ ' FUNCTIONS OF DOUBLE ARCS AND HALF ARCS. 66. If, in formulas (A), (C), (E), and (G), we make b = a, we find sin 2a = 2 sin a cos a; .... (A'.) cos 2a = cos' a — sin" a ; ... (C'.) , a 2 tan a -_,,, cot 2a = ^^^1 (G'O 2 cot a Substituting in (C) for cos' a, its value, 1 — sin' a ; and afterwards for sin' a, its value, 1 — cos' a, we have cos 2a = 1 — 2 sin' a, cos 2a = 2 cos' a — 1 ; whence, by solving these equations, /i — cos 2a ,^ > = V 2 ' • • • • ^^-^ /i + cos 2a , . We also have, from the same equations, 1 — cos 2a = 2 sin' a ; (3.) 1 + cos 2a = 2 cos' o. (4.) sm a cos TRIGONOMETBY. 71 Dividing equation (A'), first by equation (4), and then by equation (8), member by member, we have sin 2 a , ... r+^oT^^*^^**' (^-^ sin 2a , ... 3 pr- = cot a. (6.) 1 — cos 2a ^ ' Substituting |o for a, in equations (1), (2), (5), and (6), we have sin ^^ a = y/^^^; • • . . (A".) , /l + cos a ,_„ , cosia = y 2 ! • • . • (C.) i. 1 sm a /T?;#\ tan Aa = ^ ; (E .) ^ 1 + cos a ' ^ ^ X , sin a ,_„ . cotia = ^i (G .) * 1 — cos a ^ ' Taking the reciprocals of both members of the last two formulas, we have also, , , 1 + cos a J + 1 1 — cos a cot ia = — -. , and tan ia = — — r ^ sm a ' " sm a ADDITIONAL FORMULAS. 67. If formulas (A) and (B) are first added, member to member, and then subtracted, member from member, and the same operations are performed upon (C) and (D), we obtain 72 ANALYTICAL sin (a + 6) + sin {a — b) = 2 sin a cos 6 ; sin {a + b) — sin (a — 6) = 2 cos a sin 6 ; cos (a + 6) + cos (a — b) = 2 cos a cos 6 ; cos (o — 6) — cos (a + 6) = 2 sin a sin b. If in these we make a + b = p, and a — 6 = g, whence, a = Up + 2), b = iip-q); and then substitute in the aoove formulas, we obtain sin p + sin g = 2 sin |(jp + q) cos i (p — q). • (K.) sin p — sin g = 2 cos i(p + q) sin i (p — g). • (L.) cosp + cos g = 2 cos Up + q) cos ^ (p — g). • (M.) cos q — cosp = 2 sin i (p + q) sin Up — q). ■ (N.) From formulas (L) and (K), by division, we obtain sin p — sin q _ cos jjp + q) sin j (p — q) sin J) + sin g ~ sin i (p + g) cos i (p — g) _ ta n i^ (p — g) ,, . - tani(p + g) ,' ^^•■' Hence, since p and g represent any arcs whatever, the sum of the sines of two ares is to their difference, as the tangent of one half the sum of the arcs is to the tangent of one half their difference. TBIGONOMETBY. 73 Also, in like manner, we obtain sin^+sin 3 ^ sin jjp+g) cos Mp -g) ^ tan4(p+g), (2.) cosi3 + oosg cosi(i3+g) oosi(i)— g) "^^ smp-smq ^ co^ {p + q) sin i{p-q) ^ tan Hi>-3), (3-) cosjj + cosg cos^(p + g) cosi(i3— g) * sin j? + si D q ^ sin ij p+q) cosjip—q) _ cos j (p-g) ,^. sin(p + g) sin i(i>+g) cosi(jp + g) cosi(p + qy sin p— sin g _ sin jip—q) cos j {p+q) _ sin i{p—q) (5.) sin (jp+g) sin i (p+q) cos i (p+g) sin i (p+q)' sin (j>-g) _ sin j (p— g) cos j jp-q) _ cos i(p-g) ,g , in p— sing sin i (p— g) cos i (p + g) cos i(p+g)' ''' sm all of whicli give proportions analogous to that deduced from formula (1). Since the second members of (6) and (4) are the same, we have sin p — sin g _ sin (p + g) sin (p — g) sinp + sin g' (7.) that is, the sine of the difference of two arcs is to the difference of the sines, as the sum of the sines is to the sine of the sum. All of the preceding formulas may be made homo- geneous in terms of R, R being any radius, as explained in Art. 30 ; or, we may simply introduce R, as a factor, into each term as many times as may be necessary to render all of its terms of the same degree. 74 ANALYTICAL METHOD OF COMPUTING A TABLE OB NATURAL SINES. 68. Since the length of the semi-circumference of a circle whose radius is 1, is equal to the number 3.14159265..., if we divide this number by 10800, the number of minutes in 180°, the quotient, .0002908882 . . ., will be the length of the arc of one minute; and since this arc is so small that it does not differ materially from its sine or tangent, this may be placed in the table as the sine of one minute. Formula (3) of Table II., gives cosl' = Vl -sm'l' = .9999999577. • (1.) Having thus determined, to a near degree of approxi- mation, the sine and cosine of one minute, we take the first formula of Art. 67, and put it under the form, sin {a + b) = 2 sin a cos b — sin (a — b), and make in this, 6=1', and then in succession, o = 1', a = 2', a = 3', a = 4', &c., and obtain, sin 2' = 2 sin 1' cos 1' — sin = .0005817764 . . . sin 3' = 2 sin 2' cos 1' — sin 1' = .0008726646 . . . sin 4' = 2 sin 3' cos 1' - sin 2' = .0011635526-. . . sin 5' = &c., thus obtaining the sine of every number of degrees and minutes from 1' to 45°. TKIGONOMETB Y. 76 The cosines of the corresponding arcs may be com- puted by means of equation (1). Having found the sines and cosines of arcs less than 45°, those of the arcs between 45° and 90° may be de- duced, by considering that the sine of an arc is equal to the cosine of its complement, and the cosine equal to the sine of its complement. Thus, sin 50° = sin (90° - 40°) = cos 40°, cos 50° = sin 40°, in which the second members are known from the pre- vious computations. To find the tangent of any arc, divide its sine by its cosine. To find the cotangent, take the reciprocal of the corresponding tangent. As the accuracy of the calculation of the sine of any arc, by the above method, depends upon the accuracy of each previous calculation, it would be well to verify the work, by calculating the sines of the degrees separately (after having found the sines of one and two degrees), by the last proportion of Art. 67. Thus, sin 1° : sin 2° - sin 1° : : sin 2° + sin 1° : sin 3° ; sin 2° : sin 3° - sin 1° : : sin 3° + sin 1° : sin 4° ; &c. SPHERICAL TRIGONOMETRY. 69. Spherical Trigonometet is that branch of Mathe- matics which treats of the solution of spherical triangles. In every spherical triangle there are six parts : three sides and three angles. In general, any three of these parts being given, the remaining parts may be found. GENERAL PRINCIPLES. 70. For the purpose of deducing the formulas required in the solution of spherical triangles, we shall suppose the triangles to be situated on spheres whose radii are equal to 1. The formulas thus deduced may be rendered appli- cable to triangles lying on any sphere, by making them homogeneous in terms of the radius of that sphere, as explained in Art. 30. The only cases considered will be those in which each of the sides and angles is less than 180°. Any angle of a spherical triangle is the same as the diedral angle included by the planes of its sides, and its meastire is equal to that of the angle included between two right lines, one in each plane, and both perpendicular to their common intersection at the same point (B. VI., D. 4). The radius of the sphere being equal to 1, each side of the triangle will measure the angle, at the centre, sub- tended by it. Thus, in the triangle ABC, the angle at A 1: B I G O N O M E T E Y . 77 is the same as that included between the planes AOC and AOB ; and the side a is the meas- ure of the plane angle BOC, ^ being the centre of the sphere, and OB the radius, equal to 1. 71. Spherical triangles, like plane triangles, are divided into two classes, right-angled spherical tri- angles, and oblique-angled spherical triangles. will be considered in turn. Each class We shall, as before, denote the angles by the capital letters A, B, and C, and the sides opposite by the small letters a, h, and c. FORMULAS USED IN SOLVING EIGHT-ANGLED SPHEBICAL TEIANGLES. 73. Let CAB be a sperical triangle, right-angled at A, and let be the centre of the sphere on which it is situated. Denote the angles of the triangle by the letters A, B, and C, and the sides opposite by the letters a, b, and c, recollecting that B and C may change places, provided that b and c change places at the same time. Draw OA, OB, and OC, each equal to 1. From B, draw BP perpendicular to OA, and from P draw PQ perpendicu- lar to OC ; then join the points Q and B, by the line QB. The line QB will be perpendicular to OC (B. VI., P. VI.), and the anele PQB will be equal to the inclination of the 78 SPHERICAL planes OCB and OCA ; that is, it will be equal to the spherical angle C. We have, from the figure, PB = sin c, OP = cos c, QB = sin a, OQ = cos a. From the right-angled triangles OQP and QPB, we have OQ = OP cos AOC ; or, cos a = cos c cos b. ■ (1.) PB = QB sin PQB ; or, sin c = sin a sin C. • (2.) From the right-angled triangle QPB, we have OP cos PQB, or cos C = ^ ^ ; but, from the right-angled triangle PQO, we have QP = OQ tan QOP = cos a tan b ; substituting for QP and QB their values, we have _ cos a tan b ^. . r, ,o ^ cos C = -. = cot a tan o. • • (3.) sm a ' From the right-angled triangle OQP, we have OP sin QOP, or sin & = gp; but, from the right-angled triangle QPB, we have QP = PB cot PQB = sin c cot C ; substituting for QP and OP their values, we have . , sin c cot C , , _ , . , sm = = tan c cot C. • • • (4.) cos c ^ ' TRIGOJSrOMETBY. 79 If, in (2), we change c and C into b and B, we have sin 6 = sin a sin B. (5.) If, in (3), we change b and C into c and B, we have cos B — cot a tan c. (6.) If, in (4), we change b, c, and C, into c, b, and B, we have sin c = tan b cot B. (7.) Multiplying (4) by (7), member by member, we have sin b sin c = tan b tan c cot B cot C. Dividing both members by tan b tan c, we have cos b cos c = cot B cot C ; and substituting for cos b cos c, its value, cos a, taken from (1), we have cos a = cot B cot C. ...... (8.) Formula (6) may be written under the form □ cos a sin c cos B = -. sm a cos c Substituting for cos a, its value, cos b cos c, taken from (1), and reducing, we have r, cos b sin c cos B = : sm a Again, substituting for sin c, its value, sin a sin C, taken from (2), and reducing, we have 80 SPHERICAL COS B = COS 6 sin C. (9.) Changing B, 6, and C, in (9), into C, c, and B, we have cos C = cos c sin B. (10.) These ten formulas are sufficient for the solution of any right-angled spherical triangle whatever. For the purpose of classifying them under two general rules, and for con- venience in remembering them, these formulas are usually put under other forms by the use of NAPIER'S CIRCULAE, PARTS. 73. The two sides about the right angle, the complements of their opposite angles, and the complement of the hypothenuse, are called Napier's Circular Parts. If we take any three of the five parts, as shown in the figure, they will either be adjacent to each other, or one of them will be separated from each of the two others by an inter- vening part. In the first case, the one lying between the two other parts is called the middle part, and the two others, adjacent parts. In the second case, the one sepa- rated from both the other parts, is called the middle part, and the two others, opposite parts. Thus, if 90°— a is the middle part, 90° — B and 90° — C are .adjacent parts; and b and c are opposite parts; if c is the mid- dle part, b and 90° — B are adjacent parts (the right angle not being considered), and 90° — C and 90° — a are oppo- site parts : and similarly, for each of the other parts, taken as a middle part. TRIGONOMETEY. 81 74. Let us now consider, in succession, each of the five parts as a middle part, when the two other parts are opposite. Beginning with the hypothenuse, we have, from formulas (1), (2), (5), (9), and (10), Art. 72, sin (90° — a) = cos 6 cos c ; (1.) sin c = cos (90° — a) cos (90° — C) ; (2.) sin b = cos (90° — a) cos (90° — B) ; (3.) sin (90° - B) = cos b cos (90° - C) ; • • • • (4.) sin (90° - C) = cos c cos (90° _ B). • • • • (5.} Comparing these formulas with the figure, we see that The sine of the middle part is equal to the rectangle of the cosines of the opposite parts. Let us now take the same middle parts, and the other parts adjacent. Formulas (8), (7), (4), (6), and (3), Art. 72, give sin (90° - a) = tan (90° - B) tan (90° - C) sin c = tan 6 tan (90° — B) ; • • sin b = tan c tan (90° — C) ; sin (90° — B) = tan (90° - a) tan c ; ■ • sin (90° - C) = tan (90° - a) tan 6. ■ • (6.) (7.) (8.) (9.) (10.) Comparing these formulas with the figure, we see that The sine of the middle part is equal to the rectangle of +}i.P, tansSents of the adjacent parts. 82 SPHEEICAL These two rules are called Napier's rules for circular parts, and are sufficient to solve any right-angled spherical triangle. 75. In applying Napier's rules for circular parts, the part sought will be determined by its sine. Now, the same sine corresponds to two different arcs, or angles, supple- ments of each other ; it is, therefore, necessary to discover such relations between the given and the required parts, as will serve to point out which of the two arcs, or angles, is to be taken. Two parts of a spherical triangle are said to be of the same species, when they are each less than 90°, or each greater than 90°; and of different species, when one is less and the other greater than 90°. From formulas (9) and (10), Art. 72, we have, . ^ cos B J • D cos C sm C = T- , and sm B = ; cos ' cos c since the angles B and C are each less than 180°, their sines must always be positive : hence, cos B must have the same sign as cos 6, and the cos C must have the same sign as cos c. This can only be the case when B is of the same species as b, and C of the same species as c ; that is, ecu:h side about the right angle is always of the same species as its opposite angle. From formula (1), we see that when a is less than 90°, or when cos a is positive, the cosines of b and c will have the same sign ; and hence, 6 and c will be of the same species: when a is greater than 90°, or when cos a is negative, the cosines of b and c will have con- trary signs, and hence b and c will be of different species: TRIGONOMETRY. 83 therefore, when the hypothenuse is less than 90°, the two sides about the right angle, and consequently the two oblique angles, will be of the same species; when the hypothenuse is greater than 90°, the two sides about the right angle, and consequently the two oblique angles, wiU be of different species. These two principles enable us to determine the nature of the part sought, in every case, except when an oblique angle and the side opposite are given, to find the remain- ing parts. In this case, there may be two solutions, one solution, or no solution. There may be two cases : P 1°. Let there be given B and ^^^.LS:^., b, and B acute. Construct B and ^>^ / prolong its sides till they meet x^ / 1^ in B'. Then wUl BCB' and BAB' ^^ jii_ '•" be semi-circumferences of great circles, and the spherical angles B and B' will be equal to each other. As B is acute, its measure is the longest arc of a great circle that can be drawn perpendicular to the side BA and included between the sides of the angle B (B. IX., Gen. S. 2) ; hence, if the given side is greater than the measure of the given angle opposite, that is, if & > B, no triangle can be constructed, that is, there can be no solution: if 6 = 6, BC and BA' will each be a quad- rant (B. IX., P. IV.), and the triangle BA'C, or its equal B'A'C, will be birectangular (B. IX., P. XFV., C. 3), and there will be but one solution: if 6 < B, there will be two solutions, BAG and B'AC, the required parts of one being supplements of the required parts of the other. Since B < 90°, if 6 < B, 6 differs more from 90° than B does; and' if 6 > B, 6 differs less from 90° than B. 84 SPHERICAL 2d. Let B be obtuse. Construct B as before. As B is obtuse, its measure is the short- est arc of a great circle that can be drawn perpendicular to the y^ yf^ side BA and included between the sides of the angle B (B. IX., Gen. S. 2) ; hence, if 6 < B, there "^~-..._j' can be no solution: if 6 = B, the corresponding triangle, BA'C or B'A'C, will be birectangular and there will be but one solution, as before : and if 6 > B, there will be two solutions, BAC and B'AC. Since B > 90°, if 6 > B, & differs more from 90° than B does; and if & < B, h differs less from 90° than B. Hence, it appears, from both cases, that If 6 differs more from 90° than B, there will be two solutions, the required parts in the one case being supple- ments of the required parts in the other case. If 6 = B, the triangle wiU be birectangular, and there will be but one solution. If h differs less from 90° than B, the triangle can not be constructed, that is, there will be no solution. SOLUTION OF RIGHT-ANGLED SPHERICAL TRI- ANGLES. 76. In a right-angled spherical triangle, the right angle is always known. If any two of the other parts are given, the remaining parts may be found by Napier's rules for circular parts. Six cases may arise. There may be given, TBIGONOMETB Y. 85 I. The hypothenuse and one side. n. The h3rpothenuse and one oblique angle. III. The two sides about the right angle. IV. One side and its adjacent angle. V. One side and its opposite angle. VI. The two oblique angles. In any one of these cases, we select that part which is either adjacent to, or separated from, each of the other given parts, and calling it the middle part, we employ that one of Napier's rules which is applicable. Having determined a third part, the two others may then be found in a similar manner. It is to be observed, that the formulas employed are to be rendered homogeneous, in terms of R, as explained in Art. 30. This is done by simply multiplying the radius, R, into the middle part. Examples. 1. Given a = 105° 17' 29", and 6 = 38° 47' 11", to find C, c, and B. Since a > 90°, h and c must be of dif- ferent species, that is, c > 90°, and hence C > 90°. c Operation. Formula (10), Art. 74, gives for 90° — C, middle part, log cos C = log cot a + log tan 6 — 10; log cot a (105° 17' 29") 9.436811 log tan 6 (38° 47' 11") 9.905055 logcosC • • • 9.341866 .-. C = 102° 41' 33". 86 SPHERICAL Formula (2), Art. 74, gives for c, middle part, log sin c = log sin a + log sin C — 10; log sin a (105° 17' 29") 9.984346 log sin C (102° 41' 33") 9.989256 log sine • . • 9.973602 .-. c = 109° 46' 32". Formula (4) gives for 90° — B, middle part, log cos B = log sin C + log cos fe — 10 ; log sin C (102° 41' 33") 9.989256 log cos 6 (38° 47' 11") 9.891808 log cos B • • • 9.881064 .-. B = 40° 29' 50". Ans. c = 109° 46' 32", B = 40° 29' 50", C = 102° 41' 33". It is better, in all cases, to find the required parts in cerms of the two given parts. This may always be done by one of the formulas of Art. 74. Select the formula which contains the two given parts and the required part, and transform it, if necessary, so as to find the required part in terms of the given parts. Thus, let a and B be given, to find C. Regarding 90° — a as a middle part, we have, from formula (6), cos a =■ cot B cot C ; whence, cot C = — t-= ; ' cot B ' and, by the apphcation of logarithms, log cot C = log cos a + (a. c.) log cot B ; from which C may be found. In like manner, other cases may be treated. TRIGONOMETRY. 87 2. Given 6 = 61° 30', and B = 58° 35', to find o, c, d,nd C. Because 6 < B, there are two solutions. Operation. Formula (7) gives for c, middle part, log sin c = log tan 6 + log cot B — 10 ; log tan 6 (51° 30') 10.099395 log cot B (58° 35') 9.785900 log sin c • • ■ 9.885295 .• c = 50° 09' 51", and c' = 129° 50' 09". Formula (3) gives sin & = sin a sin B, sin b \kbence, sm a = ^g, and hence, log sin a = log sin 6 + (a. c.) log sin B ; log sin 6 (51° 30') 9.898544 (a. c.) log sin B (58° 35') 0.068848 log sin a • • 9.962392 .-.0= 66° 29' 53", a' = 113° 30' 07". Formula (4) gives cos B = cos b sin C, . _ cos B whence, sm C = 33^, and hence, , log sin C = log cos B + (a. c.) log cos b ; log cos B (58° 35') 9.717053 (a. c.) log cos & (51° 30') 0.205850 log sin C • • 9.922903 .: C = 56° 51' 38", C = 123° 08' 22". 88 SPHERICAL As a check, to test the accuracy of the above work, formula (2) may be used. Thus, from that formula, log sin c = log sin a + log sin C — 1 0. As found above, log sin a • • 9.962392 log sin C • • 9.922903 log sine • 9.885295 As the test is satisfied, the work is probably correct. Other cases may be treated in like manner. 3. aiven a = 86° 51', and B = 18° 08' 32", to find 6, c, and C. Ans. b = 18° 01' 50", c = 86° 41' 14", C = 88° 58' 25". 4. aiven b = 155° 27' 54", and c = 29° 46' 08", to find a, B, and C. Ans. a = 142° 09' 13", B = 137° 24' 21", C = 54° 01' 16". 5. Oiven c = 73° 41' 35", and B = 99° 17' 33", to find a, b, and C. Ans. a = 92° 42' 17", b = 99° 40' 30", C = 73° 54' 47". 6. Given b = 115° 20', and B = 91° 01' 47", to find o, c, and C. a= 64° 41' 11", c = 177° 49' 27", C = 177° 35' 36". a' = 115° 18' 49", c' = 2° 10' 33", C = 2° 24' 24". 7. Given B = 47° 13' 43", and C = 126° 40' 24", to find o, b, and c. Ans. a = 133° 32' 26", b = 32° 08' 56", c = 144° 27' 03". THIGONOMETBT. 89 QUADRANTAL SPHERICAL TRIANG-LES. 77. A QuADRANTAL SPHERICAL TRIANGLE is One in which one side is equal to 90°. To solve such a triangle, we pass to its supplemental polar triangle, by subtracting each side and each angle from 180° (B. IX., P. VI.). The re- sulting polar triangle wiU be right-angled, and may be solved by the rules already given. The supplemental polar triangle of any quadrantal triangle being solved, the parts of the given triangle may be found by subtracting each part of the supplemental triangle from 180°. Example. Let A'B'C be a quadrantal triangle, in which B'C = 90°, B' = 75° 42', and c' = 18° 37'. Passing to the supplemental polar tri- angle, we have A = 90°, b = 104° 18', and C = 161° 23'. Solving this triangle by previous rules, we find a = 76° 25' 11", c = 161° 55' 20", B = 94° 31' 21"; hence, the required parts of the given quadrantal triangle are, A' = 103° 34' 49", C = 18° 04' 40", b' = 85° 28' 39", Other quadrantal triangles may be -solved in like manner. 90 SPHEBIC AL FORMULAS USED IN SOLVING OBLIQUE-ANGLED SPHEKICAL TKIANGLE& 78. To show that, in a spherical triangle, the sines of the sides are proportional to the sines of their opposite angles. Let ABC represent an oblique-angled spherical triangle. From any vertex, as C, draw the arc of a great circle, CB', perpendicular to the opposite side. The two tri- angles ACB' and BCB' will be right- angled at B'. From the triangle ACB', we have, " formula (2) Art. 74, sin CB' = sin A sin b. From the triangle BCB', we have sin CB' = sin B sin a. Equating these values of sin CB', we have sin A sin 6 = sin B sin a ; from which results the proportion, sin a : sin 6 : : sin A : sin B. In like manner, we may deduce sin a : sin c : : sin A : sin C, sin 6 : sin c : : sin B : sin C. (1.) (2.) (3.) That is, in any spherical triangle, the sines of the sides are proportionai to the sines of their opposite angles. Had the perpendicular fallen on the prolongation of AB, the same relation would have been found. TBIOONOMETRT. 61 79. To find an expression for the cosine of any side of a spherical triangle. Let ABC rei)resent any spherical triangle, and the centre of the sphere on -which it is situated. Draw the radii OA, OB, and OC ; from C draw CP per- pendicular to the plane AOB; from P, the foot of this perpendicular, draw PD and PE respectively per- pendicular to OA and OB ; join CD and CE, these lines will be respect- ively perpendicular to OA and OB (B. VI., P. VI.), and the angles CDP and CEP will be equal to the angles A and B respectively. Draw DL and PQ, the one perpendicular, and the other parallel to OB. "We then have OE = cos a, DC = sin b, OD = cos 6. We have from the figure, OE = OL -H QP. (1.) In the right-angled triangle OLD, OL = OD cos DOL = cos b cos c. The right-angled triangle PQD has its sides respectively perpendicular to those of OLD ; it is, therefore, similar to it, and the angle QDP is equal to c, and we have QP = PD sin QDP = PD sin c. The right-angled triangle CPD gives PD = CD cos CDP = sin 6 cos A; substituting this value in (2), we have QP = sin b sin c cos A ; (2.) 92 SPHERICAL and now substituting these values of OE, OL, and QP, in (1), we have ^ cos a = cos 6 cos c + sin b sin c cos A. • • (3.) In the same way, we may deduce, cos b ~ cos a cos c + sin a sin c cos B, • • (4.) cos c = cos a cos b + sin a sin 6 cos C. • ■ (5.) That is, the cosine of any side of a spherical triangle is equal to the rectangle of the cosines of the two other sides, ■plus the rectangle of the sines of these sides into the cosine of their included angle. 80. To find an expression for the cosine of any angle of a spherical triangle. If we represent the angles of the supplemental polar triangle of ABC, by A', B', and C, and the sides by a', b', and c', we have (B. IX., P. VI.), a = 180° - A', 6 = 180° - B', c = 180° - C, A = 180° -a', B = 180° -b', C = 180° - c'. Substituting these values in equation (3), of the preceding article, and recollecting that cos (180° - A') = - cos A', sin (180° - B') = sin B', &c,, we have — cos A' = cos B' cos C — sin B' sin C cos a' ; or, changing the signs and omitting the primes (since the preceding result is true for any triangle), cos A = sin B sin C cos a — cos B cos C. • • (1.) TRIGONOMETRY. 93 In the same way, we may deduce, cos B = sin A sin C cos 6 — cos A cos C, • • (2.) cos C = sin A sin B cos c — cos A cos B. • • (3.) That is, the cosine of any angle of a spherical triangle is equal to the rectangle of the sines of the two other angles into the cosine of their included side, minus the rectangle of the cosines of these angles. The formulas deduced in Arts. 79 and 80, for cos a, cos A, etc., are not convenient for use, as logarithms can not be applied to them ; other formulas are, therefore, derived from them, to which logarithms may he applied. 81. To find an expression for the cosine of one half of any angle of a spherical triangle. From equation (3), Art. 79, we deduce, . cos a — cos h cos c ,- , cos A = — : j--: (1.) sm 6 sin c If we add this equation, member by member, to the num- ber 1, and recollect that 1 + cos A, in the first member, is equal to 2cos2iA (Art. 66), and reduce, we have , , , sin 6 sin c + cos a — cos 6 cos c 2 cos" iA = = — <--. ; * sm sm c or, formula (C), Art. 65, 2 cosHA = "°^ " r f"^ -(A+- A . . . (2.) * sm sm c ^ ' And since, formula (N), Art. 67, 94 SPHERICAL COS a — cos (6 + c) = 2 sin ^ (a + 6 + c) sin i (6 + c — a), equation (2) becomes, after dividing both members by 2, cosUA = sin i (g + & + c) sin i (b + c - o) * sin 6 sin c If in this we make J(a + 6 + c) = is; whence, ^{b + c — a) = \s — a, and extract the square root of both members, we have ^ ^ /imF^is-^o) * V sm 6 sm c ^ ' cos- That is, the cosine of one half of any angle of a spherical triangle is equal to the square root of the sine of one half of the sum of the three sides, into the sine of one half this sum minus the side opposite the angle, divided by the rect- angle of the sines of the adjacent sides. If we subtract equation (1), of this article, member by member, from the number 1, and recoUect that 1 — cos A = 2 sin^ ^A, we find, after reduction, . , ■ /sin (is — 6) sm iAs — c) ... smiA = \/ ^ — ^ir^— ■ ■ (4.) V sm sm c ^ ' Dividing equation (4) by equation (3), member by mem- ber, we obtain ** = \/' tan iA = . / sin (js - 6) sin (js - c) sin |s sin (\s — a) (5.) TRIGONOMETEY. 95 83, From the foregoing values of the functions of one half of any angle, may be deduced values of the functions of one half of any side of a spherical triangle. Representing the angles and sides of the supplenieni-a.1 polar triangle of ABC as in Art. 80, we have A = 180° -a', b = 180° - B', c = 180° - C, is = 270° -^{A' + B' + C), |s - a = 90° - i (B' 4- a — A'). Substituting these values in (3), Art. 81, and reducing by the aid of the formulas in Table III., Art. 63, we find ha' = sj- ^\^ !„' - ^ / -cosi(A' + B' + C) cos i (B' + C - A') sm ia _ \/ STB^shTC' ' Place i (A' + B' + C) = iS ; whence, i (B' + C - A') = \S - A'. Substituting and omitting the primes, we have . 1 /— cos iS cos (iS — A) ,- > ^ V sm B sm C In a similar way, we may deduce from (4), Art. 81, cos ia = . A2^S_-_B^sJi_S -^). . ^ V sm B sin C ^ ' ^ , , , , / — cos iS cos (iS— A) , and thence, tan^a = y^^^^^s^-gp^-^p— -^-y. • (3.) 96 SPHERICAL 83. To deduce Napier's Analogies. From equation (1), Art. 80, we have cos A + cos B cos C = sin B sin C cos a . ^ sm A . , ,^ , = sm C -. sm cos a ; (1. j sm a since, from proportion (1), Art. 78, we have „ sin A . , sm B = -; sm o. sm a Also, from equation (2), Art. 80, we have cos B + cos A cos C = sin A sin C cos 6 . _ sin A . , ,„ . = sm C -7 — sin a cos b. (2.} sm a ' Adding (1) and (2), and dividing by sin C, we obtain / A , D\ 1 + cos C sin A . , , ,. ,., , (cos A + cos B) •. — -=. — = -: sm (a + o). (3.) ' sm C sin a \ > / \ / The proportion, sin A : sin B : : sin a : sin b, taken first by composition, and then by division, gives sin A sin A + sin B = —. (sin a + sin b), • • (4.) sm a ^ " ^ ' sm A sin A — sin B = —. — (sin a — sin 6). • • (5.) sm a ' ^ ' Dividing (4) and (5), in succession, by (3), we obtain sin A + sin B sin C sin a + sin 6 ,. , cos A + cos B 1 + cos C sin (a + 6) ^ ■' sin A — sin B sin C _ sin a — sin b cos A + cos B 1 + cos C "" sin {a + b) ^ '' TEIGONOMETB Y. 97 But, by formiilas (2) and (4), Art, 67, and formula (E"), Art. 66, equation (6) becomes and, by the similar formulas (3) and (5), of Art. 67, equation (7) becomes ^^ tan i (A - B) tan iC = ^!^ t !" ~ S" • • (9-) As tan -JC = — rTni formulas (8) and (9) may be written t an i (A + B) _ cos jja — b) ,„, , cot iC ~ cos i (a + 6) ' ■ ■ ■ ^ •' tan HA — B) _ sin jja — b) , , cotfC "~ sin^(a + b)' ' ' ' ^ '' These last two formulas give the proportions known as the first set of Napier^s Analogies ; viz., cosi(a + 6) : cos -J (a— 6) :: cot |C : tani(A+B). (10.) sinJ(a+6) : sini(a— 6) :: ccb^-C : tani(A— B). (11.) If in these we substitute the values of a, b, C, A, and B, in terms of the corresponding parts of the supple- mental polar triangle, as expressed in Art. 80, we obtain cosi(A+B) : cosi(A— B) :: tan ic : tan^(a+6), (12.) sini(A+B) : sini(A— B) :: tan |c : tan i (a— 6), (13.) the second set of Napier's Analogies. 98 SPHEKICAL In applying logarithms to any of the preceding formu- las, they must be made homogeneous in terms of R, as explained in Art. 30. In aU the formulas, the letters may be interchanged at pleasure, provided that, when one large letter is substi- tuted for another, the like substitution is made in the corresponding small letters, and the reverse : for example, C may be substituted for A, provided that at the same time c is substituted for a, &c. Note. — It may be noted that, in formulas (10) and (12), whenever the sign of the first term of the propor- tion is minus, the sign of the last term must, also, be minus, i. e., whenever |(a + 6) is greater than 90°, |(A+B) must, also, be greater than 90°, and the reverse ; and similarly, whenever i{a + b) is less than 90°, |(A + B) must, also, be less than 90°, and the reverse. SOLUTION OF OBLIQUE-ANGLED SPHEEICAL TRI- ANGLES. 84. In the solution of oblique-angled triangles six dif- ferent cases may arise : viz., there may be given, I. Two sides and an angle opposite one of them. IL Two angles and a side opposite one of them. III. Two sides and their included angle. rV. Two angles and their included side. V. The three sides. VI. The three angles. TRIGONOMETRY. 99 CASE I. Given two sides and an angle opposite one of them. 85. The solution, in this case, is commenced by find- ing the angle opposite the second given side, for which purpose formula (1), Art. 78, is employed. As this angle is found by means of its sine, and be- cause the same sine corresponds to two different arcs, there would seem to be two different solutions. To ascer- tain when there are two solutions, when one solution, and when no solution at all, it becomes necessary to examine the relations which may exist between the given parts. Two cases may arise, viz., the given angle may be acute, or it may be obtuse. We shall consider each case separately (B. IX., Q-en. S. 1). Is* Case: A < 90°. Let A be the given acute angle, and let a and h be the given sides. Prolong the arcs AC and AB till they meet at A', forming the lune AA' ; and from C, draw the arc CB" per- pendicular to ABA'. From C, aS a pole, and with the arc a, describe the arc of a small circle BB'. If this cir- cle cuts ABA', in two points between A and A', there will be two solutions J for if C be joined with each point of intersection by the arc of a great circle, we shall have two triangles, ABC and AB'C, both of which will conform to the conditions of the problem. 100 SPHERICAL If only one point of intersection lies be- tween A and A', or if the small circle is tan- gent to ABA', there wiU be but one solution. If there is no point of intersection, or if there are points of intersection which do not lie between A and A', there wUl be no solution. From formula (2), Art. 72, we have sin CB" = sin b sin A, from which the perpendicular may be found. This per- pendicular will be less than 90°, since it can not exceed the measure of the angle A (B. IX., Gen. S. 2, 1°) ; denote its value by p. By inspection of the figure, we find the following relations : 1. W%en a is greater than p, and at the same time less than both b and 180° — b, there ivill he two solutions. 2. When a is greater than p, and intermediate in value between b and 180° — b; or, when a is equal to p, there mill be but one solution. If a = b, and is also less than 180° — 6, one of the points of intersection will be at A, and there will be but one solution. 3. When a is greater than p, and at the same time greater than both b and 180° — b; or, when a is less than p, there will be no solution. TEIGONOMETR Y. 101 2d Case: A > 90°. Adopt the same construction as before. In this case, the perpendicular will be greater than 90°, because it can not be less than the measure of the angle A (B. IX., Qen. S. 2, 2°) : it will, also, be greater than any other arc CA, CB, CA', that can be drawn from. C to ABA'. By a course of reasoning en- tirely analogous to that in the preceding case, we have the following principles : 4. WTien a is less than p, and at the same time greater than both b and 180° — b, there ndll be two solu- tions. 5. WTien a is less than p, and intermediate in value between b and 180° — b; or, when a is equal to p, there will be but one solution. 6. W%en a is less than p, and at the same time less than both b and 180° — b; or, when a is greater than p, there will be no solution. Having found the angle or angles opposite the second side, the solution may be completed by means of Napier's Analogies. Examples. 1. Given a = 43° 27' 36", h = 82° 58' 17", and A = 29° 82' 29", to find B, C, and c. We see that a> p, since p can not exceed A (B. IX., Q-en. S. 2, 1°); we see, further, that a is less than both 102 SPHEKICAL 6 and 180° — 6; hence, from the first condition there will be two solutions. Applying logarithms to formula (1), Art. 78, we have log sin B = (a. c.) log sin a + log sin 6 + log sin A — 1 ; (a. c.) log sin a • • (43° 27' 86") • • 0.162508 log sin & • • (82° 58' 17") • • 9.996724 log sin A ■ . (29° 32' 29") • • 9.692893 log sin B 9.852125 .-. B = 45° 21' 01", and B' = 134° 38' 59". From the first of Napier's Analogies (10), Art. 83, we find log cot \0 = (a. c.) log cos i{a — b) + log cos ^ (a + &) + log tan i (A + B) — 10. Taking the first value of B, we have |(A + B) = 37° 26' 45"; also, i{a + b) = 63° 12' 56"; and iia-b) = 19° 45' 20". (a. c.) log cos i (a - 6) • • (19° 45' 20") • 0.026344 log cos i (a + &) . . (63° 12' 56") • 9.653825 log tan i (A + B) • • (37° 26' 45") • 9.884130 log cot iC 9.564299 .-. iC = 69° 51' 45", and C = 139° 43' 30". The side c may be found by means of formula (12), Art. 83, or by means of formula (2), Art. 78. TBIGONOMETEY. 103 Appl3ring logarithms to the proportion, sin A : sin C : : sin a : sin c, we have log sin c = (a. c.) log sin A + log sin C + log sin a — 10 ; (a. c.) log sin A • • (29° 32' 29") • 0.307107 log sin C ■ -(139° 43' 30") • 9.810589 log sin a • • (43° 27' 36") • 9.837492 log sine 9.955138 .-. c = 115° 35' 48" We take the greater value of c, because the angle C, being greater than the angle B, requires that the side c should be greater than the side b. By using the second value of B, we may find, in a similar manner, C = 32° 20' 28", and c' = 48° 16' 18". 2. Given a = 97° 35', 6 = 27° 08' 22", and A = 40° 51' 18", to find B, C, and c. Ans. B = 17° 31' 09", C = 144° 48' 10", c = 119° 08' 25". 3. Given a = 115° 20' 10", h = 57° 30' 06", and A = 126° 37' 80", to find B, C, and c. Ans. B = 48° 29' 48", C = 61° 40' 16", c = 82° 34' 04". 4. Given 6 = 79° 14', c = 30° 20' 45", and B = 121° 10' 26", to find C, A, and a. Ans. C = 26° 06' 16", A = 49° 44' 16", a = 61° 11' 06". 104 SPHERICAL CASE n. Given two angles and a side opposite one of them. 86. The solution, in this case, is commenced by find- ing the side opposite the second given angle, by means of formula (1), Art. 78. The solution is completed as in Case I. Since the second side is found by means of its sine, there may be two solutions. To investigate this case, we pass to the supplemental polar triangle, by substituting for each part its supplement. In this triangle, there will be given two sides and an angle opposite one ; it may therefore be discussed as in the preceding case. When the supplemental triangle has two solutions, one solution, or no solution, the given triangle will, in like manner, have two solutions, one solution, or no solution. Let the given parts be A', B', and a', and let p' be the arc, CD', of a great circle drawn from the extremity of the given side perpendicular to the side opposite : we have sin^' = sin a' sin B'. There will be two cases : a' may be less than 90° ; or, a' may be greater than 90°. 1st Case: a' < 90°. Passing to the supplemental polar triangle, we shall have given a, b, A ; and since, in the given triangle, o' < 90°, in this supplemental triangle A > 90°: call the perpendicular CD, p. The conditions determining the num- TRIGONOMETB Y. 106 ber of solutions in this supplemental triangle are given in principles 4, 5, 6, Art. 85. From principle 4, Art. 85, it appears that, for two solu- tions, a. must be less than p, that is, a < p: subtracting each member of this inequality from 180°, we have 180° -a > 180° -p; but, 180° - a = A' ; and (B. IX., P. VL, C. 2), 180° -p = p' hence A' > jp' : again, it appears from principle 4, that a must be greater than 6, that is, a > ft; subtracting each member of this inequality from 180°, we have 180° - a < 180° -6; or, A' < B': it further appears from the same principle, that a must be greater than 180° — 6, that is, a > 180° -6; subtracting each, member of this inequality from 180°, we have 180° - a < 180° - (180° - b) ; or. A' < 180° - B' 106 SPHEEICAL Collecting the results, and, for convenience, omitting the primes, we have the following principle : Two angles and a side opposite one of them being given, and the given side less than 90°, i. e., A, B, a given, and a < 90° ; 1. When A is greater than p, and at the same time less than both B and 180° — B, there vdll be two solutions. In like manner, from principle 5, Art. 85, we have 2. When A is greater than p, and intermediate in value between B and 180°— B; or, when A is equal to p, there will be but one solution. And from principle 6, Art. 85, we have 3. When A is greater than p, and at the same time greater than both B and 180°— B; or, when A is less than p, there will be no solution. It is to be noted that, in this case, the perpendicular is less than 90°, and less, also, than the given side ; i. e., p < a. Id Case: a' > 90°. Passing to the supplemental polar triangle, we shall have given o, 6, A, and A < 90°. The conditions deter- mining the number of solutions in this supplemental tri- angle are given in principles 1, 2, 3, Art. 85. From principle 1, Art. 85, it appears that, for two solu tions, o must be greater than p, that is, a > p; THIGONOMETR Y. 107 subtracting each member of this inequaUty from 180°, we have 180° -a < 180° -p\ or, k' < p': in the same manner as before, "we may obtain from this principle 1, A' > B' ; and A' > 180° - B'. As before, collecting the results and omitting the primes, we have the following principle : Two angles and a side opposite one of them being given, the given side greater than 90°, i. e., A, B, a given, and a > 90°; 4. When A is less than p, and at the same time greater than both B and 180° — B, there mill be two solu- tions. In like manner, from principle 2, Art. 85, we have 6. WTien A is less than p, and intermediate in value between B and 180° — B; or, when A is equal to p, there will be but one solution. And from principle 3, Art. 85, we have 6. When A is less than p, and at the same time less than both B and 180° — B; or, when A is greater than p, there will be no solution. It is to be noted that, in this case, the perpendicular is greater than 90°, and greater, also, than the given side ; i. e., p> a. 108 SPHERICAL From the principles deduced in Articles 85 and 86, it is evident that, if the given parts of the spherical trian- gles considered are named as in the accom- panying table, we shall have the following principles, applicable to all the cases : 7. The sine of p is equal to the rectangle of the sines of the odd part and the adjacent part. Perpendimlar. Odd. Adjacent. Opposite. P A b a a B A 8. j3 is always of the same species as the odd part, and differs more from 90° than the odd part, i. e., when the odd part is less than 90°, p is still less; and when the odd part is greater than 90°, p is still greater. 9. There will be two solutions: 1°. When (odd part being less than 90°) the opposite part is greater than p, and less than the adjacent part and its supplement. 2°. When (odd part being greater than 90°) the opposite part is less than p, and greater than the adjacent part and its supplement. 10. There will be one solution: 1°. When (odd part beingj^ less than 90°) the oppo- site part is greater than p, and intermediate in value between the adjacent part and its supplement. 2°. When (odd part being greater than 90°) the TRIGONOMETRY. 109 opposite part is less than p, and intermediate in value between the adjacent part and its supplement. 3°. When the opposite part is equal to p. 11. There will be no solution: 1°. When (odd part being less than 90°) the op- posite part is either less than p, or greater than p and greater also than both the adjacent part and its supplement. 2°. When (odd part being greater than 90°) the opposite part is either greater than p, or less than p and less also than both the adjacent part and its supplement. Examples. 1. Given A = 95° 1,6', B = 80° 42' 10", and a = 57° 38', to find c, 6, and C. p might be computed from the formtda, log sinp = log sin B + log sin a — 10 ; but it is not necessary, as p < a (see principle 8). •Because A>j3, and intermediate between 80° 42' 10" and 99° 17' 50", there will, from the second condition, be but one solution. Applying logarithms to proportion (1), Art. 78, we have log sin 6 = (a. c.) log sin A + log sin B + log sin a — 10 ; (a. c.) log sin A (95° 16') 0.001837 log sin B (80° 42' 10") 9.994257 log sin a (57° 38') 9.926671 log sin 6. . • • 9.922765 .-. ft = 56° 49' 57". 110 SPHERICAL We take the smaller value of 6, for the reason that A, being greater than B, requires that a should be greater than 6. Applying logarithms to proportion (12), Art. 83, we have log tan \c = (a. c.) log cos |(A — B) + log cos i (A + B) + log tan 4 (o + 6) — 10 ; we have i(A + B) = 87° 59' 05", \{a + h) = 57° 13' 58", and i(A - B) = 7° 16' 55"; (a. c.) log cos i (A - B) • (7° 16' 55") • 0.003517 logcosi(A + B) • (87° 59' 05") • 8.546124 log tan i (a + 6) • (57° 13' 58") • 10.191352 log tan ic 8.740993 .-. ic = 3° 09' 09", and c = 6° 18' 18". Applying logarithms to the proportion, sin a : sin c : : sin A : sin C, we have log sin C = (a. c.) log sin a + log sin c + log sin A — 10 ; (a. c.) log sin a (57° 38') • • 0.073329 log sine (6° 18' 18")- 9.040685 log sin A (95° 16') • • 9.998163 log sin C 9.112177 .-. C = 7° 26' 21". The smaller value of C is taken, for the same reason as before. TEIGONOMETBT. Ill 2. Given A = 60° 12', B = 58° 08', and a = 62° 42', to find 6, c, and C. 6 = 79° 12' 10", c = 119° 03' 26", C = 130° 54' 28", 6'= 100° 47' 50", c'= 152° 14' 18", C = 156° 15' 06". 3. Given C = 115° 20', A = 57° 30', and c = 126° 38', to find a, 6, and B. Ans. a = 48° 29' 13", h= 118° 20' 44", B = 97° 35' 06". CASE m. Given two sides and their included angle. 87. The remaining angles are found by means of Napier's Analogies, and the remaining side as in the pre- ceding cases. Examples. 1. Given a = 62° 38', b = 10° 13' 19", and C = 150° 24' 12", to find c, A, and B. Applying logarithms to proportions (10) and (11), Art. 83, we have log tan i^ (A + B) = (a. c.) log cos |(a 4- 6) + log cos ^ (a — 6) + log cot iC — 10; log tan ^ (A — B) = (a. c.) log sin ^ (a + 6) + log sin i (a — 6) + log cot ^C — 10 ; we have iia-b) = 26° 12' 20", iC = 75° 12' 06", and 4(a + &) = 36° 25' 39". 112 SPHERICAL (a. c.) log COS i (a + 6) • (36° 25' 39") • 0.094415 log cos i (a -6) • (26° 12' 20") • 9.952897 log cot iC- • ■ . (72° 12' 06") . 9.421901 log tan i (A + B) 9.469213 .-. i(A + B) = 16° 24' 51". (a. c.) log sin i (a + 6) • (36° 25' 39") • 0.226356 log sin i (a -6) ■ (26° 12' 20") • 9.645022 log cot iC • • • (75° 12' 06") . 9.421901 log tan i (A - B) 9.293279 .-. i(A - B) = 11° 06' 53". The greater angle is equal to the half sum plus the half difference, and the less is equal to the half sum minus the half difference. Hence, we have A = 27° 31' 44", and B = 5° 17' 58". Applying logarithms to proportion (13), Art. 83, we have log tan ^c = (a. c.) log sin ^ (A — B) + log sin i (A + B) + log tan ^ (a — 6) — 10 ; (a. c.) log sin i (A - B) • (11° 06' 53") • 0.714952 logsini(A + B) • (16° 24' 51") • 9.451139 log tan i (a -6) • (26° 12' 20") • 9.692125 log tan ic 9.858216 .-. ic = 35° 48' 33", and c = 71° 37' 06". 2. Given a = 68° 46' 02", b = 37° 10', and C = 39° 23' 23", to find c. A, and B. Ans. A = 120° 59' 21", 8 = 33° 45' 13", c = 43° 37' 48". TRIGONOMETRY. 113 3. aiven a = 84° 14' 29", b = 44° 13' 45", and C = 36° 45' 28", to find A and B. Ans. A = 130° 05' 22", B = 32° 26' 06". 4. Given b = 61° 12', c = 131° 44', and A = 88° 40', to find B, C, and a. (See Note, Art. 83.) Ans. B = 66° 55' 69", C = 128° 25' 05", a = 72° 12' 46". CASE IV. Given two angles and their included side. 88. The solution of this case is entirely analogous to that of Case ITL Applying logarithms to proportions (12) and (13), Art. 83, and to proportion (11), Art. 83, we have log tan i (a + 6) = (a. c.) log cos ^ (A + B) + log cos ^ (A — B) + log tan ic — 10; log tan i (a — 6) = (a. c.) log sin |(A + B) + log sin | (A — B) + log tan |c — 10; log cot ^Q = (a. c.) log sin |(a — 6) + log sin \{a + b) + log tan i (A - B) — 10. The application of these formulas is sufficient for the solution of all cases. Examples. 1. Given A = 81° 38' 20", B = 70° 09' 38", and c = 59° 16' 22", to find C, a, and 6. Ans. C = 64° 46' 24", a = 70° 04' 17", b = 63° 21' 27". 114 SPHEBICAL 2. Qiven A = 34° 15' 03", B = 42° 15' 13", and c = 76° 35' 36", to find C, a, and 6. Ans. C = 121° 36' 12", a = 40° 0' 10", b = 50° 10' 30". 3. Given B = 82° 24', C = 120° 38', and a= 75° 19', to find A, b, and c. Ans. A = 73° 31' 13", b = 90° 50' 50", c = 119° 46' 22". CASE V. Given the three sides, to find the remaining parts. 89. The angles may be found by means of formula ('6), Art. 81 ; or, one angle being found by that formula, the two others may be found by means of Napier's Analogies. Examples. 1. Given a = 74° 23', b = 35° 46' 14", and c = 100° 39', to find A, B, and C. Applying logarithms to formula (3), Art. 81, we have log cos iA = 10 + i [log sin is + log sin (is — a) + (a. c.) log sin b + (a. c.) log sin c — 20] ; or, log cos ^A = ^ [log sin is + log sin Us — a) + (a. c.) log sin b + (a. c.) log sin c] ; we have |s = 105° 24' 07", and is — a == 31° 01' 07". TKIGONOMETE Y. 115 log sin|s . . . (105° 24' 07") • 9.984116 log sin {^s -a) . (31° 01' 07") • 9.712074 (a. c.) log sin b . . . (35° 46' 14") . 0.233185 (a. c.) log sin c log cos iA . (100° 39' ) ■ • • 2) 0.007546 19.936921 9.968460 .-. 4A = 21° 34' 2? and A = 43° 08' 46". Using the same formula as before, and substituting B for A, b for o, and a for b, and recollecting that is — b = 69° 37' 53", we have logsinis • • • (105° 24' 07") • 9.984116 log sin (is -6) ■ (69° 37' 53") • 9.971958 (a. c.)logsina • ■ • (74° 23') ■ • • 0.016336 (a. c.) log sine • • • (100° 39') • • • 0.007546 2) 19.979956 log cos iB 9.989978" .-. ^B = 12° 15' 43", and B = 24° 31' 26". Using the same formula, substituting C for A, c for a, and a for c, recollecting that |s — c = 4° 45' 07", we have logsinis • • • (105° 24' 07") • 9.984116 log sin (is -c) . (4° 45' 07") • 8.918250 (a. c.) log sin a • • • (74° 23') • • • 0.016336 (a. c.) log sin 6 • • • (25° 46' 14") • 9.233185 2 ) 19.151887 log cos iC 9.575943 .-. iC = 67° 52' 25", and C = 135° 44' 50". 2. Given a = 56° 40', 6 = 83° 13', and c = 114° 80', to find A, B, and C. Ans. A = 48° 81' 18", B = 62° 55' 44", C = 125° 18' '56". 116 SPHEBICAL TRIGONOMETRY. 3. Given a = 115° 15', b = 125° 30', and c = 110° 15', to find A, B, and C. Ans. A = 145° 15' 04", B = 149° 07' 52, C = 143° 45' 10". CASE VI. 2%.6 three angles being given, to find the sides. 90. The solution in this case is entirely analogous to the preceding one. Applying logarithms to formula (2), Art. 82, we have log cos la = i [log cos (iS — B) + log cos (^S — C) + (a. c.) log sin B + (a. c.) log sin C]. In the same manner as before, we change the letters, to suit each case. Examples. 1. Given A = 48° 30', B = 125° 20', ar . C = 62° 54', to find a, b, and c. Ans. a = 56° 39' 30", b = 114° 29' 58" c = 83° 12' 06". 2. Given A = 109° 55' 42", B = 116° 38' 33", and C = 120° 43' 37", to find a, b, and c. Ans. a = 98° 21' 40", b = 109° 50' 22", c = 115° 13' 28". 3. Given A ^ 160° 20', B = 135° 15', and C = 148° 25', to find o, b, and c. Ans. a = 155° 56' 10", b = 58° 32' 12", c = 140° 36' 48". MENSURATION. 91. Mensuration is that branch of Mathematics which treats of the measurement of Geometrical Magnitudes. 92. The measurement of a quantity is the operation of finding how many times it contains another quantity of the same kind, taken as a standard. This standard is called the unit of measure. 93. The unit of measure for surfaces is a square, one of whose sides is the linear unit. The unit of measure for volumes is a cube, one of whose edges is the linear unit. If the linear unit is one foot, the superficial unit is one square foot, and the unit of volume is one cubic foot. If the linear unit is one yard, the superficial unit is one square yard, and the unit of volume is one cubic yard. 94. In Mensuration, the expression product of two lines, is used to denote the product obtained by multiplying the number of linear units in one line by the number of linear units in the other. The expression product of three lines, is used to denote the continued product of the num- ber of linear units in each of the three lines. Thus, when we say that the area of a parallelogram is equal to the product of its base and altitude, we mean that the number of superficial units in the parallelogram is equal to the number of linear units in the base, mul- tiplied by the ■ number of linear units in the altitude. In 118 MENSURATION like manner, the number of units of volume, in a rectan- gular parallelopipedon, is equal to the number of super- ficial units in its base multiplied by the number of linear units in its altitude, and so on. MENSURATION OF PLANE FIGURES. To find the area of a parallelogram. 95, From the principle demonstrated in Book IV., Prop, v., we have the following Rule. — Multiply the base by the altitude; the product will be the area required. Examples. 1. Find the area of a parallelogram, whose base is 12.25, and whose altitude is 8.5. Ans. 104.125. 2. What is the area of a square, whose side is 204.3 feet ? Ans. 4 1 7 3 8.4 9 sq. ft. 3. How many square yards are there in a rectangle whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31 sq. yds. 4. What is the area of a rectangular board, whose length is 12^ feet, and breadth 9 inches? Ans. 9| sq. ft. 5. What is the number of square yards in a parallelo- gram, whose base is 37 feet, and altitude 5 feet 3 inches? Ans. 21^. To find the area of a plane triangle. 96. First Case. When the base and altitude are given. OF SURFACES. 119 From the principle demonstrated in Book IV., Prop. VI., we may write the following Rule. — Multiply the base by half the altitude ; the product will be the area required. Examples. 1. Find the area of a triangle, whose base is 625, and altitude 520 feet. Ans. 162500 sq. ft. 2. Find the area of a triangle, in square yards, whose base is 40, and altitude 30 feet. .Ans. 66|. 8. Find the area of a triangle, in square yards, whose base is 49, and altitude 25i feet. Ans. 68.7361. Second Case. When two sides and their included angle are given. Let ABC represent a plane triangle, * in which the side AB = c, BC = a, and the angle B, are given. From A draw AD perpendicular to BC ; this will be I the altitude of the triangle. From for- mula (1), Art. 37, Plane Trigonometry, we have AD = c sin B. Denoting the area of the triangle by Q, and applying the rule last given, we have Q^ac^B. ^^_ 2Q = ac sin B. Li Substituting for sin B, —- (Trig., Art. 30), and applying logarithms, we have log {2Q) = log a + log c + log sin B — 10 ; 120 MENSURATION hence, we may write the following Rule. — AddL together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract 10 ; the remainder ivill be the logarithm of double the area of the triangle. Find, from, the table, the number corresponding to this logarithm, and divide it by 2 ; the quotient will be the required area. Examples. 1. What is the area of a triangle, in which two sides, a and h, are respectively equal to 125.81, and 57.65, and whose included angle C is 57° 25'? Ans. 2Q = 6111.4, and Q = 3055.7. 2. What is the area of a triangle, whose sides are 30 and 40, and their included angle 28° 57'? Ans. 290.427. 3. What is the number of square yards in a triangle, of which the sides are 25 feet and 21.25 feet, and their included angle 45°? Ans. 20.8694. LEMMA. To find half an angle, when the three sides of a plane triangle are given. 97. Let ABC be a plane triangle, the c angles and sides being denoted as in the ^^^\\a, figure. ^^ When the angle, A, is acute, we have (B. IV., P. XIL), a2 = 62 + c2 — 2c • AD : but (Art. 37), AD = b cos A ; hence, a' = &2 + c2 — 26c cos A. cD B OF SUEFACES. 121 When the angle A is obtuse, we c have (B. IV., P. XIH.), a« = 68 + c« + 2c ■ AD : but (Art. 37), AD = 6 cos CAD : but the angle CAD is the supplement of the angle A of the given triangle, and, therefore (Art. 63), cos CAD = — cos A ; hence, AD = — 6 cos A, and, consequently, we have a^ — b^ + c^ — 26c cos A. So that whether the angle. A, is acute or obtuse, we have a^ = 62 + c' — 26ccosA; ■ • • • (1.) whence, cos A = 6^ + c^ — g" 26c (2.) If we add 1 to each member, and recollect that 1 + cos A = 2 cos'l^A (Art. 66) equation (4), we have or, 2 cos'' iA = 26c + 6^ + c^ — g^ 26c (6 + cY - a? 2bc — jb + c + a) {b + c — a) ~ 26c cos'^A = (6 + c + g) (& + c — g) 46c (3.) 122 MENSURATION If we put b + c + a = s, we have ' + ^/"-H and 6+o-a_^ a. Substituting in (3), and extracting the square root, COS the plus sign, only, being used, since ^A < 90°; hence, as A represents any angle, Hie cosine of half of any angle of a plane triangle, is equal to the square root of the product of half the sum of the three sides, and half that sum minus the side opposite the angle, divided by the rectangle of the adjacent sides. By applying logarithms, we have log cos iA = i [log is + log (Js — a) + (a. c.) log h + (a. c.) log c]. • (A.) If we subtract each member of equation (2) from 1, and recollect that 1 — cos A = 2 sin' ^A (Art. 66), we have _ g" — (b — c)8 ~ 26c _ (a + b — c) (a — 6 + c) ,, . - Wc ■ ^^-^ OF SUKFACES. Placing, as before, a + b + c = s, , a + b — c , we have — ^ — - = is — c, and a — b + c is-b. 123 Substituting in (5) and reducing, we have sin -^A V ( is - b) (i s - c) . be hence, (6.) The sine of half an angle of a plane triangle, is equal to the square root of the -product of half the sum of the three sides minus one of the adjacent sides and half that sum minus the other adjacent side, divided by the rectan- gle of the adjacent sides. Applying logarithms, we have log sin ifii = i [log (is — b) + log (is — c) + (a. c.) log b + (a. c.) log c]. • (B.) Third Case. To find the area of a triangle when the three sides are given. Let ABC represent a triangle whose sides a, b, and c are given. From the principle demonstrated in the last case, we have Q = ibc sin A. But, from formula (A'), Trig., Art. 66, we have sin A = 2 sin i^A cos -^A ; 124 MENSURATION whence, Q = 6c sin ^A cos i^A. Substituting for sin |A and cos ^A, their values, taken from Lemma, and reducing, -we have Q = Vis iis - a) ihs-l)) (is - c) ; hence, we may write the following Rule. — Find half the sum of the three sides, and from it subtract each side separately. Find the continued prod- uct of the half sum and the three remainders, and extract its square root; the result will be the area required. It is generally more convenient to employ logarithms; for this purpose, applying logarithms to the last equation, we have log Q = i [log \s + log {^s-a) + log (is-6) + log (is-c)] ; hence, we have the following Rule. — Find the half sum and the three remainders as before, then find the half sum of their logarithms; the number corresponding to the resulting logarithm will be the area required. Examples. 1. Find the area of a triangle, whose sides are 20, 80, and 40. "We have ^s = 45, -Js — a = 25, ^s — 6 = 15, ^s — c = 5. By the first rule. Q = V45x25xl5x5 = 290.4737, Ans. OF SURFACES le second rule, log is (45) . . ■ 1.658213 log iis -a) . . (25) . • 1.397940 log (is -b) . . (15) ■ • • • 1.176091 log (is -c) . . (5) . logQ . • 0.698970 2)4.926214 ■ 2.463107 .-. Q = 290.4737, Ans. 125 2. How many square yards are there in a triangle, whose sides are 30, 40, and 50 feet? Ans. 66|. To find the area of a trapezoid. 98. From the principle demonstrated in Book IV., Prop. VII., we may write the following Rule. — Mnd half the sum of the parallel sides, and multiply it by the altitude; the product will be the area required. Examples. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540 ; what is the area? Ans. 1520750. 2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 13 if. 3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet? Ans. 2053i sq. yd. 126 MENSURATION To find the area of any quadrilateral. 99. From what precedes, we deduce the following Rule. — Join the vertices of two opposite angles by a diagonal; from each of the other vertices let fall perpen- diculars upon this diagonal; multiply the diagonal by half of the sum of the perpendiculars, and the product will be the area required. Examples. 1. What is the area of the quad- rilateral ABCD, the diagonal AC being 42, and the perpendiculars Dg, B6, equal to 18 and 16 feet? Ans. 714 sq. ft. 2. How many square yards of paving are there in the quadrilateral, whose diagonal is 65 feet, and the two per- pendiculars let fall on it 28 and 33i^ feet? Ans. 222^. To find the area of any polygon. 100. From what precedes, we have the following Rule. — Draw diagonals dividing the proposed polygon into trapezoids and triangles: then find the area of these figures separately, and add them together for the area of the whole polygon. Example. 1. Let it be required to deter- mine the area of the polygon ABCDE, having five sides. Let us suppose that we have meas- ured the diagonals and perpendiculars, OF SURFACES. 127 and found AC = 36.21, EC = 39.11, B6 = 4, Dd = 7.26, Aa = 4.18 : required the area. Ans. 296.1292. To find the area of a regular polygon. lOl. Let AB, denoted by s, repre- sent one side of a regular polygon whose centre is C. Draw CA and CB, and from C draw CD perpendicular to AB. Then will CD be the apothem, and we shall have AD = BD. Denote the number of sides of the polygon by n; then will the angle ACB, at the centre, be equal to (B. Y., page 144, D. 2), and the angle ACD, 180° which is half of ACB, will be equal to In the right-angled triangle ADC, we shall have, formula (3), Art. 37, Trig., CD = \s tan CAD. But CAD, being the complement of ACD, we have tan CAD = cot ACD ; 180° hence, CD = i^s cot n a formula by means of which the apothem may be com- puted. But the area is equal to the perimeter multiplied by half the apothem (Book V., Prop. VIII.) : hence the fol- lowing Rule. — Find the apothem, by the preceding formula; inultiply the perimeter by half the apothem; the product will be the area required. 128 MENSUEATION Examples. 1. What is the area of a regular hexagon, each of whose sides is 20 ? We have CD = 10 x cot 30° ; or, log CD = log 10 + log cot 30° - 10. log is (10) 1.000000 log cot 180° n log CD (30°) • 10.238561 1.238561 .-. CD = 17.3205. The perimeter is equal to 120 : hence, denoting the area by Q, _ 120x17.3 205 .noQoo a Q = n = 1039.23, Ans. 2. What is the area of an octagon, one of whose sides is 20? Ans. 1931.37. The areas of some of the most important of the regu- lar polygons have been computed by the preceding method, on the supposition that each side is equal to 1, and the results are given in the following TABLE. ITAHES. STDES. ABEAfl. ITAUEB. SIDES. ABEAS. Triangle, . . . 3 . . 0.4330127 Octagon, . . 8 . . . 4.8284271 Square, . . . . 4 . . . 1.0000000 Nonagon, . . 9 . . . 6.1818242 Pentagon, . . . 5 . . . 1.7204774 Decagon, . . 10 . . . 7.6942088 Hexagon, . 6 . . . 2.5980762 TTndecagon, . . 11 . . . 9.3656399 Heptagon, . . . 7 . . . 3.6339124 Dodecagon, . . 12 . . . 11.1061624 OP SUBFACES. 129 The areas of similar polygons are to each other as the squares of their homologous sides (Book IV., Prop. XXVIL). Denoting the area of a regular polygon whose side is s by Q, ■ and that of a similar polygon whose side is 1 by T, the tabular area, we have Q : T : : s« : 12 ; .-. Q - Ts^ hence, the following Rule. — Multiply the corresponding tabular area by the square of the given side ; the product will be the area re- quired. Examples. 1. What is the area of a regular hexagon, each of whose sides is 20? We have T = 2.5980762, and ^ — 400 : hence, Q = 2.5980762x400 = 1039.23048, Ans. 2. Find the area of a pentagon, whose side is 25. Ans. 1075.298375. 3. Find the area of a decagon, whose side is 20. Ans. 3077.68352. To find the circumference of a circle, when the diameter is given. 103. From the principle demonstrated in Book V., Prop. XVI., we may write the following Rule. — Multiply the given diameter by 3.1416; tha product will be the circumference required. 180 MENSURATION Examples. 1. What is the circumference of a circle, whose diam- eter is 25? Ans. 78.54. 2. If the diameter of the earth is 7921 miles, what is the circvmif erence ? Ans. 24884.6136. To find the diameter of a circle, when the circumference is given. 103. From the preceding case, we may write the fol- lowing Rule. — Divide the given circumference by 3.1416 ; the quotient will be the diameter required. Examples. 1. What is the diameter of a circle, whose circumfer- ence is 11652.1944? Ans. 3709. 2. What is the diameter of a circle, whose circumfer- ence is 6850? Ans. 2180.41. To find the length of an arc containing any number of degrees. 104. The length of an arc of 1°, in a circle whose diameter is 1, is equal to the circumference, or 3.1416, divided by 360; that is, it is equal to 0.0087266.: hence, the length of an arc of n degrees will be nx 0.0087266. To find the length of an arc containing n degrees, when the diameter is d, we employ the principle demonstrated in Book V., Prop. XIII., 0. 2 : hence, we may write the following OF SURFACES. 131 Rule — Multiply the numiier of decrees in the arc by .0087266, and the product by the diameter of the circle; the result vMl be the length required. Examples. 1. What is the length of an arc of 30 degrees, the diameter being 18 feet? Ans. 4.712364 ft. 2. What is the length of an arc of 12° 10', or 12^°, the diameter being 20 feet? Ans. 2.123472 ft. To find the area of a circle. 105. From the principle demonstrated in Book V., Prop. XV., we may write the following Rule. — Multiply the square of the radius by 3.1416 ; the product vuill be the area required; Examples. 1. Find the area of a circle, whose diameter is 10 and circumference 31.416. Ans. 78.54. 2. How many square yards in a circle whose diameter is 3i feet? Ans. 1.069016. 3. What is the area of a circle whose circumference is 12 feet? Ans. 11.4595. To f,nd the area of a circular sector. 106. From the principle demonstrated in Book V., Prop. XIV., 0. 1 and 2, we may write the following Rule.— I. Multiply half the length of the arc by the ra- dius ; or. 132 MENSURATION n. Find, the area of the whole circle, by the last rule; then write the proportion, 360 is to the number of degrees in the arc of the sector, as the area of the circle is to the area of the sector. Examples. 1. Find the area of a circular sector, whose arc con- tains 18°, the diameter of the circle being 8 feet. Ans. 0.35343 sq. ft. 2. Find the area of a sector, -whose arc is 20 feet, the radius being 10. Ans. 100. 3. Eequired the area of a sector, whose arc is 147° 29' and radius 25 feet. Ans. 804.3986 sq. ft. To find the area of a circular segment. 107. Let AB represent the chord corresponding to the two segments ACB and AFB. Draw AE and BE. The seg- ment ACB is equal to the sector EACB, minus the triangle AEB. The segment AFB is equal to the sector EAFB, plus the triangle AEB. Hence, we have the following Rule. — Find the area of the corresponding sector, and also of the triangle formed by the chord of the segment and the two extreme radii of the sector; subtract the latter from the former when the segment is less than a semicir- cle, and add the latter to the former when the segment is greater than a semicircle; the result will be the area re- quired. OF SUEFACES. 133 Examples. 1. Find the area of a segment, whose chord is 12 and whose radius is 10. Solving the triangle AEB, we find the angle AEB is equal to 73° 44', the area of the sector EACB equal to 64.35, and the area of the triangle AEB equal to 48 ; hence, the segment ACB is equal to 16.85. 2. Find the area of a segment, whose height is 18, the diameter of the circle being 50. Ans. 636.4834. 3. Required the area of a segment, whose chord is 16, the diameter being 20. Ans. 44.764. To find the area of a circular ring contained between the circumferences of two concentric circles. 108. Let R and r denote the radii of the two circles, R being greater than r. The area of the outer circle is R'x 3.1416, and that of the inner circle is r^x 3.1416; hence, the area of the ring is equal to (R* — r') x 3.1416. Hence, the following Rule. — Find the difference of the squares of the radii of the two circles, and multiply it by 3.1416 ; the -product ivill be the area required. Examples. 1. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Ans. 50.2656. 2. What is the area of the ring, when the diameters of the circles are 10 and 20? Ans. 235.62. 134 MENSURATION MENSURATION OF BROKEN AND CURVED SUR- FACES. To find, the area of the entire surface of a right prism. 109. From the principle demonstrated in Book VII., Prop. L, "we may write the following Rule. — Multiply the perimeter of the base by the altitude, the product will be the area of the convex surface; to this add the areas of the two bases; the result will be the area required. Examples. 1. Find the surface of a cube, the length of each side being 20 feet. Ans. 2400 sq. ft. 2. Find the whole surface of a triangular prism, whose base is an equilateral triangle having each of its sides equal to 18 inches, and altitude 20 feet. Ans. 91.949 sq. ft. To find the area of the entire surface of a right pyramid. 110. From the principle demonstrated in Book VII., Prop. IV., we may write the following Rule. — Multiply the perimeter of the base by half the slant height; the product will he the area of the convex- surface; to this add the area of the base; the result will he the area required. Examples. 1. Find the convex surface of a right triangular pyra- mid, the slant height being 20 feet, and each side of the base 3 feet. Ans. 90 sq. ft. OF SURFACES. 186 2. What is the entire surface of a right p3rramid, whose slant height is 27 feet, and the base a pentagon of which each side is 25 feet? Ans. 2762.798 sq. ft. To find the area of the convex surface of a frustum of a right pyramid. 111. From the principle demonstrated in Book VII., Prop. IV., S., we may write the following Rule. — Multiply the half sum of the perimeters of the two bases by the slant height; the product will be the area required. Examples. 1. How many square feet are there in the convex sur- face of the frustum of a square pyramid, whose slant height is 10 feet, each side of the lower base 3 feet 4 inches, and each side of the upper base 2 feet 2 inches? Ans. 110 sq. ft. 2. What is the convex surface of the frustum of a heptagonal pyramid, whose slant height is 56 feet, each side of the lower base 8 feet, and each side of the upper base 4 feet? Ans. 2310 sq. ft. 113. Since a cylinder may be regarded as a prism whose base has an infinite number of sides, and a cone as a pjrramid whose base has an infinite number of sides, the rules just given may be applied to find the areas of the surfaces of right cylinders, cones, and frustums of cones, by simply changing the term perimeter to circumfer- ence. 136 MENSURATION Examples. 1. What is the convex surface ,of a cyhnder, the diam- eter of whose base is 20, and whose altitude 50? Ans. 3141.6. 2. What is the entire surface of a cyhnder, the alti- tude being 20, and diameter of the base 2 feet? Ans. 131.9472 sq. ft. 3. Required the convex surface of a cone, whose slant height is 50 feet, and the diameter of its base 8^- feet. Ans. 667.59 sq. ft. 4. Required the entire surface of a cone, whose slant height is 36, and the diameter of its base 18 feet. Ans. 1272.348 sq. ft. 5. Find the convex surface of the frustum of a cone, the slant height of the frustum being 12^ feet, and the circumferences of the bases 8.4 feet and 6 feet. Ans. 90 sq. ft. 6. Find the entire surface of the frustum of a cone, the slant height being 16 feet, and the radii of the bases 3 feet and 2 feet. Ans. 292.1688 sq. ft. To find the area of the surface of a sphere. 113. From the principle demonstrated in Book VIIL, Prop. X., 0. 1, we may write the following Rule. — Find the area of one of its great circles, and multiply it hy ^] the product will be the area required. Examples. 1. What is the area of the sTirface of a sphere, whose radius is 16? Ans. 3216.9984. OF SURFACES. 137 2. What is the area qt the surface of a sphere, whose radius is 27.25? Ans. 9331.3374. To find the area of a zone. 114. From the principle demonstrated in Book VIII., Prop. X., C. 2, we may write the following Rule. — Mnd the circumference of a great circle of the sphere, and -multiply it by the altitude of the zone; the product will be the area required. Examples. 1. The diameter of a sphere being 42 inches, what is the area of the surface of a zone whose altitude is 9 inches ? Ans. 1187.5248 sq. in. 2. If the diameter of a sphere is 12^ feet, what will be the surface of a zone whose altitude is 2 feet? Ans. 78.54 sq, ft. To find the area of a spherical polygon. 115. From the principle demonstrated in Book IX., Prop. XIX., we may write the following Rule. — From the sum of the angles of the polygon, sub- tract 180° taken as many tim,6S, less two, as the polygon has sides, and divide the remainder by 90°; the quotient will be the spherical excess. Find the area of a great cir- cle of the sphere, and divide it by 2 ; the quotient will be the area of a tri-rectangulnr triangle. Multiply the area of the tri-rectangular triangle by the spherical excess, and the product will be the area required. 138 MENSURATION This rule applies to the spherical triangle, as well as to any other spherical polygon. Examples. 1. Required the area of a triangle, described on a sphere whose diameter is 30 feet, the angles being 140°, 92°, and 68°. Ans. 471.24 sq. ft. 2. What is the area of a polygon of seven sides, de- scribed on a sphere whose diameter is 17 feet, the sum of the angles being 1080°? Ans. 226.98. 3. What is the area of a regular polygon of eight sides, described on a sphere whose diameter is 30 yards, each angle of the polygon being 140°? Ans. 157.08 sq. yds. MENSURATION OF VOLUMES. To find the volume of a prism. 116. From the principle demonstrated in Book VII., Prop. XrV., we may write the following Rule. — Multiply the area of the hose by the altitude; the product will be the volume required. Examples. 1. What is the volume of a cube, whose side is 24 inches? Ans. 13824 cu. in. 2. How many cubic feet in a block of marble, of which the length is 3 feet 2 inches, breadth 2 feet 8 inches, and height or thickness 2 feet 6 inches? Ans. 21i cu. ft OF VOLUMES. 189 3. Required the volume of a triangular prism, whose height is 10 feet, and the three sides of its triangular base 3, 4, and 5 feet. Ans. 60. To find the volume of a pyramid. 117. From the principle demonstrated in Book VII., Prop. XVn., we may write the following Rule. — Multiply the area of the base by one third of the altitude; the product will be the volum^e required. Examples. 1. Required the volume of a square pyramid, each side of its base being 30, and the altitude 25. Ans. 7500. 2. Find the volume of a triangular pyramid, whose altitude is 30, and each side of the base 3 feet. Ans. 38.9711 cu. ft. 3. What is the volume of a pentagonal pyramid, its altitude being 12 feet, and each side of its base 2 feet? Ans. 27.5276 cu. ft. 4. What is the volume of a hexagonal p3rramid, whose altitude is 6.4 feet, and each side of its base 6 inches? Ans. 1.38564 cu. ft. To find the volume of a frustum of a pyramid. 118. From the principle demonstrated in Book VII., Prop. XVIIL, C, we may write the following Rule. — Find the sum of the upper base, the lower base, and a mean proportional between them,; multiply the re- sult by one third of the altitude; the product will be the volume required. 140 MENSURATION Examples. 1. Find the number of cubic feet in a piece of timber, whose bases are squares, each side of the lower base being 15 inches, and each side of the upper base 6 inches, the altitude being 24 feet. Ans. 19.5. 2. Required tne volume of a pentagonal frustum, whose altitude is 5 feet, each side of the lower base 18 inches, and each side of the upper base 6 inches. Ans. 9.31925 cu. ft. 119. Since cylinders and cones are limiting cases of prisms and pyramids, the three preceding rules are equally applicable to them. Examples. 1. Required the volume of a cylinder whose altitude is 12 feet, and the diameter of its base 15 feet. Ans. 2120.58 cu. ft. 2. Required the volume of a cylinder whose altitude is 20 feet, and the circumference of whose base is 5 feet 6 inches. Ans. 48.144 cu. ft. 3. Required the volume of a cone whose altitude is 27 feet, and the diameter of the base 10 feet. Ans. 706.86 cu. ft. 4. Required the volume of a cone whose altitude is 10^ feet, and the circumference of its base 9 feet. Ans. 22.56 cu. ft. 5. Find the volume of the frustum of a cone, the altitude being 18, the diameter of the lower base 8, and that of the upper base 4. Ans. 527.7888. or VOLUMES. 141 6. What is the volume of the frustum of a cone, the altitude being 25, the circumference of the lower base 20, and that of the upper base 10? Ans. 464.216. 7. If a cask, which is composed of two equal conic frustums joined together at their larger bases, have its bung diameter 28 inches, the head diameter 20 inches, and the length 40 inches, how many gallons of wine will it contain, there being 231 cubic inches in a gallon? Ans. 79.0613. To find the volume of a sphere. 120. From the principle demonstrated in Book VIIL, Prop. XIV., we may write the following Rule. — Cube the diameter of the sphere, and multiply the result by ^n, that is, by 0.5236 ; the product will be the volum,e required. Examples. 1. What is the volume of a sphere, whose diameter is 12? Ans. 904.7808. 2. What is the volume of the earth, if the mean diameter is taken equal to 7918.7 miles? Ans. 259992792082 cu. miles. To find the volume of a wedge. 131. A Wedge is a volume bounded by a rectangle A BCD, called the hack, two trapezoids ABHG, DCHG, called faces, and two triangles ADG, CBH, called ends. The line GH, in which the faces meet, is called the edge. 142 MENSURATION There are three cases ; 1st, When the length of the edge is equal to the length of the back; 2d, When it is less; and 3d, When it is greater. In the first case, the wedge is equal in volume to a right prism, whose base is the triangle ADG, and altitude GH or AB : hence, its volume is equal to ADG multiplied by AB. In the second case, through H, a point of the edge, pass a plane HCB perpendicular to the back, and intersecting it in the line BC parallel to AD. This plane will divide the wedge into two parts, one of which is represented by the figure. Through G, draw the plane GNM parallel to HCB, and it wiU divide the part of the wedge represented by the figure into the right triangular prism GNM-B, and the quadrangular pyramid ADNM-G. Draw GP perpendicular to N M : it win also be perpendicular to the back of the wedge (B. VI., P. XVII.), and hence, will be equal to the altitude of the wedge. Denote AB by L, the breadth AD by 6, the edge GH by /, the altitude by h, and the volume by V; then, AM = L - Z, MB = GH = I, and area NGM = ^bh: OF VOLUMES. 143 then Prism = ibhl; Pyramid = b{L-l)ih = ^bh (L - I), and V = ^bhl + ^bh (L — Z) = ibhl + ^bhL - :^hl = ibh(l + 2L). We can find a similar expression for the remaining part of the wedge, and by adding, the factor within the parenthesis becomes the entire length of the edge plus twice the length of the back. G H In the third case, I is greater M than L; the volume of each part //"MX is equal to the difference of the //; 1\ prism and pyranaid, and is of the N^y-i-ifiV , same form as before. Hence, in 'a/ ^ either case, we have the following M A Rule. — Add twice the length of the bach to the length of the edge; multiply the sum by the breadth of the back, and that result by one sixth of the altitude; the final product will be the volume required. Examples. 1. If the back of a wedge is 40. by 20 feet, the edge 35 feet, and the altitude 10 feet, what is the volume? Ans. 3833.83 cu. ft. 2. What is the volume of a wedge, whose back is 18 feet by 9, edge 20 feet, and altitude 6 feet? Ans. 504 cu. ft. 144 MENSURATION To find the volume of a prismoid. 133. A Prismoid is a frustum of a wedge. Let L and B denote the length and breadth of the lower base, I and b the length and breadth of the upper base, M and m the length and breadth of the section equidistant from the bases, and h the altitude of the prismoid. Through the edges L and I', let a plane be passed, and it will divide the prismoid into two wedges, having for bases the bases of the prismoid, and for edges the lines L and I'. The volume of the prismoid, denoted by V, will be equal to the sum of the volumes of the two wedges; hence, I' l'\b . \ 1 V._._.... _J \ l\m] . \ 1 ' * ^/ . M or, V = iBh{l + 2L) +ibh(L+ 21); V = -JA (2BL + 2bl + Bl + 6L) ; which may be written under the form, y ^ ih [(BL + bl + Bl + 6L) + BL + bl]. • (A.) Because the auxiliary section is midway between the bases, we have 2M = L + I, and 2m = B + 6; hence, 4Mm = (L + I) {B + b) = BL + Bl + bL + bl. Substituting in (A), we have y = ih (BL + bl + 4Mm). OF VOLUMES. 145 But BL is the area of the lower base, or lower section, bl is the area of the upper base, or upper section, and Mm is the area of the middle section ; hence, the fol- lowing Rule. — To find the volume of a prismoid, find the sum of the areas of the extreme sections and four times the middle section; multiply the result by one sixth of the dis- tance between the extreme sections; the result will he the volume required. This rule is used in computing volumes of earth-work in railroad cutting and embankment, and is of very ex- tensive application. It may be shown that the same rule holds for every one of the volumes heretofore discussed in this work. Thus, in a pyramid, we may regard the base as one extreme section, and the vertex (whose area is 0), as the other extreme ; their sum is equal to the area of the base. The area of a section midway between them is equal to one fourth of the base : hence, four times the middle section is equal to the base. Multiply- ing the sum of these by one sixth of the altitude, gives the same result as that already found. The application of the rule to the case of cylinders, frustums of cones, spheres, &c., is left as an exercise for the student. Examples. 1. One of the bases of a rectangular prismoid is 25 feet by 20, the other 15 feet by 10, and the altitude 12 feet: required the volume. Ans. 3700 cu. ft. 2. What is the volume of a stick of hewn timber, whose ends are 30 inches by 27, and 24 inches by 18, ivc lAntrth heinff 24 feet? Ans. 102 cu. ft. 146 MENSURATION MENSURATION OF REQULAR POLYEDRONS. 133. A Regular Polyedron is a polyedron bounded by- equal regular polygons. The polyedral angles of any regular polyedron are all equal. 134. There are five regular polyedrons (Book VII., page 219). To find the diedral arigle contained between two consecutivp- faces of a regular -polyedron. 135. As in the figure, let the ver- tex, 0, of a polyedral angle of a tetraedron be taken as the centre of a sphere whose radius is 1 : then -will the three faces of this polyedral angle, by their intersections with the surface of the sphere, determine the spherical triangle FAB. The plane angles FOA, FOB, and AOB, being equal to each other, the arcs FA, FB, and AS, which meas- ure these angles, are also equal to each other, and the spherical triangle FAB is equilateral. The angle FAB of the triangle is equal to the diedral angle of the planes FOA and AOB, that is, to the diedral angle between the faces of the tetraedron. In like manner, if the vertex of a polyedral angle of any one of the regular polyedrons be taken as the centre of a sphere whose radius is 1, the faces of this polyedral angle will, by their intersections with the surface of the sphere, determine a regular spherical polygon ; the number of sides of this spherical polygon will be equal to the OF POLTEDEONS. 147 number of faces of the polyedral angle ; each side of the polygon will be the measure of one of the plane angles formed by the edges of the polyedral angle ; and each angle of the polygon will be equal to the diedral angle contained between two consecutive faces of the regular polyedron. To find the required diedral angle, therefore, it only remains to deduce a formula for finding one angle of a regular spherical polygon when the sides are given. Let ABCDE represent a regular spherical polygon, and let P be the pole of a small cu'cle passing through its vertices. Suppose P to be connected with each of the vertices by arcs of great circles ; there will thus be formed as many equal isosceles triangles as the polygon has sides, the vertical angle in each being equal to 360° divided by the number of sides. Through P draw the arc of a great circle, PQ, perpendicular to AB: then will AQ be equal to BQ, and the angle APQ to the angle QPB (B. IX., P. XI., C). If we denote the number of sides of the spherical polygon by n', the angle APQ will be equal to 360° 180° or 2»' ' n' In the right-angled spherical triangle AQP, we know the base AQ, and the vertical angle APQ; hence, by Napier's rules for circular parts, we have sin (90° - APQ) = cos (90° - PAQ) cos AQ, or, cos APQ = sin PAQ cos AQ ; denoting the side AB of the polygon by s', and the angle PAQ, which is half the angle EAB of the polygon, by ^A, we have 148 MENSURATION 180° ... COS — — = sm iA cos is ; 180° cos whence, sin ^A = In the Tetraedron, 180° cos |s' Examples. 60°, and ^ = 30°; .". A = 70° 31' 42". n In the Hexaedron, ^- = 60°, and Js' = 45° ; .-. A = 90°. In the Octaedron, ^- = 45°, and K = 30°; .-. A = 109° 28' 19", In the Dodecaedron, i^ = 60°, and is' = 54°; .-. A = 116° 68' 54". In the Icosaedron, 1 80° ^- = 36° and is' = 30°; .". A = 138° 11' 23". To find the volume of a regular polyedron. 136i If planes be passed through the centre of the polyedron and each of the edges, they will divide the polyedron into as many equal right pyramids as the poly- edron has faces. The common vertex of these pyramids will be at the centre of the polyedron, their bases will be the faces of the polyedron, and their lateral faces will bisect the diedral angles of the polyedron. The volume of each pyramid will be equal to the product of its base and one third of its altitude, and this product multiplied ^ HS^ lU^- •\ ■■ .A 0... " ./ — - yy OF POLTEDBONS. 149 by the number of faces, will be the volume of the poly- edron. It only remains to deduce a formula for finding the altitude of the several pyramids, i. e., the distance from the centre to one face of the polyedron. Conceive a perpendicular OC to be drawn from 0, the centre of the poly- edron, to one face ; the foot of this per- pendicular will be the centre of the face. From C, the foot of this perpendicular, draw a perpendicular to one side of the face in which it lies, and connect the point D with the centre of the polyedron. There will thus be formed a right-angled triangle, OCD, whose base, CD, is the apothem of the face, whose angle ODC is half the angle CDL con- tained between two consecutive faces of the polyedron, and whose altitude OC is the required altitude of the pyramid, or, in other words, the radius of the inscribed sphere. This will be true for any one of the regular polyedrons — the hexaedron is taken here for simplicity of illustration. Denote the line CD by j3, the angle ODC by |A, and the perpendicular OC by R. p may be found by the formula, given in Art. 101, for finding the apothem of a regular polygon ; ^A may be found from the formula for sin ^A, given in Art. 125; then, in the right-angled triangle OCD, we have, formula (3), Art. 37, R = p tan |A. Compute the area of one of the faces of the given polyedron and multiply it by ^R, as determined by the formula just given, and multiply the result thus obtained by the number of faces of the polyedron ; the final product will be the volume of the given regular polyedron. 150 MENS UE ATION. The volumes of all the regular polyedrons have been computed on the supposition that their edges are each equal to 1, and the results are given in the following TABLE. HAMXa. NO. or FAOBB. VOLITMEB. Tetraedron, • • . . 4 . . • ■ 0.1178513 Hexaedron, • • . . 6 • • • • 1.0000000 Octaedron, • • . . 8 • • • • 0.4714045 Dodecaedron, • • • ■ 12 ■ ■ • . 7.6631189 Icosaedron, • ■ . . 20 • • • • 2.1816950 From the principles demonstrated in Book VII., we may write the following Rule. — To find the volume of any regular polyedron, multiply the cube of its edge by the corresponding tabular volume; the product will he the volume required. Examples. 1. What is the volume of a tetraedron, whose edge is 15? Ans. 397.75. 2. What is the volume of a hexaedron, whose edge is 12? Ans. 1728. 3. What is the volume of an octaedron, whose edge is 20? A-ns. 3771.236. 4. What is the volume of a dodecaedron, whose edge is 25? Ans. 119736.2328. 5. What is the volume of an icosaedron, whose edge is 20? Ans. 17453.56. A TABLE LOGARITHMS OF NUMBERS FROM 1 TO 10,000. N. JjOg. N. Log. N. Log. N. Log. 1 0-000000 26 1-414973 51 1-707670 76 1-880814 2 0-301030 27 1-431364 62 1-716003 77 1-886491 3 0-477121 28 1-447158 53 1-724276 78 1-892095 4 0-602060 29 1-462398 64 1-732394 79 1-897627 6 0-698970 SO 1-477121 55 1-740363 80 1-903090 6 0-778151 31 1-491362 66 1-748188 81 1-908485 7 0-845098 82 1-505150 57 1-755876 82 1-913814 8 0-903090 33 1-518514 58 1-763428 83 1-919078 9 0-954243 34 1-531479 59 1-770852 84 1-924279 10 1-000000 35 1-544068 60 1-778151 85 1-929419 11 1-041393 36 1-556303 61 1-786330 86 1-934498 12 1-079181 87 1-568202 62 1-792392 87 1-939319 13 1-113948 38 1-579784 63 1-799341 88 1-944483 14 1-146128 39 1-591065 64 1-806181 89 1-949390 15 1-176091 40 1-602060 65 1-812913 90 1-964243 16 1-204120 41 1-612784 66 1-819544 91 1-969041 17 1-230449 42 1-623249 67 1-826076 92 1-963788 18 1-255273 43 1-633468 68 1-832509 93 1-968483 19 1-278754 44 1-643453 69 1-838849 94 1-973128 20 1-301030 45 1-653213 70 1-845098 95 1-977724 21 1-322219 46 1-662758 71 1-861268 96 1-982271 22 1-342423 47 1-672098 72 1-857333 97 1-986772 23 1-361728 48 1-681241 73 1-863323 98 1-991226 24 1-880211 49 1-690196 74 1-869232 99 1-995635 25 1-397940 50 1-698970 75 1-875061 100 2-000000 Remarks. In the following table, in the nine right-hand columns of each page, where the first or leading figures change from 9's to O's, points or dots are introduced instead of the O's, to catch the eye, and to indicate that from thence the two figures of the Logarithm to be taken from the second column, stand in the next line below. A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. 1 1 2 3 4 ' G 7 .8 9 D. 100 000000 0434 0868 1301 1734 2166 2598 3029 3461 3891 432 101 4321 4751 5181 6609 6038 6466 6894 7321 7748 8174 428 102 8600 9026 9451 9876 ♦ 800 ♦ 724 1147 1570 1993 2415 424 103 012837 3259 3680 4100 4521 4940 5360 5779 6197 6616 419 104 7083 7451 7868 8284 8700 9116 9532 9947 ♦ 361 ♦ 775 416 105 021189 1603 2016 2428 2841 3252 3664 4075 4486 4896 412 106 5306 5715 6125 6533 6942 7350 7757 8164 8571 8978 408 107 9384 9789 ♦ 195 ♦ 600 1004 1408 1812 2216 2619 3021 404 108 033424 3826 4227 4628 5029 5480 5830 6230 6629 7028 400 109 7426 7825 8223 8620 9017 9414 9811 ♦ 207 ♦ 602 ♦ 998 396 110 041393 1787 2182 2576 2969 3362 3755 4148 4540 4932 393 111 5323 5714 6105 6495 6885 7275 7664 8053 8442 8830 389 112 9218 9606 9993 ♦ 380 ♦ 766 1153 1538 1924 2309 2694 386 113 053078 3463 3846 4230 4613 4996 5378 5760 6142 6524 382 114 6905 7286 7666 8046 8426 8805 9185 9563 9942 ♦ 320 379 115 060698 1075 1452 1829 2206 2582 2958 3333 3709 4083 376 116 4458 4832 5206 5580 5953 6326 6699 7071 7443 7815 372 117 8186 8557 8928 9298 9668 ♦♦38 ♦407 ♦776 1145 1514 369 118 071882 2250 2617 2985 3352 3718 4085 4451 4816 5182 366 119 5547 5912 6276 6640 7004 7368 7731 8094 8457 8819 363 120 079181 9543 9904 ♦266 ♦ 626 ♦ 987 1347 1707 2067 2426 360 121 082785 3144 3503 3861 4219 4576 4984 5291 5647 6004 357 122 6360 6716 7071 7426 7781 8136 8490 8845 9198 9552 355 123 9905 ♦ 258 ♦ 611 ♦ 963 1315 1667 2018 2370 2721 3071 351 124 093422 3772 4122 4471 4820 5169 5518 5866 6215 6562 349 125 6910 7257 7604 7951 8298 8644 8990 9335 9681 ♦♦26 346 126 100371 0715 1059 1403 1747 2091 2434 2777 3119 3462 343 127 3804 4146 4487 4828 5169 5510 5851 6191 6531 6871 340 128 7210 7549 7888 8227 8565 8903 9241 9579 9916 ♦ 253 338 129 110590 0926 1263 1599 1934 2270 2605 2940 3275 3609 385 130 113943 4277 4611 4944 5278 5611 5943 6276 6608 6940 333 131 7271 7603 7934. 8265 8595 8926 9256 9586 9915 ♦ 245 330 132 120574 0903 1231 1560 1888 2216 2544 2871 3198 3525 328 133 3852 4178 4504 4830 5156 5481 5806 6131 6456 6781 825 134 7105 7429 7753 8076 8399 8722 9045 9368 9690 ♦ ♦12 323 135 130334 0655 0977 1298 1619 1939 2260 2580 2900 3219 321 136 3539 3858 4177 4496 4814 5133 5451 5769 6086 6403 318 137 6721 7037 7354 7671 7987 8303 8618 8934 9249 9564 315 138 9879 ♦ 194 ♦ 508 ♦ 822 1136 1450 1763 2076 2389 2702 314 139 143015 3327 3639 3951 4263 4574 4885 5196 5507 5818 311 140 1461^8 6438 6748 7058 7867 7676 7985 8294 8603 8911 309 141 9219 9527 9835 ♦ 142 ♦449 ♦756 1063 1370 1676 1982 307 142 152288 2594 2900 3205 3510 3815 4120 4424 4728 5032 305 143 5336 5640 5948 6246 6549 6852 7154 7457 7759 8061 303 144 8362 8664 8965 9266 9567 9868 ♦168 ♦469 ♦ 769 1068 301 145 161368 1667 1967 2268 2564 2863 3161 3460 3758 4055 299 146 4353 4650 4947 5244 5541 5838 6134 6430 6726 7022 297 147 7317 7613 7908 8203 8497 8792 9086 9380 9674 9968 295 148 170262 0555 0848 1141 1434 1726 2019 2311 2603 2895 293 149 3186 3478 3769 4060 4351 4641 4932 5222 5512 5802 291 150 176091 6381 6670 6959 7248 7536 7825 8113 8401 8689 289 151 8977 9264 9552 9839 ♦ 126 ♦ 413 ♦ 699 ♦ 985 1272 1558 287 152 181844 2129 2415 2700 2985- 3270 3555 3839 4123 4407 285 153 4691 4975 5259 5542 5825 6108 6391 6674 6956 7239 283 154 7521 7803 8084 8366 8647 8928 9209 9490 9771 ♦♦51 281 155 190332 0612 0892 1171 1451 1730 2010 2289 2567 2846 279 156 3125 3403 3681 3959 4237 4514 4792 5069 5346 5623 278 157 5899 6176 6453 6729 7005 7281 7556 7832 8107 8382 276 158 8657 8932 9206 9481 9755 ♦ ♦29 ♦ 303 ♦ 577 ♦ 850 1124 274 159 201397 1 1670 1943 2216 2488 2761 8033 3305 8577 3848 272 ]V. 1 1 2 3 4 5 6 7 8 9 D. A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. I 2 3 4 5 6 "^ 1 8 9 D. 160 204120 4391 4663 4934 5204 5475 5746 6016 6286 6556 271 161 6826 7096 7365 7634 7904 8173 8441 8710 8979 9247 269 162 9515 9783 ♦♦51 ♦319 ♦686 ♦ 853 1121 1388 1654 1921 267 163 212188 2464 2720 2986 8252 8518 3783 4049 4314 4579 266 164 4844 5109 5373 6638 5902 6166 6430 6694 6957 7221 264 165 7484 7747 8010 8273 8636 8798 9060 9323 9685 9846 262 166 220108 0370 0631 0892 1153 1414 1675 1936 2196 2456 261 167 2716 2976 3236 3496 3755 4015 4274 4533 4792 5051 259 168 B309 5568 5826 6084 .6342 6600 6858 7115 7372 7630 268 169 7887 8144 8400 8657 8913 9170 9426 9682 9938 ♦ 193 266 170 230449 0704 0960 1215 1470 1724 1979 2234 2488 2742 264 171 2996 3250 3504 3757 4011 4264 4517 4770 5023 5276 253 172 5528 5781 6033 6285 6537 6789 7041 7292 7544 7795 252 173 8046 8297 8548 8799 9049 9299 9550 9800 ♦♦50 ♦300 250 174 240549 0799 1048 1297 1546 1795 2044 2293 2541 2790 249 175 3038 3286 3534 3782 4030 4277 4525 4772 5019 5266 248 176 5513 5759 6006 6252 6499 6745 6991 7237 7482 7728 246 177 7973 8219 8464 8709 8954 9198 9443 9687 9932 ♦176 245 178 250420 0664 0908 1151 1395 1638 1881 2125 2368 2610 243 179 2853 3096 3338 3580 3822 4064 4806 4548 4790 5031 242 180 255273 5514 5755 5996 6237 6477 6718 6968 7198 7439 241 181 7679 7918 8158 8898 8637 8877 9116 9355 9594 9833 239 182 260071 0310 0548 0787 1025 1263 1501 1739 1976 2214 288 183 2451 2688 2925 3162 8399 3636 3873 4109 4346 4682 237 184 4818 5054 5290 5525 6761 6996 6232 6467 6702 6937 235 185 7172 7406 7641 7875 8110 8344 8578 8812 9046 9279 234 186 9513 9746 9980 ♦213 ♦446 ♦ 679 ♦912 1144 1377 1609 233 187 271842 2074 2306 2538 2770 3001 3233 8464 8696 3927 282 188 4158 4889 4620 4850 6081 6311 5542 5772 6002 6232 230 189 6462 6692 6921 7151 7380 7609 7838 8067 8296 8525 229 190 278754 8982 9211 9439 9667 9895 ♦ 123 ♦ 351 ♦578 ♦806 228 191 281033 1261 1488 1715 1942 2169 2396 2622 2849 3075 227 192 3301 3527 3753 3979 4205 4431 4656 4882 5107 5332 226 193 5557 5782 6007 6232 6456 6681 6905 7130 7354 7578 225 194 7802 8026 8249 8473 8696 8920 9143 9366 9589 9812 223 195 290035 0257 0480 0702 0925 1147 1369 1591 1813 2034 222 196 2256 2478 2699 2920 3141 3363 3584 3804 4026 4246 221 197 4466 4687 4907 5127 5347 5567 5787 6007 6226 6446 220 198 6665 6884 7104 7323 7542 7761 7979 8198 8416 8635 219 199 8853 9071 9289 9507 9725 9943 ♦ 161 ♦ 378 ♦ 595 ♦813 218 200 301030 1247 1464 1681 1898 2114 2331 2547 2764 2980 217 201 3196 3412 3628 3844 4059 4275 4491 4706 4921 6136 216 202 5351 5566 5781 5996 6211 6425 6639 6854 7068 7282 216 203 7496 7710 7924 8137 8851 8564 8778 8991 9204 9417 213 204 9630 9843 ♦♦56 ♦268 ♦481 ♦ 693 ♦906 1118 1330 1642 212 205 311754 1966 2177 2389 2600 2812 3023 3234 3446 8666 211 206 3867 4078 4289 4499 4710 4920 5130 5340 5651 5760 210 207 5970 6180 6390 6599 6809 7018 7227 7436 7646 7854 209 : 208 8063 8272 8481 8689 8898 9106 9314 9622 9730 9938 208 209. 820146 0354 0562 0769 0977 1184 1391 1698 1805 2012 207 210 322219 2426 2633 2839 3046 3262 3468 3665 8871 4077 206 211 4282 4488 4694 4899 5105 6310 5516 6721 5926 6131 205 212 6336 6541 6745 6950 7155 7359 7563 7767 7972 8176 204 213 8380 8588 8787 8991 9194 9398 9601 9805 ♦♦♦8 ♦211 203 214- 330414 0617 0819 1022 1225 1427 1630 1832 2034 2236 202 215 .2438 2640 2842 3044^ 3246 3447 3649 3860 4051 4263 202 216 4454 4655 4856 5057 6257 5458 6658 5859 6059 6260 201 217 6460 6660 6860 7060 7260 7459 7659 7858 8058 8257 200 218 8456 8656 8855 9054 9253 9451 9650 9849 ♦ ♦47 ♦ 246 199 219 340444 0642 0841 1039 1237 1435 1632 1830 2028 2225 198 N. 1 3 3 4 5 6 7 8 9 D. A TABLE OF LOGAKITHMS FEOM 1 TO 10,000. N. 1 2 3 4 5 6 r 8 9 D. 220 342423 2620 2817 3014 3212 3409 3606 3802 3999 4196 197 221 4392 4589 4785 4981 6178 5374 6570 5766 5962 6157 196 222 6353 6649 6744 6939 7185 7330 7625 7720 7916 8110 195 223 8305 8500 8694 8889 9083 9278 9472 9666 9860 ♦ ♦54 194 224 350248 0442 0636 0829 1028 1216 1410 1603 1796 1989 193 225 2183 2375 2568 2761 2954 3147 3339 3532 3724 3916 193 226 4103 4301 4493 4685 4876 6068 6260 5452 6643 5834 192 227 6026 6217 6408 6599 6790 6981 7172 7363 7554 7744 191 228 7935 8125 8316 8506 3696 8886 9076 9266 9456 9646 190 229 9835 **25 ♦215 ♦404 ♦ 593 ♦ 783 ♦972 1161 1350 1639 189 230 361728 1917 2105 2294 2482 2671 2859 3048 3236 3424 188 231 3612 3800 8988 4176 4363 4551 4739 4926 5113 5301 188 232 5488 5675 5862 6049 6236 6423 6610 6796 6983 7169 187 233 7356 7542 7729 7915 8101 8287 8473 8659 8845 9030 186 234 9216 9401 9587 9772 9958 ♦ 143 ♦328 ♦613 ♦ 698 ♦883 185 235 371068 1253 1437 1622 1806 1991 2176 2360 2544 2728 184 236 2912 3096 3280 3464 3647 3831 4015 4198 4382 4565 184 237 4748 4932 5115 6298 5481 5664 5846 6029 6212 6394 183 238 6577 6759 6942 7124 7306 7488 7670 7852 8034 8216 182 239 8398 8580 8761 8943 9124 9306 9487 9668 9849 ♦ ♦30 181 240 380211 0392 0573 0754 0934 1115 1296 1476 1656 1837 181 241 2017 2197 2377 2557 2737 2917 3097 3277 8456 3636 180 242 8815 3995 4174 4353 4533 4712 4891 5070 5249 5428 179 243 6600 5785 6964 6142 6321 6499 6677 6856 7034. 7212 178. 244 7390 7568 7746 7923 8101 8279 8456 8634 8gli 8989 178 245 9166 9343 9520 9698 9875 ♦ ♦51 ♦ 228 ♦405 ♦ 582 ♦759 177 246 390935 1112 1288 1464 1641 1817 1993 2169 2345 2521 176 247 2697 2873 8048 3224 3400 3575 3751 3926 4101 4277 176 248 4452 4627 4802 4977 5152 5326 6601 5676 5850 6025 175 249 6199 6374 6548 6722 6896 7071 7246 7419 7592 7766 174 250 397940 8114 8287 8461 8634 8808 8981 9164 9328 9501 173 251 9674 9847 ♦♦20 ♦ 192 ♦ 365 ♦ 538 ♦ 711 ♦ 883 1066 1228 173 252 401401 1573 1745 1917 2089 2261 2433 2606 2777 2949 172 253 3121 8292 3464 3635 3807 3978 4149 4320 4492 4663 171 254 4834 5005 5176 5846 5517 6688 5868 6029 6199 6370 171 256 6540 6710 6881 7051 7221 7391 7661 7731 7901 8070 170 256 8240 8410 8579 8749 8918 9087 9267 9426 9595 9764 169 257 9933 ♦102 ♦ 271 ♦440 ♦609 ♦777 ♦ 946 1114 1283 1451 169 258 411620 1788 1966 2124 2293 2461 2629 2796 2964 8132 168 259 3300 3467 3635 3803 3970 4137 4306 4472 4639 4806 167 260 414973 5140 5307 5474 6641 5808 6974 6141 6308 6474 167 261 6641 6807 6973 7139 7306 7472 7638 7804 7970 8185 166 262 8301 8467 8633 8798 8964 9129 9295 9460 9625 9791 165 263 9956 ♦121 ♦ 286 ♦461 ♦616 ♦ 781 ♦945 1110 1276 1439 165 264 421604 1788 1933 2097 2261 2426 2590 2764 2918 3082 164 265 3246 3410 8574 3737 3901 4066 4228 4392 4555 4718 164 266 4882 6045 5208 5371 5634 6697 5860 6023 6186 6349 163 267 6511 6674 6836 6999 7161 7324 7486 7648 7811 7973 162 268 8135 8297 8459 8621 8783 8944 9106 9268 9429 9591 162 269 9752 9914 ♦ ♦75 ♦ 236 ♦398 ♦ 569 ♦ 720 ♦881 1042 1203 161 270 431364 1526 1685 1846 2007 2167 2328 2488 2649 2809 161 271 2969 8130 3290 3460 3610 3770 3930 4090 4249 4409 160 272 4569 4729 4888 6048 5207 5367 5526 5685 5844 6004 159 273 6163 6322 6481 6640 6798 6957 7116 7276 7433 7592 169 274 7761 7909 8067 8226 8384 8542 8701 8859 9017 9175 158 275 9333 9491 9648 9806 9964 ♦ 122 ♦279 ♦437 ♦ 594 ♦752 158 276 440909 1066 1224 1381 1638 1696 1852 2009 2166 2328 157 277 2480 2637 2793 2950 3106 3263 3419 3576 8732 3889 157 278 4045 4201 4357 4513 4669 4826 4981 6137 5293 6449 156 279 5604 5760 5915 6071 6226 6382 6537 6692 6848 7003 9 155 ^7 N. 1 2 3 4 5 6 T 8 A TABLE OP LOGARITHMS FROM 1 TO 10,000. N. 1 a 3 4 5 e 7 8 9 D. 280 447158 7313 7468 7623 7778 7983 8088 8242 8397 8552 155 281 8706 8861 9015 9170 9324 9478 9633 9787 9941 ♦'♦95 154 282 450249 0403 0557 0711 0865 1018 1172 1326 1479 1633 154 283 1786 1940 2093 2247 2400 2553 2706 2869 3012 3165 153 284 3318 3471 8624 3777 3930 4082 4235 4387 4540 4692 153 285 4845 4997 6160 6302 5454 5606 6758 6910 6062 6214 152 286 6366 6518 6670 6821 6973 7125 7276 7428 7579 7731 152 287 7882 8083 8184 8336 8487 8638 8789 8940 9091 9242 151 288 9392 9543 9694 9845 9995 ♦ 146 ♦ 296 ♦447 ♦ 597 ♦ 748 151 289 460898 1048 1198 1348 1499 1649 1799 1948 2098 2248 150 290 462398 2548 2697 2847 2997 3146 3296 3445 3594 3744 150 291 8893 4042 4191 4340 4490 4639 4788 4936 5085 5234 149 292 5383 5532 5680 5829 5977 6126 6274 6423 6571 6719 149 293 6368 7016 7164 7312 7460 7608 7756 7904 8052 8200 148 294 8347 8495 8643 8790 8938 9085 923B 9380 9527 9675 148 295 9822 9969 ♦ 116 ♦ 263 ♦410 ♦ 557 ♦ 704 ♦ 851 ♦ 998 1145 147 296 471292 1438 1585 1732 1878 2026 2171 2318 2464 2610 146 297 2756 2903 8049 3195 3341 3487 8633 3779 3925 4071 146 298 4216 4362 4608 4653 4799 4944 5090 5235 6381 5526 146 299 5671 5816 6962 6107 6252 6397 6542 6687 6832 6976 145 300 477121 7266 7411 7565 7700 7844 7989 8133 8278 8422 145 301 8566 8711 8855 8999 9143 9287 9431 9575 9719 9863 144 302 480007 0151 0294 0438 0582 0725 0869 1012 1156 1299 144 303 1443 1586 1729 1872 2016 2159 2302 2446 2588 2731 143 304 2874 3016 3159 3302 3445 3587 3730 3872 4016 4167 143 805 4300 4442 4585 4727 4869 5011 5153 6295 5437 5579 142 306 5721 5863 6005 6147 6289 6430 6572 6714 6855 6997 142 307 7138 7280 7421 7563 7704 7845 7986 8127 8269 8410 141 308 8551 8692 8833 8974 9114 9255 9396 9537 9677 9818 141 309 9958 ♦ ♦99 ♦ 239 ♦380 ♦ 520 ♦ 661 ♦ 801 ♦ 941 1081 1222 140 310 491362 1502 1642 1782 1922 2062 2201 2341 2481 2621 140 311 2760 2900 3040 3179 3319 3458 3597 8737 3876 4015 139 312 4155 4294 4433 4572 4711 4850 4989 5128 5267 5406 139 313 5544 5683 5822 5960 6099 6238 6376 6515 6653 6791 139 314 6930 7068 7206 7344 7483 7621 7759 7897 8035 8173 138 315 8311 8448 8586 8724 8862 8999 9137 9275 9412 9550 138 316 9687 9S24 9962 ♦ ♦99 ♦ 236 ♦ 374 ♦ 511 ♦ 648 ♦ 785 ♦ 922 137 817 501059 1196 1333 1470 1607 1744 1880 2017 2154 2291 137 318 2427 2564 2700 2837 2973 3109 3246 3382 3518 3655 136 319 3791 3927 4063 4199 4335 4471 4607 4743 4878 5014 136 320 505160 5286 5421 5557 5693 5828 6964 6099 6234 6370 136 321 6606 6640 6776 6911 7046 7181 7316 7451 7586 7721 135 322 7856 7991 8126 8260 8395 8530 8664 8799 8934 9068 135 323 9208 9337 9471 9606 9740 9874 ♦♦♦9 ♦ 143 ♦277 ♦411 134 824 610545 0679 0813 0947 1081 1216 1849 1482 1616 1750 134 325 1883 2017 2151 2284 2418 2561 2684 2818 2951 3084 133 326 3218 3351 3484 3617 3750 3883 4016 4149 4282 4414 133 327 4548 4681 4813 4946 6079 5211 5344 5476 5609 5741 133 328 5874 eoo6 6139 6271 6403 6535 6668 6800 6932 7064 132 329 7196 7328 7460 7592 7724 7855 7987 8119 8251 8382 132 330 B18514 8646 8777 8909 9040 9171 9303 9434 9566 9697 131 331 9828 9959 ♦ ♦90 ♦ 221 ♦ 353 ♦ 484 ♦ 616 ♦ 746 ♦ 876 1007 131 332 521138 1269 1400 1530 1661 1792 1922 2063 2183 2314 131 333 2444 2576 2705 2835 2966 3096 3226 3356 3486 3616 130 834 3746 3876 4006 4136 4266 4396 4526 4666 4785 4915 130 835 5045 5174 6304 5434 6563 5693 6822 5961 6081 6210 129 836 6339 6469 6598 6727 6856 6985 7114 7243 7372 7501 129 337 7630 7759 7888 8016 8145 8274 8402 8631 8660 8788 129 338 8917 9045 9174 9302 9430 9559 9687 9815 9943 ♦♦72 128 839 530200 0328 0466 0584 0712 0840 0968 1096 1223 1351 128 N. 1 S 3 4 5 6 7 8 9 D. A TABLE OF LOGAEITHMS FROM 1 TO 10,000. N. 1 2 3 4 5 6 7 8 1 9 D. 310 531479 1607 1734 1862 1990 2117 2245 2872 2500 2627 128 841 2754 2882 3009 3136 3264 3391 3518 3645 3772 3899 127 842 4026 4163 4280 4407 4584 4661 4787 4914 5041 5167 127 843 5294 5421 5547 5674 5800 5927 6053 6180 6306 6432 126 844 6558 6685 6811 6937 7063 7189 7315 7441 7567 7693 126 845 7819 7945 8071 8197 8822 8448 8574 8699 8825 8951 i26 846 9076 9202 9327 9452 9578 9703 9829 9954 ♦ ♦79 ♦ 204 125 347 540329 0455 0580 0705 0830 0955 1080 1205 1330 1454 125 348 1579 1704 1829 1953 2078 2203 2327 2452 2576 2701 125 349 2825 2950 3074 3199 3323 3447 3571 3696 3820 3944 124 850 544068 4192 4316 4440 4564 4688 4812 4936 5060 5183 124 851 5307 5431 5555 5678 5802 5925 6049 6172 6296 6419 124 352 6543 6666 6789 6913 7036 7159 7282 7405 7529 7652 123 353 7775 7898 8021 8144 8267 8389 8512 8635 8758 8881 128 854 9003 9126 9249 9371 9494 9616 9739 9861 9984 ♦ 106 128 855 550228 0351 0473 0595 0717 0840 0962 1084 1206 1328 122 356 1450 1572 1694 1816 1938 2060 2181 2303 2425 2547 122 357 2668 2790 2911 3033 8155 8276 3398 3519 3640 3762 121 358 3888 4004 4126 4247 4368 4489 4610 4731 4852 4973 121 359 5094 5215 5336 5457 5578 5699 5820 5940 6061 6182 121 860 556803 6423 6544 6664 6785 6905 7026 7146 7267 7387 120 361 7507 7627 7748 7868 7988 8108 8228 8349 8469 8589 120 862 8709 8829 8948 9068 9188 9308 9428 9548 9667 9787 120 363 9907 ♦ ♦26 ♦ 146 ♦ 265 ♦385 ♦ 504 ♦ 624 ♦ 748 ♦ 863 ♦ 982 119 864 561101 1221 1340 1459 1578 1698 1817 1936 2055 2174 119 865 2293 2412 2581 2650 2769 2887 3006 8125 3244 8862 119 366 3481 3600 3718 3837 3955 4074 4192 4311 4429 4548 119 867 4666 4784 4903 5021 5139 5257 5376 5494 5612 5730 118 368 5848 5966 6084 6202 6320 6437 6555 6673 6791 6909 118 369 7026 7144 7262 7379 7497 7614 7732 7849 7967 8084 118 870 568202 8319 8436 8554 8671 8788 8905 9023 9140 9257 117 371 9874 9491 9608 9725 9842 9959 ♦ ♦76 ♦ 193 ♦ 309 ♦ 426 117 372 570543 0660 0776 0893 1010 1126 1243 1359 1476 1592 117 373 1709 1825 1942 2058 2174 2291 2407 2523 2639 2755 116 874 2872 2988 3104 3220 8386 8452 3568 3684 3800 3915 116 375 4031 4147 4263 4379 4494 4610 4726 4841 4957 5072 116 376 5188 6803 5419 5534 5650 5765 6880 5996 6111 6226 115 377 6841 6457 6572 6687 6802 6917 7032 7147 7262 7877 115 378 7492 7607 7722 7836 7951 8066 8181 8295 8410 8525 115 879 8639 8754 8868 8988 9097 9212 9826 9441 9555 9669 114 880 579784 9898 ♦ ♦12 ♦ 126 ♦241 ♦ 355 ♦ 469 ♦ 583 ♦ 697 ♦ 811 114 381 580925 1089 1153 1267 1881 1495 1608 1722 1836 1950 114 382 2068 2177 2291 2404 2518 2681 2745 2858 2972 8085 114 383 3199 8312 3426 3589 8652 3765 3879 3992 4105 4218 113 384 4881 4444 4557 4670 4783 4896 5009 5122 5235 5348 113 385 5461 5574 5686 5799 5912 6024 6137 6250 6362 6475 113 386 6587 6700 6812 6925 7037 7149 7262 7874 7486 7599 112 387 7711 7823 7935 8047 8160 8272 8384 8496 8608 8720 112 388 8832 8944 9056 9167 9279 9891 9508 9615 9726 9838 112 889 9950 ♦ ♦61 ♦173 ♦ 284 ♦396 ♦ 507 ♦ 619 ♦ 780 ♦ 842 ♦ 953 112 890 591065 1176 1287 1399 1510 1621 1732 1843 1955 2066 111 391 2177 2288 2399 2510 2621 2732 2843 2954 3064 ?175 111 392 8286 3397 3508 3618 3729 3840 3950 4061 4171 4282 111 398 4893 4503 4614 4724 4884 4945 5055 5165 5276 5386 110 394 5496 5606 5717 5827 5987 6047 6157 6267 6377 6487 110 395 6597 6707 6817 6927 7037 7146 7256 7866 7476 7586 110 896 7695 7805 7914 8024 8134 8243 8353 8462 8572 8681 110 397 8791 8900 9009 9119 9228 9837 9446 9556 9665 9774 109 398 9883 9992 ♦ 101 ♦210 ♦ 319 ♦428 ♦ 537 ♦ 646 ♦ 755 ♦ 864 109 399 600973 1082 1191 1299 1408 1517 1625 1784 1843 1951 109 N. 1 2 3 4 5 6 r 8 9 D. A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. 1 2 3 4 5 G 7 8 9 D. 1 400 602060 2169 2277 2386 2494 2603 2711 1 2819 2928 3036 108 401 3144 3253 8861 8469 8577 3686 8794 3902 4010 4118 108 402 4226 4384 4442 4550 4658 4766 4874 4982 6089 6197 108 403 5305 5418 5521 5628 5736 5844 5951 6059 6166 6274 108 404 6881 6489 6596 6704 6811 6919 7026 7133 7241 7348 107 405 7455 7562 7669 7777 7884 7991 8098 8205 8312 8419 107 406 8526 8688 8740 8847 8954 9061 9167 9274 9381 9488 107 407 9594 9701 9808 9914 ♦ ♦21 ♦ 128 ♦ 234 ♦ 341 ♦ 447 ♦ 554 107 408 610660 0767 0873 0979 1086 1192 1298 1405 1511 1617 106 409 1723 1829 1936 2042 2148 2264 2360 2466 2572 2678 106 410 612784 2890 2996 3102 8207 3313 3419 8525 3630 3736 106 411 3842 8947 4058 4159 4264 4370 4475 4581 4686 4792 106 412 4897 5003 5108 5213 6319 5424 5529 5634 6740 5845 105 413 5950 6055 6160 6265 6370 6476 6581 6686 6790 6895 105 414 7000 7105 7210 7315 7420 7525 7629 7734 7839 7943 105 415 8048 8153 8257 8362 8466 8571 8676 8780 8884 8989 105 416 9093 9198 9302 9406 9511 9615 9719 9824 9928 ♦ ♦32 104 417 620136 0240 0844 0448 0552 0656 0760 0864 0968 1072 104 418 1176 1280 1384 1488 1692 1695 1799 1903 2007 2110 104 419 2214 2318 2421 2525 2628 2732 2835 2989 8042 3146 104 420 623249 3358 3456 3559 3663 3766 3869 3973 4076 4179 103 421 4282 4385 4488 4591 4695 4798 4901 5004 5107 5210 103 422 5312 5415 5518 5621 5724 5827 6929 6082 6135 6238 103 423 6340 6443 6546 6648 6751 6853 6956 7058 7161 7268 103 424 7366 7468 7571 7673 7775 7878 7980 8082 8186 8287 102 425 8389 8491 8593 8695 8797 8900 9002 9104 9206 9308 102 426 9410 9512 9613 9715 9817 9919 ♦ ♦21 ♦ 123 ♦ 224 ♦ 326 102 427 630428 0530 0631 0733 0835 0936 1038 1139 1241 1342 102 428 1444 1545 1647 1748 1849 1951 2052 2153 2255 2356 101 429 2457 2559 2660 2761 2862 2963 3064 8165 3266 3367 101 430 633468 3569 3670 3771 3872 8973 4074 4175 4276 4376 100 431 4477 4578 4679 4779 4880 4981 5081 6182 5283 5383 100 432 5484 5584 5685 6785 5886 6986 6087 6187 6287 6388 100 433 6488 6588 6688 6789 6889 6989 7089 7189 7290 7390 100 484 7490 7590 7690 7790 7890 7990 8090 8190 8290 8389 99 435 8489 8589 8689 8789 8888 8988 9088 9188 9287 9887 99 436 9486 9586 9686 9785 9885 9984 ♦♦84 ♦ 183 ♦ 283 ♦ 382 99 487 640481 0581 0680 0779 0879 0978 1077 1177 1276 1375 99 438 1474 1573 1672 1771 1871 1970 2069 2168 2267 2366 99 439 2465 2563 2662 2761 2860 2959 3058 3156 3255 3364 99 440 648458 3551 8650 3749 3847 3946 4044 4143 4242 4340 98 441 4439 4537 4636 4734 4832 4931 6029 6127 5226 5824 98 442 5422 6621 5619 5717 5815 5913 6011 6110 6208 6306 98 443 6404 6602 6600 6698 6796 6894 6992 7089 7187 7285 98 444 7383 7481 7579 7676 7774 7872 7969 8067 8165 8262 98 445 8360 8458 8555 8653 8750 8848 8945 9043 9140 9237 97 446 9335 9432 9530 9627 9724 9821 9919 ♦ ♦16 ♦ 113 ♦ 210 97 447 650308 0405 0502 0599 0696 0793 0890 0987 1084 1181 97 448 1278 1375 1472 1569 1666 1762 1859 1956 2053 2150 97 449 2246 2343 2440 2536 2633 2780 2826 2923 3019 8116 97 450 653213 3309 3405 3502 3598 3695 8791 3888 3984 4080 96 451 4177 4273 4369 4465 4562 4658 4754 4850 4946 5042 96 452 5138 5235 5831 5427 5523 6619 5715 5810 5906 6002 96 453 6098 6194 6290 6386 6482 6577 6673 6769 6864 6960 96 454 7056 7152 7247 7343 7438 7534 7629 7725 7820 7916 96 455 8011 8107 8202 8298 8393 8488 8584 8679 8774 8870 95 456 8965 9060 9155 9250 9346 9441 9536 9631 9726 9821 95 457 9916 ♦ ♦11 ♦ 106 ♦ 201 ♦ 296 ♦ 391 ♦ 486 ♦ 581 ♦ 676 ♦771 95 458 660865 0960 1055 1150 1245 1389 1434 1529 1623 1718 96 459 1813 1907 2002 2096 2191 2286 2380 2475 2669 2663 95 D. N. 1 3 3 4 5 6 7 8 » A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. 1 2 3 4 5 6 7 8 9 D. 460 662758 28B2 2947 3041 3135 3230 3324 8418 3512 3607 94 461 3701 3795 3889 3983 4078 4172 4266 4360 4454 4548 94 462 4642 4786 4830 4924 B018 5112 5206 5299 5398 5487 94 463 5581 5675 5769 5862 5956 6050 6143 6237 6331 6424 94 464 6518 6612 6705 6799 6892 6986 7079 7173 7266 7360 94 465 7453 7546 7640 7733 7826 7920 8013 8106 8199 8293 93 466 8386 8479 8572 8665 8759 8862 8945 9038 9131 9224 93 467 9317 9410 9503 9596 9689 9782 9875 9967 ♦ ♦60 ♦ 153 93 468 670246 0339 0431 0324 0617 0710 0802 0895 0988 1080 93 469 1173 1265 1358 1451 1543 1636 1728 1821 1913 2005 93 470 672098 2190 2283 2375 2467 2560 2652 2744 2836 2929 92 471 3021 3113 3205 8297 3390 3482 8574 3666 3768 3850 92 472 3942 4034 4126 4218 4310 4402 4494 4686 4677 4769 92 473 4861 4953 5045 5137 5228 5320 5412 5503 5595 6687 92 474 5778 B870 5962 6053 6145 6236 6328 6419 6511 6602 92 475 6694 6785 6876 6968 7059 7151 7242 7333 7424 7516 91 476 7607 7698 7789 7881 7972 8063 8154 8245 8336 8427 91 477 8518 8609 8700 8791 8882 8973 9064 9155 9246 9337 91 478 9428 9519 9610 9700 9791 9882 9973 ♦ ♦63 ♦ 154 ♦ 245 01 479 680336 0426 0517 0607 0698 0789 0879 0970 1060 1151 91 480 G81241 1332 1422 1513 1603 1693 1784 1874 1964 2065 90 481 2145 2235 2326 2416 2506 2396 2686 2777 2867 2957 90 482 3047 3137 3227 3317 3407 3497 3587 3677 3767 3857 00 483 3947 4037 4127 4217 4307 4396 4486 4676 4666 4766 90 484 4845 4935 5025 5114 5204 5294 6383 5473 6563 6652 90 485 5742 5831 5921 6010 6100 6189 6279 6368 6458 6647 89 486 6636 6726 6815 6904 6994 7083 7172 7261 7351 7440 89 487 7529 7618 7707 7796 7886 7975 8064 8153 8242 8331 89 488 8420 8509 8598 8687 8776 8865 8953 9042 9131 9220 89 489 9309 9398 9486 9575 9664 9753 9841 9930 ♦ ♦19 ♦ 107 89 490 690196 0285 0373 0462 0550 0639 0728 0816 0905 0993 89 491 1081 1170 1258 1347 1435 1524 1612 1700 1789 1877 88 492 1965 2053 2142 2230 2318 2406 2494 2583 2671 2759 88 493 2847 2035 8023 3111 3199 3287 3375 8463 3551 3639 88 494 3727 3815 3903 3991 4078 4166 4254 4342 4430 4517 88 495 4605 4693 4781 4868 4956 5044 5131 5219 6307 5394 83 496 5482 5569 5657 5744 5882 6919 6007 6094 6182 6269 87 497 6356 6444 6531 6618 6706 6793 6880 6968 7055 7142 87 498 7229 7317 7404 7491 7578 7665 7752 7839 7926 8014 87 499 8101 8188 8275 8362 8449 8535 8622 8709 8796 8883 87 BOO 698970 9057 9144 9231 9317 9404 9491 9578 9664 9761 87 501 9838 9924 .♦11 ♦ ♦98 ♦ 184 ♦271 ♦ 353 ♦ 444 ♦ 631 ♦ 617 87 502 700704 0790 0877 0963 1050 1136 1222 1309 1395 1482 86 503 1568 1654 1741 1827 1913 1999 2086 2172 2268 2344 86 504 2431 2517 2603 2689 2775 2861 2947 8033 3119 3205 86 503 3291 3377 3463 8549 3635 3721 8807 3893 3979 4066 86 506 4151 4236 4322 4408 4494 4579 4665 4751 4837 4922 86 507 5008 5094 5179 5265 5350 5436 6522 5607 6693 5778 86 508 5861 5949 6035 6120 6206 6291 6376 6462 6547 6632 85 509 6718 0803 6888 6974 7059 7144 7229 7315 7400 7485 85 BIO 707570 76BB 7740 7826 7911 7996 8081 8166 8251 8336 85 511 8421 8506 8591 8676 8761 8846 8931 9015 9100 9186 83 512 9270 9355 9440 9524 9609 9694 9779 9863 9948 ♦•♦33 85 513 710117 0202 0287 0371 0456 0540 0625 0710 0794 0879 83 514 0903 1048 1132 1217 1301 1385 1470 1554 1639 1723 84 515 1807 1892 1976 2060 2144 2229 2313 2397 2481 2566 84 516 2650 2734 2818 2902 2986 3070 3154 3238 3323 3407 84 517 3491 3575 3659 3742 3826 8910 8994 4078 4162 4246 84 518 4330 44li 4497 4581 4665 4749 4883 4916 6000 5084 84 619 5167 5251 5335 B418 BB02 5586 5669 5753 5836 5920 84 N. 1 2 3 4 5 6 7 8 9 D. A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. 1 2 3 4 5 6 7 8 9 6754 D. 620 716003 6087 6170 6254 6887 6421 6504 6588 6671 83 521 6838 6921 7004 7088 7171 7254 7338 7421 7604 7587 88 522 7671 7754 7887 7920 8003 8086 8169 8253 8336 8419 83 523 8502 8585 8668 8751 8884 8917 9000 9083 9165 9248 83 524 9331 9414 9497 9580 9668 9745 9828 9911 9994 ♦♦77 88 625 720169 0242 0325 0407 0490 0678 0655 0738 0821 0903 88 626 -0986 1068 1151 1288 1316 1398 1481 1563 1646 1728 82 627 1811 1893 1976 2058 2140 2222 2305 2887 2469 2552 82 528 2634 2716 2798 2981 2963 3046 3127 3209 3291 3374 82 529 3456 3538 8620 3702 3784 8866 3948 4030 4112 4194 82 530 724276 4358 4440 4522 4604 4686 4767 4849 4981 5013 82 531 5095 6176 5258 5840 6422 5503 5585 5667 5748 5830 82 582 5912 5993 6076 6156 6288 6320 6401 6483 656i 6646 82 533 6727 6809 6890 6972 7053 7184 7216 7297 7379 7460 81 634: 7541 7623 7704 "7785 7866 7948 8029 8110 8191 8273 81 635 8364 8485 8516 8597 8678 8759 8841 8922 9003 9084 81 636 9165 9246 9827 9408 9489 9570 9651 9732 9813 9893 81 537 9974 ♦ ♦55 ♦ 136 ♦ 217 ♦ 298 ♦ 378 ♦ 459 ♦ 640 ♦ 621 ♦ 702 81 538 730782 0863 0944 1024 1105 1186 1266 1347 1428 1508 81 539 1589 1669 1750 1830 1911 1991 2072 2152 2233 2313 81 540 732394 2474 2555 2685 2715 2796 2876 2956 d087 3117 80 541 3197 3278 3358 8488 3518 8598 3679 8759 3839 3919 80 642 3999 4079 4160 4240 4320 4400 4480 4560 4640 4720 80 543 4800 4880 4960 6040 5120 5200 5279 5859 5439 5519 80 644 5599 5679 5759 6838 5918 5998 6078 6157 6237 6317 80 545 6397 6476 6556 6685 6715 6795 6874 6954 7034 7118 80 546 7193 7272 7352 7431 7511 7590 7670 7749 7829 7908 79 547 7987 8067 8146 8225 8305 8384 8463 8543 8622 8701 79 648 8781 8860 8939 9018 9097 9177 9256 9335 9414 9493 79 549 95T2 9651 9731 9810 9889 9968 ♦ ♦47 ♦ 126 ♦ 206 ♦284 79 550 740363 0442 0521 0600 0678 0767 0886 0915 0994 1073 79 551 1132 1230 1309 1388 1467 1546 1624 1703 1782 1860 79 552 1939 2018 2096 2175 2254 2332 2411 2489 2568 2647 79 553 2725 2804 2882 2961 8089 3118 3196 3275 8858 8431 78 664 3610 3588 3667 8745 3823 3902 3980 4058 4136 4215 78 655 4293 4371 4449 4628 4606 4684 4762 . 4840 4919 4997 78 556 5075 5153 5231 6309 5887 5465 5543 5621 6699 5777 78 557 5855 5933 6011 6089 6167 6245 6823 6401 6479 6556 78 558 6634 6712 6790 6868 6945 7023 7101 7179 7256 7334 78 669 7412 7489 7567 7646 7722 7800 7878 7956 8033 8110 78 560 748188 8266 8343 8421 8498 8576 8653 8731 8808 8885 77 561 8963 9040 9118 9195 9272 9850 9427 9504 9682 9659 77 562 9736 9814 9891 9968 ♦ ♦45 ♦ 123 ♦ 200 ♦ 277 ♦ 364 ♦ 431 77 563 750508 0586 0663 0740 0817 0894 0971 1048 1125 1202 77 564 1279 1366 1483 1510 1587 1664 1741 1818 1895 1972 77 665 2048 2125 2202 2279 2356 2433 2509 2686 2663 2740 77 666 2816 2893 2970 3047 3123 3200 8277 3358 8430 3506 77 667 3683 3660 3736 3818 8889 3966 4042 4119 4195 4272 77 568 4848 4425 4501 4578 4654 4730 4807 4888 4960 5036 76 669 6112 6189 5265 6841 6417 5494 5570 5646 5722 5799 76 670 755875 5951 6027 6108 6180 6266 6382 6408 6484 6560 76 571 6636 6712 6788 6864 6940 7016 7092 7168 7244 7320 76 572 7396 7472 7548 7624 7700 7775 7851 7927 8008 8079 76 573 8155 8230 8306 8882 8458 8583 8609 8685 8761 8836 76 574 8912 8988 9068 9139 9214 9290 9366 9441 9517 9592 76 576 9668 9743 9819 9894 ,9970 ♦ ♦45 ♦ 121 ♦ 196 ♦ 272 ♦ 347 75 576 760422 0498 0573 0649 0724 0799 0875 0950 1025 1101 75 577 1176 1251 1826 1402 1477 1552 1627 1702 1778 1853 75 578 1928 2003 2078 2153 2228 2803 2378 2458 2529 2604 75 579 2679 2754 2829 2904 2978 3053 3128 3203 3278 8353 75 N. 1 2 3 4 5 6 7 8 9 D. 10 A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. i T'l 2 3 4 S e i 7 8 9 D. 1 j 580 763428 3503 3578 3653 3727 3802 3877 1 3952 4027 4101 T5i 681 4176 4251 4326 4400 4475 4550 4624 4699 4774 4848 76 582 4923 4998 5072 5147 6221 5296 5370 5445 6520. 5594 75 583 5669 5743 5818 6892 6966 6041 6115 6190 6264 6338 74 584 6418 0487 6562 6636 6710 6785 6869 6933 7007 7082 74 585 7156 7230 7304 7379 7453 7527 7601 7675 7749 7823 74 586 7898 7972 8046 8120 8194 8268 8342 8416 8490 8564 74 587 8638 8712 8786 8860 8934 9008 9082 9156 9230 9303 74 583 9377 9451 9525 9599 9673 9746 9820 9894 9968 ♦ ♦42 74 589 770115 0189 0263 0336 0410 0484 0557 0631 0705 0778 74 590 770852 0926 0999 1073 1146 1220 1293 1367 1440 1614 74 1 591 1587 1661 1734 1808 1881 1955 2028 2102 2175 2248 73 592 2322 2395 2468 2542 2615 2688 2762 2835 2908 2981 73 593 3055 3128 3201 3274 3348 3421 8494 3567 3640 3713 73 594 3786 3860 3933 4006 4079 4152 4225 4268 4371 4444 73 595 4517 4590 4663 4736 4809 4882 4955 5028 5100 6173 73 596 5246 5319 5392 5465 6538 5610 6683 5756 5829 5902 73 597 5974 6047 6120 6193 6265 6338 6411 6483 6556 6629 73 598 6701 6774 6846 6919 6992 7064 7137 7209 7282 7354 73 599 7427 7499 7572 7644 7717 7789 7862 7934 8006 8079 72 600 778151 8224 8296 8368 8441 8513 8685 8653 8730 8802 72 601 8874 8947 9019 9091 9163 9236 9308 9380 9452 9524 72 602 9596 9669 9741 9813 9885 9957 **29 ♦ 101 ♦ 173 ♦ 245 72 603 780317 0389 0461 0533 0605 0677 0749 0821 0893 0965 72 604 1037 1109 1181 1253 1324 1396 1468 1640 1612 1684 72 605 1755 1827 1899 1971 2042 2114 2186 2258 2329 2401 72 606 2478 2544 2616 2688 2759 2831 2902 2974 3046 3117 72 607 3189 3260 8332 3403 3475 3546 3618 3689 3761 3832 71 608 3904 3975 4046 4118 4189 4261 4332 4403 4475 4546 71 609 4617 4689 4760 4881 4902 4974 6045 5116 5187 5259 71 610 785330 5401 5472 5543 5615 5686 5757 5828 5899 5970 71 611 6041 6112 6183 6254 6325 6396 6467 6538 6609 6680 71 612 6751 6822 6893 6964 7035 7106 7177 7248 7319 7390 71 613 7460 7531 7602 7673 7744 7815 7885 7956 8027 8098 71 614 8168 8239 8310 8381 8451 8522 8593 8663 8734 8804 71 615 8875 8946 9016 9087 9157 9228 9299 9369 9440 9510 71 616 9581 9651 9722 9792 9863 9933 ♦ «*4 ♦ ♦74 ♦ 144 ♦ 215 70 617 790285 0356 0426 0496 0567 0637 0707 0778 0848 0918 70 618 0988 1059 1129 1199 1269 1340 1410 1480 1550 1620 70 619 1691 1761 1831 1901 1971 2041 2111 2181 2252 2322 • 70 620 792392 2462 2532 2602 2672 2742 2812 2882 2952 3022 70 621 3092 3162 3231 3301 3371 3441 3511 3581 3661 3721 70 622 3790 3860 3930 4000 4070 4139 4209 4279 4349 4418 70 623 4488 4558 4627 4697 4767 4836 4906 4976 6045 5115 70 624 5185 5254 5324 5393 5463 5532 5602 .5672 6741 5811 70 625 5880 5949 6019 6088 6158 6227 6297 6366 6436 6505 69 628 6574 6644 6713 6782 6852 6921 6990 7060 7129 7198 69 627 7268 7337 7406 7475 7545 7614 7683 7752 7821 7890 69 628 7960 8029 8098 8167 8236 8305 8374 8443 8513 8582 69 1 629 8651 8720 8789 8858 8927 8996 9065 9134 9203 9272 69 630 799341 9409 9478 9547 9616 9685 9754 9823 9892 9961 69 631 800029 0098 0167 0236 0305 0373 0442 0511 0580 0648 69 632 0717 0786 0854 0923 0992 1061 1129 1198 1266 1335 69 633 1404 1472 1541 1609 1678 1747 1815 1884 1952 2021 69 634 2089 2158 2226 2295 2363 2432 2500 2568 2637 2705 69 ■ 635 2774 2842 2910 2979 3047 8116 3184 3252 3321 3389 68 636 3457 3525 3594 3662 3730 3798 3867 3935 4003 4071 68 637 4139 4208 4276 4344 4412 4480 4548 4616 4685 4753 68 638 4821 4889 4957 5025 5093 5161 5229 5297 5365 5433 68 639 5501 5569 5637 5705 5773 5841 6908 5976 6044 6112 68 N. 1 2 3 4 5 6 7 8 9 D. A TABLE OF LOGARITHMS FROM 1 TO 10,000. 11 s. 1 2 3 4 5 6 7 8 9 D. 68 640 806180 6248 6316 6884 6451 I 6519 6587 6655 6723 6790 041 6868 6926 6994 7061 7129 7197 7264 7332 7400 7467 68 642 7535 7603 7670 7738 7806 7873 7941 8008 8076 8148 68 643 8211 8279 8346 8414 8481 8549 8616 8684 8751 8818 1 67 644 8886 8953 9021 9088 9156 9223 9290 9858 9425 9492 67 645 9560 9627 9694 9762 9829 9896 9964 ♦ ♦31 ♦ ♦98 ♦ 165 67 648 810233 0300 0367 0484 0501 0569 0636 0703 0770 0837 67 647 0904 0971 1039 1106 1173 1240 1307 1374 1441 1508 67 648 1575 1642 1709 1776 1843 1910 1977 2044 2111 2178 67 649 2245 2312 2879 2445 2512 2579 2646 2713 2780 2847 67 050 812913 2980 3047 8114 3181 3247 3814 3381 3448 3514 67 651 3581 8648 8714 3781 3848 3914 3981 4048 4114 4181 67 652 4248 4314 4381 4447 4514 4581 4647 4714 4780 4847 67 653 4918 4980 5046 5113 5179 5246 5312 5378 5445 5511 66 654 5578 5644 5711 5777 5843 5910 5976 6042 6109 6175 66 655 6241 6308 6874 6440 6506 6573 6639 6705 6771 6838 66 656 6904 6970 7036 7102 7169 7235 7301 7367 7433 7499 66 657 7565 7631 7698 7764 7830 7896 7962 8028 8094 8160 66 658 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 66 659 8885 8951 9017 9083 9149 9215 9281 9346 9412 9478 66 660 819544 9610 9676 9741 9807 9873 9939 ♦ ♦♦4 ♦ ♦70 ♦ 136 66 661 820201 0267 0883 0399 0464 0530 0595 0661 0727 0792 66 662 0858 0924 0989 1055 1120 1186 1251 1317 1383 1448 66 663 1514 1579 1645 1710 1775 1841 1906 1972 2037 2103 65 664 2168 2288 2299 2364 2430 2495 2560 2626 2691 2756 63 665 2822 2887 2952 3018 3083 3148 8218 3279 8344 8409 65 666 3474 8539 3605 3670 3735 3800 3865 3980 3996 4061 65 667 4126 4191 4256 4321 4386 4451 4516 4581 4646 4711 65 668 4776 4841 4906 4971 5036 5101 5166 5231 5296 5861 65 669 5426 5491 5556 5621 5686 5751 5815 5880 5945 6010 65 670 826075 6140 6204 6269 6384 6399 6464 6528 6593 6658 63 671 6723 6787 6852 6917 6981 7046 7111 7175 7240 7805 65 672 7369 7434 7499 7563 7628 7692 7757 7821 7886 7951 65 673 8015 8080 8144 8209 8273 8338 8402 8467 8531 8595 64 674 8660 8724 8789 8853 8918 8982 9046 9111 9175 9289 64 675 9304 9868 9432 9497 9561 9625 9690 9754 9818 9882 64 676 9947 ♦ .11 ♦ ♦75 ♦ 139 ♦ 204 ♦ 268 ♦ 332 ♦ 396 ♦ 460 ♦ 525 64 677 830589 0653 0717 0781 0845 0909 0973 1037 1102 1166 64 678 1230 1294 1358 1422 1486 1550 1614 1678 1742 1806 64 679 1870 1984 1998 2062 2126 2189 2253 2317 2381 2445 64 680 832509 2573 2637 2700 2764 2828 2892 2956 3020 3083 64 681 3147 3211 8275 3338 8402 3466 8530 8593 3657 3721 64 682 3784 3848 8912 3975 4039 4108 4166 4230 4294 4357 64 683 4421 4484 4548 4611 4675 4739 4802 4866 4929 4993 64 684 5056 5120 5183 5247 5310 5873 5437 5500 5564 5627 63 685 5691 5754 5817 5881 5944 6007 6071 6134 6197 6261 63 686 6324 6387 6451 6514 6577 0641 6704 6767 6830 6894 63 687 6957 7020 7088 7146 7210 7273 7386 7399 7462 7525 68 688 7588 7652 7715 7778 7841 7904 7967 8030 8093 8156 63 689 8219 8282 8345 8408 8471 8534 8597 8660 8723 8786 63 690 838849 8912 8975 9038 9101 9164 9227 9289 9352 9415 63 691 9478 954J. 9604 9667 9729 9792 9855 9918 9981 ♦ ♦43 63 692 840106 0169 0232 0294 0357 0420 0482 0545 0608 0671 63 693 0783 0796 0859 0921 0984 1046 1109 1172 1234 1297 63 694 1359 1422 1485 1547 1610 1672 1735 1797 1860 1922 63 695 1985 2047 2110 2172 2235 2297 2360 2422 2484 2547 62 696 2609 2672 27S4 2796 2850 2921 2983 8046 1 8108 3170 62 697 3233 3295 3857 3420 3482 3544 8606 3669 i 3731 3793 62 698 3855 3918 8980 4042 4104 4166 4229 4291 1 4353 4415 62 699 4477 4539 4601 4664 4726 4788 4850 4912 1 4974 5036 62 1 • 1 1 2 3 4 1 5 1 6 7 1 8 1 » D. 12 A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. X 2 3 4 5 6 7 8 9 6656 D. 700 845098 6160 6222 6284 B846 5408 5470 5532 6594 62 701 5718 6780 6842 6904 6966 6028 6090 6161 6213 6275 62 702 6337 6399 6461 6523 6585 6646 6708 6770 6832 6394 62 703 6955 7017 7079 7141 7202 7264 7326 7388 7449 7511 62 704 7573 7634 7696 7758 7819 7881 7943 8004 8066 8L28 62 705 8189 8251 8312 8374 8435 8497 8559 8620 8682 8743 62 706 8805 8866 8928 8989 9051 9112 9174 9235 9297 9S58 61 707 9419 9481 9642 9604 9665 9726 9788 9849 9911 f J72 61 708 850033 0095 0166 0217 0279 0340 0401 0462 0624 0585 61 709 0646 0707 0769 0830 0891 0962 1014 1075 1136 1197 61 710 851258 1320 1381 1442 1503 1564 1625 1686 1747 j.809 61 711 1870 1931 1992 2053 2114 2175 2236 2297 2358 2419 61 712 2480 2541 2602 2663 2724 2785 2846 2907 2968 3029 61 713 3090 3150 3211 3272 3333 3394 3455 3516 3577 3637 61 714 3698 3759 3820 3881 3941 4002 4063 4124 4186 4245 61 715 4306 4367 4428 4488 4549 4610 4670 4731 4792 4852 61 716 4913 4974 6034 5095 5156 5216 6277 5337 6398 5459 61 717 5519 5580 5640 6701 5761 5822 5882 6943' 6003 6064 61 718 6124 6185 6245 6306 6366 6427 6487 6548 6608 6668 60 719 6729 6789 6850 6910 6970 7031 7091 7152 7212 7272 60 720 857332 7393 7453 7513 7574 7634 7694 7755 7815 7875 60 721 7935 7995 8066 8116 8176 8236 8297 8357 8417 8477 60 722 8537 8597 8657 8718 8778 8838 8898 8958 9018 9078 60 723 9138 9198 9268 9318 9379 9439 9499 9659 9619 9679 60 724 9739 9799 9859 9918 9978 **38 ♦ ♦98 ♦ 158 ♦ 218 ♦ 278 60 725 860338 0398 0458 0618 0578 0637 0697 0767 0817 0877 60 726 0937 0996 1066 1116 1176 1236 1296 1355 1415 1475 60 727 1534 1594 1654 1714 1773 1833 1893 1952 2012 2072 60 728 2131 2191 2251 2310 2370 2430 2489 2549 2608 2668 60 729 2728 2787 2847 2906 2966 3025 3085 3144 8204 3263 60 730 863323 3382 8442 3601 8561 3620 3680 3739 3799 3858 59 731 3917 3977 4036 4096 4L55 4214 4274 4333 4392 4452 59 732 4511 4570 4630 4689 4748 4808 4867 4926 4985 5045 59 733 5104 6163 6222 5282 5341 6400 6459 6619 6578 5637 69 734 5696' 5755 5814 5874 5933 5992 6051 6110 6169 6228 59 735 6287 6346 6405 6466 6524 6583 6642 6701 6760 6819 59 736 6878 6937 6996 7066 7114 7173 7232 7291 7350 7409 69 737 7467 7526 7586 7644 7703 7762 7821 7880 7939 7998 69 738 8056 8115 8174 8233 8292 8360 8409 8468 8627 8686 59 739 8644 8703 8762 8821 8879 8938 8997 9056 9114 9173 69 740 869232 9290 9349 9408 9466 9625 9584 9642 9701 9760 59 741 9818 9877 9936 9994 *«53 ♦ 111 ♦ 170 ♦ 228 ♦ 287 ♦ 345 59 742 870404 0462 0621 0579 0638 0696 0755 0813 0872 0930 58 743 0989 1047 1106 1164 1223 1281 1339 1398 1466 1515 58 744 1573 1631 1690 1748 1806 1865 1923 1981 2040 2098 68 745 2156 2215 2273 2331 2389 2448 2506 2564 2622 2681 58 746 2739 2797 2856 2913 2972 3030 3088 3146 3204 3262 68 747 3321 3379 3437 3495 3553 3611 3669 3727 3785 3844 68 748 8902 3960 4018 4076 4134 4192 4250 4308 4366 4424 58 749 4482 4540 4598 4656 4714 4772 4830 4888 4945 5003 58 750 875061 5119 5177 5235 5293 5351 6409 6466 5624 5582 58 1 751 5640 6698 5756 5813 5871 5929 5987 6045 6102 6160 68 ' 752 6218 6276 6333 6391 6449 6507 6564 6622 6680 6737 68 ' 753 6795 6863 6910 6968 7026 7083 7141 7199 7256 7314 58 754 7871 7429 7487 7544 7602 7659 7717 7774 7832 7889 68 755 7947 8004 8062 8119 8177 8234 8292 8349 8407 8464 57 756 8522 8579 8637 8694 8752 8809 8866 8924 8981 9039 57 757 9096 9153 9211 9268 9325 9383 9440 9497 9555 9612 67 758 9669 9726 9784 9841 9898 9956 ♦ ♦13 ♦ ♦70 ♦ 127 ♦ 186 57 759 880242 0299 0366 0413 0471 0528 0585 0642 0699 0756 57 N. 1 8 3 4 5 C 7 8 9 D. A TABLE OF LOGARITHMS FBOM 1 TO 10,000. 13 N. 1 2 3 4 5 G 7 8 9 D. 760 880814 0871 0928 0985 1042 1099 1156 1213 1271 1328 57 761 1385 1442 1499 1556 1613 1670 1727 1784 1841 1898 57 762 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 57 763 2525 2581 2688 2695 2752 2809 2866 2923 2980 3037 57 764 3098 3150 3207 3264 3321 3377 3434 3491 3548 3605 67 765 3661 3718 3775 3832 3888 3945 4002 4059 4115 4172 67 766 42'29 4285 4342 4399 4455 4512 4569 4625 4682 4739 57 767 4795 4852 4909 4965 5022 5078 5135 5192 5248 5305 57 768 5361 5418 5474 5531 5587 5644 5700 5757 5813 5870 57 769 5926 5983 6039 6096 6152 6209 6265 6321 6378 6434 56 770 886491 6547 6604 6660 6716 67^3 6829 6885 6942 6998 66 771 7054 7111 7167 7223 7280 7336 7392 7449 7505 7561 56 772 7617 7674 7730 7786 7842 7898 7955 8011 8067 8123 66 773 8179 8236 8292 8348 8404 8460 8516 8573 8629 8685 56 774 8741 8797 8853 8909 8965 9021 9077 9134 9190 9246 56 775 9302 9358 9414 9470 9526 9582 9638 9694 9750 9806 56 776 9862 9918 9974 ♦ ♦80 ♦ ♦86 ♦141 ♦ 197 ♦ 253 ♦ 309 ♦ 365 66 777 890421 0477 0533 0589 0645 0700 0756 0812 0868 0924 66 778 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 56 779 1537 1593 1649 1705 1760 1816 1872 1928 1983 2039 56 780 892095 2150 2206 2262 2317 2373 2429 2484 2540 2595 66 781 2651 2707 2762 2818 2873 2929 2985 3040 8096 3151 56 782 3207 3262 3318 3373 3429 3484 3540 3595 3651 3706 56 783 3762 3817 3873 8928 3984 4039 4094 4150 4205 4261 65 784 4816 4371 4427 4482 4538 4593 4648 4704 4769 4814 55 785 4870 4925 4980 5036 5091 5146 5201 5257 5312 5367 55 786 5423 5478 5538 5588 5644 5699 5754 5809 5864 5920 55 787 5975 6030 6085 6140 6195 6251 6306 6361 6416 6471 56 788 6526 6581 6636 6692 6747 6802 6857 6912 6967 7022 55 789 7077 7132 7187 7242 7297 7352 7407 7462 7617 7572 55 790 897627 7682 7737 7792 7847 7902 7957 8012 8067 8122 55 791 8176 8231 8286 8341 8396 8451 8506 8561 8615 8670 56 792 8725 8780 8835 8890 8944 8999 9054 9109 9164 9218 55 793 9273 9328 9383 9437 9492 9547 9602 9656 9711 9766 55 794 9821 9875 9930 9985 ♦ ♦39 ♦ ♦94 ♦ 149 ♦ 203 ♦ 258 ♦ 312 55 795 900367 0422 0476 0531 0586 0640 0695 0749 0804 0859 56 796 0913 0968 1022 1077 1131 1186 1240 1295 1349 1404 55 797 1458 1513 1567 1622 1676 1731 1785 1840 1894 1948 54 798 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 54 799 2547 2601 2655 2710 2764 2818 2873 2927 2981 3036 54 800 903090 3144 3199 3253 3307 3361 3416 3470 3524 3578 54 801 3633 3687 3741 3795 3819 3904 3958 4012 4066 4120 54 802 4174 4229 4283 4337 4391 4445 4499 4553 4607 4661 54 803 4716 4770 4824 4878 4932 4986 5040 5094 5148 5202 54 804 5256 5310 5364 5418 6472 5526 5580 5634 5688 5742 54 805 5796 5830 5904 5958 6012 6066 6119 6173 6227 6281 54 806 6335 6389 6443 6497 6551 6604 6658 6712 6766 6820 64 807 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 54 808 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 54 809 7949 8002 8056 8110 8163 8217 8270 8324 8378 8431 54 810 908485 8539 8592 8646 8699 8753 8807 8860 8914 8967 54 811 9021 9074 9128 9181 9235 9289 9842 9396 9449 9503 54 812 9556 9610 9663 9716 9770 9823 9877 9930 9984 ♦ ♦37 53 813 910091 0144 0197 0251 0304 0358 0411 0464 0518 0571 63 814 0624 0678 0731 0784 0838 0891 0944 0998 1051 1104 63 815 1158 1211 1264 1317 1371 1424 1477 1530 1584 1637 63 816 1690 1743 1797 1850 1903 1956 2009 2063 2116 2169 53 817 2222 2275 2328 2381 2435 2488 2541 2594 2647 2700 63 818 2753 2806 2859 2913 2966 3019 3072 3125 3178 3231 53 819 3284 3337 3390 3443 3496 3549 3602 3655 3708 3761 63 N. 1 2 3 4 5 6 7 8 9 D. 14 A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. 1 2 1 3 4 5 6 7 \ 8 9 D. 820 918814 3867 3920 3973 4026 1079 1132 1181 1237 4290 53 821 4343; 1396 4449 4502 4555 4608 1660 1713 4766 1819 53 822 4872 4925 4977 5030 5083 5136 5189 6211 5294 5347 53 823 5400 5453 5505 5568 5611 5661 5716 5769 5822 5875 53 821 5927 5980 6033 6085 6138 6191 6248 6296 6349 6401 53 825 6451 6507 6559 6612 6664 6717 6770 6822 6875 6927 53 826 6980 7033 7085 7138 7190 7213 7295 7348 7400 7453 53 827 7506 7558 7611 7663 7716 7768 7820 7873 7925 7978 62 828 8030 8033 8135 8188 8240 8293 8315 8397 8150 8602 62 829 8555 8607 8659 8712 8764 8816 8869 8921 8973 9026 52 830 919078 9130 9183 9235 9287 9310 9392 9444 9196 9549 52 831 9601 9653 9706 9758 9810 9862 9911 9967 ♦ ♦19 ♦♦71 52 832 920123 0176 0228 0280 0332 0384 0136 0489 0541 0593 62 833 0615 0697 0749 0801 0853 0906 0958 1010 1062 1114 62 831 1166 1218 1270 1322 1374 1426 1478 1530 1582 1634 52 835 1686 1738 1790 1842 1894 1946 1998 2050 2102 2151 52 836 2206 2258 2310 2362 2414 2166 2518 2570 2622 2674 52 837 2725 2777 2829 2881 2933 2985 3037 3089 3140 3192 52 838 8244 3296 3348 3399 3151 3503 3555 3607 3658 3710 52 839 3762 3814 3865 3917 3969 4021 4072 1121 4176 4228 52 840 924279 4331 4383 4431 4486 1538 4589 1641 4693 4744 52 811 4796 4818 4899 4951 5003 5054 5106 5157 6209 5261 52 812 5312 5361 5415 5167 5518 5570 5621 5673 5725 5776 52 813 5828 5879 6931 5982 6034 6085 6137 6188 6210 6291 61 811 6342 6394 6115 6497 6548 6600 6651 6702 6754 6805 51 815 6857 6908 6959 7011 7062 7114 7165 7216 7268 7319 51 816 7370 7422 7173 7524 7576 7627 7678 7730 7781 7832 61 817 7883 7935 7986 8037 8088 8140 9191 8212 8293 8346 51 818 8396 8447 8198 8549 8601 8652 8703 8751 8805 8857 51 819 8908 8959 9010 9061 9112 9163 9215 9266 9317 9368 51 850 929419 9470 9521 9572 9623 9674 9725 9776 9827 9879 51 851 9930 9981 ♦ ♦32 ♦ ♦83 ♦ 134 ♦ 185 ♦ 236 ♦ 287 ♦ 338 ♦ 389 51 852 930110 0491 0542 0592 0643 0694 0745 0796 0847 0898 51 853 0919 1000 1051 1102 1153 1204 1254 1305 1356 1107 51 851 1158 1509 1560 1610 1661 1712 1763 1811 1865 1915 51 855 1966 2017 2068 2118 2169 2220 2271 2322 2372 2423 51 856 2474 2524 2575 2626 2677 2727 2778 2829 2879 2930 51 857 2981 3031 3082 3133 3183 3234 3285 3335 3386 3137 51 858 3487 3538 3589 8639 3690 3710 3791 8811 3892 3943 61 859 3993 1011 4091 4145 4195 1216 4296 4347 1397 4448 51 860 931198 1519 4599 4650 4700 1751 4801 4852 1902 4963 50 861 5003 5051 5104 5151 5205 5255 5306 5356 6106 5467 50 862 5507 5558 5608 5658 5709 6759 5809 5860 6910 6960 50 863 6011 6061 6111 6162 6212 6262 6313 6363 6413 6463 50 861 6514 6561 6611 6665 6715 6765 6815 6865 6916 6966 60 865 7016 7066 7117 7167 7217 7267 7317 7367 7118 7468 60 866 7518 7568 7618 7668 7718 7769 7819 7869 7919 7969 50 867 8019 8069 8119 8169 8219 8269 8320 8370 8120 847,0 60 868 8520 8570 8620 8670 8720 8770 8820 8870 8920 8970 50 869 9020 9070 9120 9170 9220 9270 9320 9369 9119 9469 50 870 939519 9569 9619 9669 9719 9769 9819 9869 9918 9968 60 871 940018 0068 0118 0168 0218 0267 0317 0367 0417 0467 50 872 0516 0566 0616 0666 0716 0765 0815 0865 0915 0964 50 873 1014 1061 1114 1163 1213 1263 1313 1862 1412 1462 50 871 1511 1561 1611 1660 1710 1760 1809 1859 1909 1958 50 875 2008 2058 2107 2157 2207 2256 2306 2355 2405 2455 50 876 2504 2554 2603 2653 2702 2752 2801 2851 2901 2950 50 877 3000 3049 3099 3148 3198 3217 3297 3846 3396 3445 19 878 3195 8541 3593 3613 3692 8712 3791 3841 3890 3939 49 879 3989 4038 4088 4137 4186 1236 4285 4335 4384 4433 49 N. 1 2 3 4 5 6 7 8 9 D. A TABLE OF LOGABITHMS FROM 1 TO 10,000. 15 .N. 1 2 3 4 5 1 6 7 8 9 D. 880 944488 4532 4581 4631 4680 4729 4779 4828 4877 4927 49 881 4976 5025 5074 5124 5173 5222 5272 5321 5370 5419 49 882 6469 B518 5667 5616 5665 5715 5764 6813 5862 5912 49 888 5961 6010 6059 6108 6157 6207 6256 6305 6354 6403 49 884 6452 6501 6551 6600 6649 6698 6747 6796 6845 6894 49 885 6943 6992 7041 7090 7140 7189 7238 7287 7836 7385 49 886 74^4 7483 7532 7581 7630 7679 7728 7777 7826 7875 49 887 7924 7978 8022 8070 8119 8168 8217 8266 8315 8364 49 888 8413 8462 8511 8560 8609 8657 8706 8755 8804 8853 49 889 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 49 890 949390 9439 9488 9536 9585 9634 9683 9731 9780 9829 49 891 9878 9926 9975 ♦ ♦24 ♦ ♦73 ♦ 121 ♦ 170 ♦ 219 ♦ 267 ♦ 316 49 892 950365 0414 0462 0511 0560 0608 0657 0706 0754 0803 49 893 0851 0900 0949 0997 1046 1095 1143 1192 1240 1289 49 894 1338 1386 1435 1483 1532 1580 1629 1677 1726 1775 49 895 1823 1872 1920 1969 2017 2066 2114 2163 2211 2260 48 896 2308 2356 2405 2453 2502 2550 2599 2647 2696 2744 48 897 2792 2841 2889 2938 2986 3034 3083 3131 3180 8228 48 898 3276 3325 3373 3421 3470 3518 8566 3615 3663 3711 48 899 3760 3808 3856 3905 3953 4001 4049 4098 4146 4194 48 900 954243 4291 4389 4387 4435 4484 4582 4580 4628 4677 48 901 4725 4773 4821 4869 4918 4966 5014 5062 5110 5158 48 902 5207 5255 5303 5351 5399 5447 5495 5543 5592 5640 48 903 5688 6736 5784 5832 5880 5928 5976 6024 6072 6120 48 904 6168 6216 6265 6313 6361 6409 6457 6505 6553 6601 48 905 6649 6697 6745 6793 6840 6888 6936 6984 7032 7080 48 906 7128 7176 7224 7272 7320 7368 7416 7464 7512 7559 48 907 7607 7655 7703 7751 7799 7847 7894 7942 7990 8038 48 908 8086 8134 8181 8229 8277 8325 8373 8421 8468 8516 48 909 8564 8612 8659 8707 8755 8803 8850 8898 8946 8994 48 910 959041 9089 9137 9185 9232 9280 9328 9375 9423 9471 48 911 9518 9566 9614 9661 9709 9757 9804 9852 9900 9947 48 912 9995 **42 **90 ♦ 138 ♦ 185 ♦ 233 ♦ 280 ♦ 328 ♦ 376 ♦423 48 913 960471 0518 0566 0613 0661 0709 0756 0804 0851 0899 48 914 0946 0994 1041 1089 1136 1184 1281 1279 1326 1874 47 915 1421 1469 1516 1563 1611 1658 1706 1753 1801 1848 47 916 1895 1943 1990 2038 2085 2132 2180 2227 2275 2322 47 917 2869 2417 2464 2511 2559 2606 2653 2701 2748 2795 47 918 2843 2890 2937 2985 3032 3079 3126 3174 3221 3268 47 919 8316 3363 3410 3457 3504 3552 3590 3646 3693 3741 47 920 963788 3835 3882 3929 3977 4024 4071 4118 4165 4212 47 921 4260 4307 4354 4401 4448 4495 4542 4590 4637 4684 47 922 4731 4778 4825 4872 4919 4966 5013 5061 5108 5155 47 923 5202 5249 5296 5343 6390 5437 5484 5531 5578 5625 47 924 5672 5719 5766 5813 5860 5907 5954 6001 6048 6095 47 925 6142 6189 6236 6283 6329 6376 6423 6470 6517 6564 47 926 6611 6658 6705 6752 6799 6845 6892 6939 6986 7033 47 927 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 928 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 I 929 8016 8062 8109 8156 8203 8249 8296 8343 8390 8436 47 930 968483 8530 8576 8623 8670 8716 8763 8810 8856 8903 47 931 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 47 932 9416 9463 9509 9556 9602 9649 9695 9742 9789 9335 47 933 9882 9928 9975 ♦ ♦21 ♦ ♦68 ♦ 114 ♦ 161 ♦ 207 ♦ 254 ♦ 300 47 934 970347 0393 0440 0486 0533 0579 0626 0672 0719 0765 46 i 935 0812 0858 0904 0951 0997 1044 1090 1137 1183 1229 4« 936 1276 1822 1369 1415 1461 1508 1554 1601 1647 1693 46 937 1740 1786 1832 1879 1925 1971 2018 2064 2110 2157 46 988 2203 2249 2295 2342 2388 2434 2481 2527 1 2573 2619 46 939 2666 2712 2758 2804 2851 2897 2943 2989 1 3035 3082 46 1 1 1 a 3 4 5 1 c ! 7 8 ' 9 D. 16 A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. 1 2 3 4 5 C 7 8 9 D. 940 973128 3174 8220 3266 3313 3359 3405 3461 3407 3543 46 941 3590 3636 3682 3728 3774 8820 3866 3913 3959 4005 46 942 4031 4097 4143 4189 4235 4281 4327 4374 4420 4466 46 943 4512 4558 4604 4630 4696 4742 4788 4834 4880 4026 46 944 4972 5018 5064 3110 5156 5202 6248 5294 3340 5386 46 943 3432 5478 6524 5370 6616 5662 5707 6763 5790 6846 46 946 5891 5937 5983 6029 6075 6121 6167 6212 6258 6304 46 947 6330 6396 6442 6488 6533 6579 6625 6671 6717 6763 46 948 6808 6854 6900 6946 6992 7037 7083 7129 7173 7220 46 949 7266 7312 7358 7403 7449 7495 7541 7586 7632 7678 46 950 077724 7769 7815 7861 7906 7952 7998 8043 8089 8135 46 931 8181 8226 8272 8317 8363 8409 8454 8500 8546 8501 46 032 8637 8683 8728 8774 8819 8865 8911 8066 9002 9047 46 953 9093 9138 9184 9230 9275 9321 0366 0412 9457 0503 46 954 9548 9594 9639 9685 9730 0776 0821 0867 0012 0958 46 955 980008 0049 0094 0140 0185 0231 0276 0322 0367 0412 45 956 0458 0503 0349 0594 0640 0685 0730 0776 0821 0867 46 957 0912 0957 1003 1048 1098 1130 1184 1229 1275 1320 45 938 1366 1411 1436 1501 1347 1592 1637 1683 1728 1773 45 959 1819 1864 1909 1954 2000 2045 2090 2185 2181 2226 45 960 982271 2316 2362 2407 2432 2407 2543 2588 2633 2678 43 961 2723 2769 2814 2839 2904 2949 2904 3040 3085 3130 45 962 3175 3220 3265 3310 3356 3401 3446 3491 3536 3581 43 963 3626 3671 3716 3762 3807 3852 3897 3942 3087 4033 45 964 4077 4122 4167 4212 4257 4302 4347 4392 4437 4482 46 965 4327 4572 4617 4662 4707 4732 4797 4842 4887 4932 46 966 4977 5022 5067 5112 3157 5202 5247 3292 5337 6382 45 967 5426 5471 5516 5561 5606 6631 5696 6741 6786 6830 45 968 5875 5920 5965 6010 6055 6100 6144 6189 6234 6270 45 969 6324 6869 6413 6458 6503 6348 6593 6637 6682 6727 46 970 986772 6817 6861 6906 6951 6996 7040 7086 7130 7175 45 971 7219 7264 7309 7333 7398 7443 7488 7532 7677 7622 46 973 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 45 973 8113 8137 8202 8247 8291 8336 8381 8425 8470 8514 46 974 8559 8604 8648 8693 8737 8782 8826 8871 8916 8060 45 975 9005 9049 9094 9138 9183 9227 9272 9316 9361 9405 45 976 9450 9494 9539 9583 9628 9672 9717 9761 9806 9850 44 977 9895 9939 9983 ♦ ♦28 ♦ ♦72 ♦ 117 ♦ 161 ♦ 206 ♦ 260 ♦ 294 44 978 090339 0383 0428 0472 0516 0561 0605 0660 0604 0738 44 979 0783 0827 0871 0916 0960 1004 1049 1093 1137 1182 44 980 091226 1270 1315 1359 1403 1448 1492 1586 1580 1626 44 981 1669 1713 1758 1802 1846 1890 1935 1979 2023 2067 44 982 2111 2156 2200 2244 2288 2333 2377 2421 2466 2509 44 983 2354 2398 2642 2686 2780 2774 2810 2863 2007 2951 44 984 2993 3039 8083 3127 3172 3216 3260 3304 8348 3392 44 985 3436 3480 3524 3568 3613 3637 8701 8745 3780 8833 44 986 3877 3921 3963 4009 4053 4097 4141 4185 4229 4273 44 987 4317 4361 4403 4449 4493 4537 4581 4626 4669 4713 44 988 4757 4801 4845 4889 4933 4977 5021 6065 5108 5152 44 989 5196 5240 5284 6328 5372 5416 6460 5504 6647 6591 44 €9-) 095635 5679 3723 5767 5811 6854 6808 6942 5986 6030 44 991 6074 6117 6161 6205 6249 6293 6337 6380 6424 6468 44 992 6312 6335 6599 6643 6687 6731 6774 6818 6862 6906 44 993 6949 6993 7037 7080 7124 7168 7212 7253 7290 7348 44 994 7386 7430 7474 7517 7561 7605 7648 7692 7736 7779 44 995 7823 7867 7910 7934 7008 8041 8085 8129 8172 8216 44 996 8239 8303 8347 8390 8434 8477 8521 8564 8608 8652 44 997 8695 8739 8782 8826 8869 8913 8066 9000 9043 9087 44 998 9131 9174 9218 9261 9305 9348 0392 9435 9479 9522 44 999 9563 9609 9652 0696 9730 0783 9826 9870 9913 9957' 43 Ti. ! . ! 1 1 1 1 2 3 4 5 6 7 8 9 D. A TABLE LOGARITHMIC SINES AND TANGENTS FOR EVERY DEGREE AND MINUTE OF THE QUADRANT. Remark. The minutes in the left-hand column of each page, increasing downward, belong to the degrees at the top ; and those increasing upward, in the right-hand column, belong to the degrees below. (0 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. i Tang. D. Cotang. 0-000000 10 000000 0-000000 Inflnite. 60 1 6-463726 5017-17 000000 ■00 ; 6-463726 5017-17 13-536274 59 2 764756 2934-85 000000 1 •00 7647B6 2934-83 235244 58 I 3 940847 2082-31 000000 -00 ■ 940847 2082-31 059153 57 d: 7-065786 1615-17 000000 -00 . 7-065786 1616-17 12-934214 56 5 162696 1319-68 000000 -00 1 162696 1319 • 69 837304 65 6 241877 1115-75 9-999999 •01 241878 1115 • 78 758122 54 7 308824 966-53 999999 •01 308825 996 • 53 691175 53 8 866816 852-54 999999 •01 366817 852^54 633183 62 9 417968 762-63 999999 •01 417970 762^63 582030 51 10 463725 689-88 999998 •01 463727 689 •88 536273 60 11 7-505118 629-81 9-999998 •01 7-505120 629-81 12 •494880 49 12 542906 579-36 999997 •01 542909 579-33 457091 48 13 577668 536-41 999997 •01 577672 536-42 422328 47 14 609853 499-38 999996 •01 609857 499-39 390143 46 15 639816 467-14 999996 •01 639820 467 15 360180 45 16 667845 438-81 999995 •01 667849 438-82 332151 44 17 694173 413-72 999995 -01 694179 413-73 305821 48 18 718997 391-35 999994 -01 719004 391-36 280997 42 19 742477 371-27 999993 -01 742484 371-28 257516 41 20 764754 353-15 999993 ■01 764761 351-36 235239 40 21 7-785943 336-72 9-999992 -01 7-785951 336-73 12^214049 39 22 806146 321-75 999991 •01 806155 321-76 193845 38 23' 825451 308-05 999990 -01 825460 308-06 174540 37 24 843984 295-47 999989 •02 843944 295-49 156056 36 25 861662 283-88 999988 •02 861674 283-90 138326 35 26 878695 273-17 999988 •02 878708 273-18 121292 34 27 895085 263-23 999987 •02 895099 263-25 104901 33 28 910879 253-99 999986 •02 910894 254-01 089106 32 29 - 926119 245-38 999985 -02 926134 245-40 073866 31 30 940842 237-33 999983 -02 940858 237-35 059142 30 81 7-955082 229-80 9-999982 -02 7-955100 229-81 12-044900 29 32 968870 222-73 999981 -02 968889 222-75 031111 28 33 982233 216-08 999980 -02 982253 216-10 017747 27 34 995198 209-81 999979 -02 995219 209-83 004781 26 35 8-007787 203-90 999977 ■02 8-007809 203-92 11-992191 25 36 020021 198-31 999976 -02 030045 198-33 979955 24 37 031919 193-02 999975 -02 031945 193-05 968055 23 38 043501 188-01 999973 -02 043527 188-03 956473 22 89 054781 183-25 999972 ■02 054809 183-27 945191 21 40 065776 178-72 999971 •02 065806 178-74 934194 20 41 8-076500 174-41 9-999969 •02 8'-076531 174-44 11-923469 19 42 086965 170-31 999968 •02 086997 170-34 913003 18 43 097183 166-39 999966 •02 097217 166^42 902783 17 44 107167 162-65 999964 •03 107202 162-68 892797 16 45 116926 159-08 999963 •03 116963 159-10 883037 15 46 126471 155-66 999961 •03 126510 155-68 873490 14 47 135810 152-88 999959 •03 135851 152-41 864149 13 48 144953 149-24 999958 •03 144996 149-27 855004 12 49 153907 146-22 999956 •03 153952 146-27 846048 11 50 162681 143-33 999954 •03 162727 143-36 837273 10 51 8-171280 140-54 9-999952 •03 8-171328 140-57 11-828672 9 52 179713 137-86 999950 •03 179768 137-90 820237 8 53 187985 135-29 999948 •03 188036 185-32 811964 7 54 196102 132-80 999946 -03 196156 132-84 803844 6 55 204070 130-41 999944 -08 204126 130-44 795874 5 56 211895 128-10 999942 •04 211953 128-14 788047 4 57 219581 125-87 999940 •04 219641 125-90 780359 3 58 227134 123-72 999938 •04 227195 123-76 772805 2 59 234557 121-64 999936 •04 234621 121-68 765379 1 60 241855 119-63 999934 -04 ; 241921 119-67 758079 Cosine. D. Sine. 1 Cotang. i D. Tang. JH. (89 DEGREES.) SINES AND TANGENTS . (1 DEGREE.) 19 M. ' Sine. D. CoRine. 1 D. ! Tang. D. i Cotang. 1 8-241855 119-63 9-999934 ! -04 8-241921 , 119-67 11-758079 | 60 1 249033 117-68 999982 -04 249102 ; 117^72 750898 69 2 256094 116-80 999929 •04 256165 1 115 ■ 84 743835 58 8 263042 113-98 999927 •04 268115 114^02 736885 57 4 269881 112-21 999925 -04 269956 U2^25 730044 56 6 276614 110-50 999922 -04 276691 110-54 723309 55 ! 6 288248 108-83 999920 •04 288328 108-87 716677 64 7 289778 107-21 999918 •04 289856 107-26 710144 58 8 296207 106-65 999915 -04 296292 105-70 703708 52 9 302546 104-18 999913 -04 302634 104-18 697866 51 10 808794 102-66 999910 -04 808884 102-70 691116 50 11 8-314954 101-22 9-999907 •04 8-815046 101-26 LI ■684954 49 12 821027 99-82 999905 -04 321122 99-87 678878 48 13 327016 98-47 999902 •04 327114 98-51 672886 47 14 332924 97-14 999899 -05 333025 97-19 666975 46 15 388758 95-86 999897 -05 338856 95-90 661144 45 16 344504 94-60 999894 -05 344610 94-65 655390 44 17 350181 93-38 999891 •05 350289 93-43 649711 48 18 855783 92-19 999888 -05 855895 92-24 644105 42 19 361315 91-03 999885 -05 361480 . 91-08 *• 638570 41 20 366777 89-90 999882 -05 866895 89-95 633105 40 21 8 -'372171 88-80 9-999879 -05 8 ■ 372292 88-85 11^ 627708 39 22 877499 87-72 999876 -05 377622 87-77 622878 38 23 382762 86-67 999873 -05 382889 86-72 617111 87 24 887962 85-64 999870 -05 388092 85-70 611908 36 25 893101 84-64 999867 •05 393234 84-70 606766 36 26 398179 83-66 999864 -05 398315 B3^71 601685 34 27 408199 82-71 999861 -05 403388 82^76 596662 83 28 408161 81-77 999858 -05 408304 81^82 591696 32 29 413068 80-86 999864 -05 418213 80^91 68678T 81 BO 417919 79-96 999851 -06 418068 80^02 581932 30 31 8-422717 79-09 9-999848 •06 8^422869 79^14 11-677181 29 82 427462 78-23 999844 -06 427618 78^30 572382 28 33 482156 77-40 999841 -06 482315 77-45 567685 27 34 486800 76-57 999888 -06 436962 76-68 563038 26 85 441894 75-77 999834 -06 441660 75-83 558440 25 86 445941 74-99 999831 •06 446110 75-05 563890 24 37 450440 74-22 999827 -06 450613 74-28 549387 23 88 454893 78-46 999828 •06 455070 73-52 544980 22 39 459301 72-78 999820 •06 459481 72-79 540519 21 40 463665 72-00 999816 -06 463849 72-06 536151 20 41 8-467985 71-29 9-999812 •06 8-468172 '71-85 11 ■531828 19 42 472263 70-60 999809 ■06 472454 70-66 627646 ^f 43 476498 69-91 999805 •06 476693 69-98 523307 17 i 44 480693 69-24 999801 •06 480892 69-31 519108 16 i 45 484848 68-59 999797 •07 485050 68-66 514950 ' 15 ' 46 488963 67-94 999793 •07 489170 68-01 510830 14 47 493040 67-31 999790 ■07 493250 67-38 506750 , 13 48 497078 66-69 999786 •07 497298 66-76 502707 ! 12 49 501080 66-08 999782 •07 501298 66-16 498702 1 11 50 605045 66-48 999778 •07 505267 65-55 494733 ■ 10 51 8-508974 64-89 9-999774 •07 8^509200 64-96 11-490800 ! 9 62 612867 64-31 '999769 ■07 -618098 64-89 486902 ! 8 53 516726 68-75 999765 -07 516961 63-82 483039 1 7 54 520551 63 -It 999761 -07 520790 63-26 479210 ! 6 65 524843 62-64 999757 -07 , 524586 62-72 475414 ■ 5 66 628102 62-11 999753 -07 1 D28349 62^18 471651 j 4 67 631828 61-58 999748 -07 1 532080 61-65 467920 1 3 68 685523 61-06 999744 •07 585779 61-13 464221 1 2 59 ' 639186 60-55 999740 •07 539447 60-62 460663 ! 1 60 642819 60-04 999785 ■07 543084 60- 12 456916 1 Cosine. D. Sine. Cotang. D. Tang. JH. (88 DEGREES.) 20 (2 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 8-542819 60-04 9-999735 -07 8 •643084 60^12 11-456916 60 1 546422 59-55 999731 -07 646691 59^62 453309 59 2 549995 59-06 999726 •07 550268 59^14 449732 58 3 553539 58-58 999722 -08 558817 58 ■66 446183 67 4 557054 68-11 999717 ■08 567336 58^19 442664 56 5 560540 67-65 999713 •08 660828 57^73 439172 55 6 563999 57-19 999708 •08 564291 57^27 435709 64 7 567431 56-74 999704 •08 567727 56^82 432273 53 8 570886 56-30 999699 •08 571137 56^38 428863 52 9 574214 55-87 999694 -08 574520 55^95 426480 61 10 577566 55-44 999689 -08 577877 65^52 422123 60 11 8-580892 55-02 9-999686 -08 8-581208 55^10 11-418792 49 12 584193 64-60 999680 -08 584514 64-68 415486 48 13 587469 54-19 999676 -08 587795 54-27 412206 47 14 590721 53-79 999670 -08 691051 53-87 408949 46 15 593948 53-39 999665 -08 594283 63 -47 405717 45 16 597152 53-00 999660 -08 597492 63-08 402508 44 17 600332 52-61 999655 -08 600677 52-70 399323 43 18 603489 52-23 999650 -08 603839 52-32 396161 42 19 606623 61-86 999646 •09 606978 51-94 393022 41 20 609734 51-49 999640 -09 610094 51-58 389906 40 21 8-612823 51-12 9-999635 -09 8-613189 51-21 11-386811 39 22 615891 50-76 999629 -09 616262 50-85 383738 38 23 618937 60-41 999624 •09 619313 50-50 380687 87 24 621962 50- 06 999619 -09 622343 60-15 377657 36 25 624965 49-72 999614 -09 625352 49-81 374648 35 26 627948 49-38 999608 -09 628340 49-47 371660 84 27 630911 49-04 999603 -09 631308 49-13 368692 33 28 633854 48-71 999597 -09 634256 48-80 865744 32 29 636776 48-39 999592 •09 637184 48-48 362816 31 30 639680 48-06 999586 •09 640093 48-16 359907 30 81 8-642563 47-75 9-999581 -09 8-642982 47-84 11-867018 29 82 645428 47-43 999576 -09 645863 47-53 854147 28 33 648274 47-12 999570 -09 648704 47-22 351296 27 34 651102 46-82 999564 ■09 661637 46-91 848463 26 35 653911 46-52 999658 ■10 654352 46-61 345648 25 36 656702 46-22 999553 -10 657149 46-31 342851 24 37 659475 45-92 999647 -10 659928 46-02 840072 28 38 662230 45-63 999641 -10 662689 45-73 337311 22 39 664968 45-35 999585 ■10 665433 45-44 334567 21 40 667689 45-06 999529 •10 668160 45-26 331840 20 41 8-670393 44-79 9-999524 ■10 8-670870 44-88 11-829130 19 42 673080 44-51 999518 -10 673563 44-61 326437 18 43 675751 44-24 999512 -10 676239 44-34 323761 17 44 678405 43-97 999506 -10 678900 44-17 321100 16 45 681043 48-70 999500 -10 681644 43-80 318456 15 46 683665 43-44 999493 -10 684172 43-54 315828 14 47 686272 43-18 999487 ■10 686784 43-28 313216 13 48 688863 42-92 999481 -10 689381 43-03 310619 12 49 691438 42-67 999475 -10 691963 42-77 308037 11 50 698998 42-42 999469 -10 694529 42-52 305471 10 51 8-696543 42-17 9-999463 -11 8-697081 42-28 11-302919 9 52 699073 41-92 999466 -11 699617 42-03 800383 8 53 701589 41-68 999450 -11 702139 41-79 297861 7 54 704090 41-44 999443 •11 704646 41-55 295354 6 65 706577 41-21 999437 -11 707140 41-32 292860 5 56 709049 40-97 999431 -11 709618 41-08 290382 4 57 711507 40-74 999424 -11 712083 40-85 287917 3 58 713952 40-51 999418 -11 714634 40-62 285465 2 S9 716383 40-29 999411 •11 716972 40-40 283028 1 60 718800 40-06 999404 •11 719396 40-17 280604 Cosine. D. Sine. I Cotang. D. Tang. M. (87 DEGREES.) SINES AND TANGENTS. (3 DEGREES.) 21 M. Sine. D. Cosine. D. Tang. 1 D. Cotang. 8-718800 40-06 9-999404 - 11 8-719396 40-17 11-280604 60 1 721204 39-84 999398 - 11 721806 39-95 278194 59 2 723595 39-62 999391 11 724204 39-74 275796 58 3 725972 39-41 999384 • 11 726588 39-52 278412 57 4 728337 39-19 999378 11 728959 39-80 271041 56 5 730688 38-98 999371 11 731317 39-09 268683 55 6 733027 38-77 999364 12 733663 38-89 266337 54 7 735354 38-57 999357 12 735996 38-68 264004 53 8 737667 38-36 999350 12 738317 38-48 261683 52 9 739969 38-16 999343 12 740626 38-27 259374 51 10 742259 37-96 999336 12 742922 38-07 257078 50 11 8-744536 37-76 9-999329 • 12 8-745207 37-87 11-254793 49 12 746802 37-56 999322 12 747479 37-68 252521 48 13 749055 37-37 999315 12 749740 37-49 250260 47 14 751297 37-17 999308 12 751989 37-29 248011 46 15 753528 36-98 999301 12 754227 37-10 245773 45 16 755747 36-79 999294 12 756453 86-92 243547 44 17 757965 36-61 999286 12 758668 36-73 241332 43 18 • 760151 36-42 999279 12 760872 36-55 239128 42 19 762337 36-24 999272 12 763065 36-36 236935 41 20 764511 36-06 999265 12 765246 36-18 234754 40 21 8-766675 35-88 9-999257 12 8-767417 36-00 11-232583 39 22 768828 35-70 999250 13 769578 35-83 230422 38 23 770970 35-53 999242 13 771727 35-65 228273 37 24 773101 35-35 999235 13 773866 35-48 226134 36 25 775223 35-18 999227 13 775995 35-31 224005 35 26 777333 35-01 999220 13 778114 35-14 221885 34 27 779434 34-84 999212 13 780222 34-9r 219778 33 28 781524 34-67 999205 13 782320 34-80 217680 32 29 783605 34-51 999197 13 784408 34-64 215592 31 30 785675 34-31 999189 13 786486 34-47 213514 30 31 8-787786 34-18 9-999181 13 8-788554 34-31 11-211446 29 32 789787 34-02 999174 13 790613 34-15 209387 28 33 791828 33-86 . 999166 13 792662 33-99 207338 27 34 793859 33-70 999158 13 794701 33-83 205299 26 35 795881 33-54 999150 13 796731 33-68 203269 25 36 797894 33-39 999142 13 798752 33-52 201248 24 37 799897 33-23 999134 13 800763 33-37 199237 23 38 801892 33-08 999126 13 802765 33-22 197235 22 39 803876 32-98 999118 13 804758 33-07 195242 21 40 805852 32-78 999110 13 806742 32-92 193258 20 41 8-807819 32-63 9-999102 13 8-808717 32-78 11-191283 19 42 809777 32-49 999094 14 810683 32-62 189317 18 43 811726 32-34 999086 14 812641 32-48 187359 17 44 813667 32-19 999077 14 814589 32-33 185411 16 45 815599 32-05 999069 14 816529 32-19 183471 15 1 46 817522 31-91 999061 14 818461 32-05 181539 14 1 47 819436 31-77 999053 14 820384 31-91 179616 ! 13 ! 48 821343 31-63 999044 14 822298 31-77 177702 12 1 49 823240 31-49 999036 14 824205 31-63 175795 11 1 50 825130 31-35 999027 14 826103 31-50 173897 10 ' 51 8-827011 31-22 9-999019 14 8-827992 31-36 11-172008 9 i 52 828884 31-08 999010 14 829874 31-23 170126 8 1 53 830749 30-95 999002 14 831748 31-10 168252 7 ! 54 832607 30-82 998993 -14 833613 30-96 166387 6 55 834456 30-69 998984 -14 835471 80-83 164529 5 56 836297 80-56 998976 -14 837321 30-70 162679 4 57 838130 30-43 998967 -15 839163 30-57 160837 3 58 839956 30-30 998958 -15 840998 30-45 159002 2 59 841774 30-17 998950 ■15 842825 30-32 157175 1 60 843585 30-00 998941 -15 844644 30- 19 155356 Cosine. D. Sine. 1 Cotang. D. Tang. Itl. (86 DEGREES.) 22 (4 DEGREES.) A TABLE OP LOGARITHMIC M. Sine. D. Cosine. | ». Taug. D. Cotang. , 8-843585 30-05 9-998941 -15 8-844644 i 30 19 11 •155356 60 1 845387 29-92 998932 -15 846455 1 30-07 153545 69 2 847183 29-80 998923 -15 848260 29-95 151740 58 3 848971 29-67 998914 -15 850057 29-82 149943 57 4 850751 29-55 998905 -15 851846 29-70 148154 56 6 852525 29-43 998896 -15 853628 29-58 146372 55 6 854291 29-31 998887 •15 855403 29-46 144597 54 7 856049 29-19 998878 •15 857171 29-35 142829 53 8 857801 29-07 998869 •15 858932 29-23 141068 52 9 859546 28-96 998860 •15 860686 29-11 139314 51 10 861283 28-84 998851 -15 862433 29-00 137567 50 11 8-863014 28-73 9-998841 -15 8-864173 28-88 11 •135827 49 12 864738 28-61 998832 -15 865906 28-77 134094 48 13 866455 28-50 998823 •16 867632 28^66 132368 47 14 868165 28-39 998813 •16 869351 28-54 130649 46 15 869868 28-28 998804 •16 871064 28-43 128936 45 16 871565 28-17 998795 •16 872770 28-32 127230 44 17 873255 28-06 998785 •16 874469 28-21 125531 43 18 874938 27-95 998776 •16 876162 28-11 123838 42 19 876615 27-84 998766 •16 877849 28-00 122151 41 20 878285 27-73 998757 •16 879529 27-89 120471 40 21 8-879949 27-63 9-998747 •16 8-881202 27-79 11-118798 39 22 881607 .27-52 998738 •16 882869 27-68 117131 38 23 883258 ■ 27-42 998728 •16 884530 27-58 115470 37 24 884903 27-31 998718 •16 886185 27-47 113815 36 25 886542 37-21 998708 •16 887833 27-37 112167 35 26 888174 27-11 998699 •16 889476 27-27 110524 34 27 889801 2-T-OO 998689 •16 891112. 27-17 108888 38 28 891421 £6-90 998679 •16 892742 27-07 107258 32 29 893035 26-80 998669 •17 894366 26-97 105634 31 30 894643 26-70 998659 •17 895984 26-87 104016 30 31 8-896246 26*60 9-998649 •17 8-897596 26^77 11-102404 29 32 897842 26-51 998639 -17 899203 26-67 100797 28 33 899432 26-41 998629 -17 900803 26-58 099197 27 34 901017 26-31 998619 -17 902398 26-48 097602 26 35 902596 26-22 998609 •17 903987 26-38 096013 25 36 904169 26-12 998599 •17 905570 26-29 094430 24 37 905736 26-03 998589 •17 907147 26-20 092853 23 38 907297 25-93 998578 •17 908719 26-10 091281 22 39 908853 25-84 998568 •17 910285 26-01 089715 21 40 910404 25-75 998558 •17 911846 25^92 088154 20 41 8-911949 25-66 9-998548 •17 8-913401 25^83 11^086599 19 42 913488 25-56 998537 ■17 914951 25-74 085049 18 43 915022 25-47 998527 •17 916495 25-65 083505 17 44 916550 25-38 998516 •18 918034 25-56 081966- 16 45 9180V3 25-29 998506 •18 919568 25-47 080432 15 46 910591 2o:J0 998495 •18 921096 25-38 076304 14 47 9i;iM« 25 12 998485 •18 922619 25-80 077381 13 48 922610 25-03 998474 -18 924136 25-21 075864 12 49 924 U 2 24-94 998464 •18 925649 25-12 074351 11 50 92o60fl" 24-86 998453 •18 927156 25-03 072844 10 di 8-927100 24-77 9-998442 •18 8-928658 24-95 11-071842 9 52 92S587 24-69 998431 -18 930155 24-86 069845 8 53 930068 24-60 998421 ■18 931647 24-78 ()68853 7 54 931544 24-52 998410 -18 933134 24-70 066866 6 55 933015 24-43 998399 •18 934616 24-61 065384 5 56 934481 24-35 998888 •18 936093 24-53 068907 4 57 935942 24-27 998377 •18 937565 24-45 062435 8 58 937898 24-19 998366 •18 939032 24-37 060968 2 59 938850 24-11 998355 •18 940494 24-30 059506 1 60 1 940296 24-03 998344 ■18 i 941952 24-21 058048 Cosine. D. Sine. 1 Cotang. D. Tang. m. (85 DBJGBEES.'i SINES AND TANGENTS. (5 DEGREES.) 23 M. Sine. D. Cosine. D. Tang. D. Cotang. 8-940296 24-08 9-998844 19 8-941952 24-21 11-058048 60 1 941738 23-94 998333 19 943404 24-13 056596 59 2 943174 23-87 998322 19 944852 24-05 066148 58 8 944606 23-79 998311 19 946295 28-97 058705 57 4 946084 23-71 998800 19 947784 23-90 062266 56 5 947456 28-63 998289 19 949168 23-82 050832 55 6 948874 23-66 998277 19 950597 23-74 049403 54 7 950'287 28-48 998266 19 952021 23-66 047979 53 8 951696 28-40 998255 19 953441 23-60 046559 52 9 953100 23-32 998243 19 954856 23-61 045144 51 10 954499 23-25 998232 19 956267 23-44 043733 60 11 8-955894 23-17 9-998220 19 8-957674 23-37 11-042326 49 12 967284 23-10 998209 19 959075 23-20 040925 48 13 958670 23-02 998197 19 960473 23-23 089527 47 14 960052 22-96 998186 19 961866 23-14 038184 46 15 961429 22-88 998174 19 963255 23-07 036745 46 16 962801 22-80 998163 19 964639 23-00 036361 44 17 964170 22-73 998151 19 966019 22-93 033981 43 18 966634 22-66 998189 20 967394 22-86 032606 42 19 966893 22-59 998128 20 968766 22-79 031234 41 20 968249 22-52 998116 20 970133 22-71 029867 40 21 8-969600 22-44 9-998104 20 8-971496 22-65 11-028504 89 22 970947 22-38 998092 20 972856 22-67 027146 38 23 972289 22-31 998080 20 974209 22-51 025791 37 24 973628 22-24 998068 20 975560 22-44 024440 86 25 974962 22-17 998056 20 976906 22-37 028094 35 26 976293 22-10 998044 20 978248 22-30 021752 34 27 9T7619 22-08 998082 20 979586 22-23 020414 88 28 978941 21-97 998020 20 980921 22-17 019079 32 29 980259 21-90 998008 20 982251 22-10 017749 31 SO 981578 21-83 997996 20 983677 22-04 016423 30 31 8-982883 21-77 9-997985 20 8-984899 21-97 11-015101 29 32 984189 21-70 997972 20 986217 21-91 013783 28 33 985491 21-63 997959 20 987532 21-84 012468 27 34 986789 21-67 997947 20 988842 21-78 011158 26 35 988088 21-50 997935 21 990149 21-71 009851 26 86 989374 21-44 997922 21 991461 21-65 008549 24 37 990660 21-38 997910 21 992750 21-68 007260 23 38 991948 21-81 997897 21 994046 21-52 005955 22 39 993222 21-25 997886 21 995337 21-46 004668 21 40 994497 21-19 997872 21 996624 21-40 003376 20 41 8-995768 21-12 9-997860 21 8-997908 21-34 11-002092 19 42 997086 21-06 997847 21 999188 21-27 000812 18 43 998299 21-00 997835 21 9 •000465 21-21 10-999535 17 44 999660 20-94 997822 21 001738 21-15 998262 16 45 9-000816 20-87 997809 21 003007 21-09 996993 15 46 002069 20-82 997797 21 004272 21-03 995728 14 47 003318 20-76 997784 21 005534 20-97 994466 13 48 004568 20-70 997771 21 006792 20-91 993208 12 49 005805 20-64 997758 21 008047 20-85 991953 11 50 007044 20-68 997745 21 009298 20-80 990702 10 51 9-008278 20-52 9-997732 21 9-010546 20-74 10-989454 9 52 009610 20-46 997719 21 011790 20-68 988210 8 53 010737 20-40 997706 21 013031 20-62 986969 7 54 011962 20-34 997693 22 014268 20-56 985732 6 55 013182 20-29 997680 22 015502 20-51 984498 5 66 014400 20-23 997667 22 016732 20-45 983268 4 57 015613 20-17 997654 22 017959 20-40 982041 3 68 016824 20-12 997641 22 019188 20-33 980817 2 59 018031 20-06 997628 -22 020403 20-28 979597 1 60 019285 20-00 997614 -22 021620 20-23 978380 M. Cosine. D. Sine. Cotang. D. Tang. (84 DEGREES.) 24 (6 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tansf. D. Cotang. 60 9-019235 20-00 9-997614 22 9-021620 20-23 10-978380 1 020435 19-95 997601 22 022834 20-17 977166 59 2 021632 19-89 997588 22 024044 20-11 975956 58 3 022825 19-84 997574 22 025251 20-06 974749 57 i 024016 19-78 997561 22 026455 20-00 973545 56 5 025203 19-73 997547 22 027655 19-95 972345 55 6 026386 19-67 997534 23 028852 19-90 971148 54 7 027567 19-62 997520 23 030046 19-85 969954 53 8 028744 19-57 997507 23 031237 19-79 968763 52 9 029918 19-51 997493 23 032425 19-74 967575 51 10 031089 19-47 997480 28 033609 19-69 966391 50 11 9-032257 19-41 9-997466 23 9-034791 19-64 10-965209 49 12 033421 19-36 997452 23 035969 19-58 964081 48 13 034582 19-30 997439 23 037144 19-53 962856 47 14 035741 19-25 997425 23 038316 19-48 961684 46 15 036896 19-20 997411 23 039485 19-43 960515 45 16 038048 19 15 997397 23 040651 19-38 959349 44 17 039197 19-10 997383 23 041813 19-33 958187 43 18 040342 19-05 997S69 23 042973 19-28 957027 42 19 041485 18-99 997355 23 044130 19-23 955870 41 20 042625 18-94 997341 23 045284 19-18 954716 40 21 9-043762 18-89 9-997327 24 9-046434 19-13 10-953566 39 22 044895 18-84 997313 24 047582 19-08 952418 38 23 046026 18-79 997299 24 048727 19-03 951278 37 24 047154 18-75 997285 24 049869 18-98 950131 36 25 048279 18-70 997271 24 051008 18-93 948992 35 26 049400 18:65 997257 24 052144 18-89 947856 34 27 050519 18-60 997242 24 053277 18-84 946723 33 28 051635 18-55 997228 24 054407 18-79 945593 32 29 052749 18-50 997214 24 055535 18-74 944465 31 30 053859 18-45 997199 24 056659 18-70 943341 30 31 9-054966 18-41 9-997185 24 9-057781 18-65 10-942219 29 32 056071 18-86 997170 24 058900 18-60 941100 28 33 057172 18-31 997156 24 060016 18-55 939984 27 34 058271 18-27 997141 24 061130 18-51 938870 26 35 059367 18-22 997127 24 062240 18-46 937760 25 36 060460 18-17 997112 24 063348 18-42 936652 24 37 061551 18-13 997098 24 064453 18-37 936547 23 88 062639 18-08 997083 25 065556 18-33 934444 22 39 063724 18-04 997068 25 066655 18-28 933345 21 40 064806 17-99 997053 25 067752 18-24 932248 20 41 9-065885 17-94 9-997039 25 9-068846 18-19 10-931154 19 42 066962 17-90 997024 25 069938 18-15 930062 18 43 068086 17-86 997009 25 071027 . 18-10 928973 17 44 069107 17-81 996994 25 072113 18-06 927887 16 45 070176 17-77 996979 25 073197 18-02 926803 15 46 071242 17-72 996964 25 074278 17-97 925722 14 47 072306 17-68 996949 25 075356 17-93 924644 13 48 073366 17-63 996934 25 076432 17-89 923568 12 49 074424 17-59 996919 25 077505 17-84 922495 11 50 075480 17-55 996904 25 078576 17-80 921424 10 51 9-076533 17-50 9-996889 25 9-079644 17-76 10-920356 9 52 077583 17-46 996874 25 080710 17-72 919290 8 53 078631 17-42 996858 25 081773 17-67 918227 7 54 079676 17-38 996843 25 082833 17-63 917167 6 55 080719 17-33 996828 25 083891 17-59 916109 5 56 081759 17-29 996812 26 084947 17-55 915053 4 57 082797 17-25 996797 26 086000 17-51 914000 3 58 083832 17-21 996782 26 087050 17-47 912950 2 69 084864 17-17 996766 26 088098 17-43 911902 1 60 085894 17-13 996751 26 089144 17-38 910856 Cosine. D. Sine. Cotang. D. Tang. (83 DEGREES.'^ SINES AND TANGENTS. (7 DEGREES.) 25 M. Sine. D. Cosine. D. Tang. D. Cotang. 60 9-085894 17-13 9-996751 -26 9-089144 17-38 10-910856 1 086922 17-09 996735 -26 090187 17-34 909813 59 2 087947 17-04 996720 -26 091228 17-30 908772 58 3 088970 17-00 996704 •26 092266 17-27 907734 57 4 089990 16-96 996688 -26 093302 17-22 906698 56 6 091008 16-92 996673 -26 094336 17-19 905664 55 6 092,024 16-88 996657 -26 095367 17-15 904633 54 7 093037 16-84 996641 -26 096395 17-11 903605 53 8 094047 16-80 996625 -26 097422 17-07 902578 52 9 095056 16-76 996610 -26 098446 17-03 901554 51 10 096062 16-73 996594 -26 099468 16-99 900532 50 11 9-097065 16-68 9-996578 -27 9-100487 16-95 10-899513 49 12 098066 16-65 996562 •27 101504 16-91 898496 48 13 099065 16-61 996546 -37 102519 16-87 897481 47 14 100062 16-57 996530 -27 103532 16-84 896468 46 IB 101056 16-53 996514 -27 104542 16-80 895458 45 16 102048 16-49 996498 -27 105550 16-76 894450 44 17 103037 16-45 996482 -27 106556 16-72 893444 43 18 104025 16-41 996465 -27 107559 16-69 892441 42 19 105010 16-38 996449 -27 108560 16-65 891440 41 20 105992 16-34 996433 -27 109559 16-61 890441 40 21 9-106973 16-30 9-996417 -27 9-110556 16-58 10-889444 39 22 107951 16-27 996400 -27 111551 16-54 888449 38 23 108927 16-23 9.96384 -27 112543 16-50 887457 37 24 109901 16-19 996368 -27 113533 16-46 886467 86 25 110873 16-16 996351 -27 114521 16-43 885479 35 26 111842 16-12 996336 -27 115507 16-89 884493 34 27 112809 16-08 996318 -27 116491 16-36 883509 33 28 118774 16-05 996302. -28 117472 16-32 882528 32 i 29 114737 16-01 996285 -28 118452 16-29 881548 31 ; 80 115698 15-97 996269 -28 119429 16-25 880571 30 81 9-116656 15-94 9-996252 ■28 9-120404 16-22 10-879596 29 82 117613 15-90 996235 -28 121377 16-18 878623 28 38 118567 15-87 996219 -28 122348 16-15 877652 27 34 119519 15-83 996202 -28 123317 16-11 876683 26 35 120469 15-80 996185 -28 124284 16-07 875716 25 86 121417 15-76 996168 •28 125249 16-04 874751 24 37 122362 15-73 996151 -28 126211 16-01 873789 23 38 123306 15-69 996134 -28 127172 15-97 872828 22 39 124248 15-66 996117 -28 128130 15-94 871870 21 40 125187 15-62 996100 •28 129087 15-91 870913 20 41 9-126125 15-59 9-996083 •29 9-130041 15-87 10^869959 19 42 127060 15-56 996066 -29 130994 15-84 869006 18 43 127993 15-52 996049 -29 131944 15-81 868056 17 44 128925 15-49 996032 -29 132893 15-77 867107 16 45 129854 15-45 996015 -29 133839 15-74 866161 15 1-6 130781 15-42 995998 -29 134784 15-71 865216 14 47 131706 15-39 995980 -29 135726 15-67 864274 13 48 132630 15-35 995963 -29 136667 15-64 863333 12 49 133551 15-32 995946 -29 137605 15-61 862395 11 50 134470 15-29 995928 -29 138542 15-58 861458 10 51 9-135387 15-25 9-995911 •29 9-139476 15-55 10^860524 9 52 136303 15-22 995894 •29 140409 15-51 859591 8 53 137216 15-19 995876 •29 141340 15-48 858660 7 54 138128 15-16 995859 •29 142269 15-45 857731 6 55 139037 15-12 995841 •29 143196 15-42 856804 5 56 139944 15-09 995828 -29 144121 15-39 855879 4 57 140850 15-06 995806 ■29 145044 15-35 854956 3 58 141754 15-03 995788 -29 145966 15-32 854034 2 59 142655 15-00 995771 -29 146885 15-29 853115 1 60 143555 14-96 995753 -29 147803 15-26 852197 Cosine> D. Sine. { Cotang. D. 1 Tang. M. (82 DEGE EES.) 2 6 (H DEGREES.) A TABLE OF LOGARITHMIC M. Sine. I D. 1 Cosine, j D" Tsms. D. Cotang. 1 9-143555 ■ 14-96 9-995753 -30 9-147803 13-26 i 10-852197 60 1 144453 14-93 995735 •30 148718 15-23 851232 : 59 2 145349 14-90 995717 ■30 149632 : 15-20 850368 : 58 i 3 146243 14-87 995699 -30 150544 ! 15^17 849456 : 37 ' 4 147136 14-84 995681 -30 151454 15^14 848546 56 5 148026 14-81 995664 -30 152363 i 13-11 847637 53 ! 6 148915 14-78 995646 •30 153269 15-08 846781 54 i 7 149802 14-75 995628 -30 154174 15-05 845826 53 1 8 150686 14-72 995610 •30 135077 15-02 844923 52 ! 9 151569 14-69 995591 •30 155978 14-99 844022 51 1 10 152451 14-66 995573 -30 156877 14-96 843123 50 i 11 9-153330 14-63 9-995555 •30 9-157775 i 14-93 10-842225 49 1 12 154208 14-60 995537 -30 158671 ' 14-90 841339 48 ! 13 155083 14-57 995519 •30 159563 i 14-87 840435 47 1 14 155957 14-54 995501 •31 160437 14-84 889543 46 i 15 156830 14-51 995482 •31 161347 ; 14-81 888653 45 i 16 157700 14-48 995464 •31 162236 ■■ 14^79 837764 44 17 158569 14-45 995446 ■31 168123 14^76 836877 48 13 159435 14-42 995427 •31 164008 14^73 835992 42 19 160301 14-39 995409 •31 164892 14^70 835108 41 20 161164 14-36 995390 •31 165774 14-67 834226 40 21 9-162025 14-33 9-995372 •31 9-166654 14-64 10-838846 39 22 16288S 14-30 995353 •31 167532 : 14-61 832468 38 23 163743 14-27 993334 •31 168409 14-58 831591 37 24 164600 14-24 995316 •81 169284 14-55 880716 36 25 165454 14-22 995297 •31 170137 14-53 829843 35 26 166307 14-19 99327S •31 171029 14-50 828971 i 34 27 167159 14-16 995260 •31 171899 14-47 828101 1 33 28 168008 14-13 995241 •32 172767 14-44 827233 ! 32 29 168856 14-10 993222 •32 173684 14-42 826366 31 30 169702 14-07 995203 •32 174499 14-89 825501 30 31 9-170547 14-05 9-995184 •32 9-175362 14-36 10-824638 1 29 32 171389 14-02 995165 •32 176224 14-33 823776 1 28 33 172230 13-99 995146 •32 1770S4 14-81 82293 6 27 34 173070 13-96 995127 •32 177942 14-28 822058 ! 26 i 35 173908 13-94 995108 •32 178799 14-25 821201 25 ; 36 174744 13-91 995089 •32 179653 14-23 820345 24 ! 87 175578 13-88 995070 •32 130308 14-20 819492 23 1 38 176411 13-86 995051 -32 181360 14-17 818640 22 1 39 177242 13-83 995032 -32 182211 14-15 817789 21 ; 40 178072 13-80 995013 -32 183059 14-12 816941 20 41 9-178900 13-77 9-994993 •32 9-183907 14-09 10-816098 19 42 179726 13-74 994974 •32 184752 14-07 813248 18 43 180551 13-72 994955 •32 183597 14^04 814408 17 44 181374 13-69 994935 •32 186439 14^02 813561 16 45 132196 13-66 994916 •33 187280 13^99 812720 15 46 183016 13-64 994896 •83 188120 13-96 811880 14 : 47 183834 13-61 994877 •33 188958 13-93 811042 13 : 48 184631 13-59 994837 •33 189794 18-91 810206 12 i 49 185466 13-56 994838 •33 190629 13-89 809371 11 50 186280 13-53 994818 •38 191462 13^86 808588 10 51 9-187092 13-51 9-994798 •33 9-192294 13-84 10-807706 9 52 187903 13-48 994779 -33 193124 18-81 806876 8 53 188712 13-46 994759 -33 193958 13-79 806047 T 54 189519 13-43 994739 -33 194780 13-76 803220 6 55 190325 13-41 994719 -38 195606 18-74 804394 5 56 191130 13-38 994700 •33 196430 13-71 803570 4 57 191933 13-36 994680 •33 197253 13-69 802747 3 58 192734 13-33 994660 •33 198074 13-66 801926 2 59 193334 13-30 994640 •33 198894 , 13-64 801106 1 60 194332 Cosine. 13-28 994620 -33 199713 1 13-61 800287 1 1 »• Sine. Cotang. D. 1 Tansr. M. .(81 'DEGE l5^l!;,s^ SINES AND TANGENTS. (9 DEGREES.) 27 JH. Sine< D. Cosine. ; D. Tang. D. 1 Cotang. 10-800287" 60 9-194332 13-28 9-994620 •33 9-199713 13-61 1 195129 13-26 994600 •33 200529 13-59 : 799471 59 2 195925 13-28 994580 -33 201345 13-56 i 798655 58 3 196719 13-21 994560 -34 202159 13-54 ; 797841 57 4 197511 13-18 994540 -34 202971 13-52 ; 797029 56 5 198302 13-16 994519 -34 203782 13-49 796218 55 6 199091 K-13 994499 -34 204592 13-47 795408 54 7 199879 13-11 994479 -34 205400 13-45 1 794600 53 8 200666 13-08 994459 -34 206207 13-42 ■ 793793 52 9 201451 13-06 994438 -34 207013 13-40 1 792987 51 10 202234 13-04 994418 -34 207817 13-38 ; 792183 50 11 9-203017 13-01 9-994397 -34 9-208619 13-35 10-791381 49 12 203797 12-99 994377 -34 209420 13-33 790580 48 13 204577 12-96 994357 •34 210220 13-31 789780 47 14 205354 12-94 994336 •34 211018 13^28 788982 46 15 206131 12-92 994316 •34 211815 13^26 788185 45 16 206906 12-89 994295 -34 212611 13^24 787389 44 17 207679 12-87 994274 -35 213405 13^21 786595 43 18 208452 12-85 994254 •35 214198 13^19 785802 42 19 209222 12-82 994233 ■35 214989 13^17 785011 41 20 209992 12-80 994212 ■35 215780 13^15 784220 40 21 9-210760 12-78 9-994191 •35 9-216568 13^12 10-783432 39 22 211526 12-75 994171 •35 217356 13 -10 782644 38 23 212291 12-73 994150 •35 218142 13^08 781858 37 24 213055 12-71 994129 •35 218920 13^05 781074 36 25 213818 12-68 994108 •35 219710 13 03 780290 35 26 214579 12-66 994087 •35 220492 13^01 779508 34 27 215338 12-64 994066 •35 221272 12-99 778728 33 28 216097 12-61 994045 ■35 222052 12^97 777948 32 29 216864 12-59 994024 •35 222830 12^94 777170 31 30 217609 12-57 994003 ■35 223606 12^92 776394 30 1 31 9-218363 12-55 9-993981 -35 9-224382 12-90 10-775618 29 j 32 219116 12-53 993960 -35 225156 12-88 774844 28 33 219868 12-50 993939 -35 225«29 12-86 774071 27 34 220618 12-48 993918 -35 226700 i2-84 773300 26 35 221367 12-46 993896 -36 227471 i?,-8;i 772529 25 36 222115 12-44 993875 •36 228239 12 --^9 771761 24 37 222861 12-42 993854 •36 229O0fi 12-77 770993 23 38 223606 12-39 993832 •36 229773 12-75 770227 22 39 224349 12-37 993811 •36 230539 12-73 769461 21 40 225092 12-35 993789 •36 231302 12-71 768698 20 41 9-225833 12-33 9-993768 •36 9-232065 12-69 10-767935 19 42 226573 12-31 993746 •36 232826 12^67 767174 18 43 227311 12-28 993725 ■36 233586 12-65 766414 17 44 228048 12-26 993703 ■36 234345 12-62 765655 16 45 228784 12-24 993681 •36 235103 12-60 764897 16 46 229518 12-22 998660 •36 235859 12-58 764141 14 1 47 230252 12-20 993638 •36 236614 12-56 763386 13 1 48 230984 12-18 993616 •36 237368 12-54 762632 12 \ 49 231714 12-16 993594 ■37 238120 12-52 761880 11 50 232444 12-14 993572 ■37 238872 12-50 761128 10 51 9-233172 12-12 9-993550 •37 9-239622 12-48 10-760378 9 52 233899 12-09 993528 •37 240371 12-46 759629 8 63 234625 12-07 993506 •37 241118 12-44 758882 7 64 235349 12-05 993484 •37 241865 12-42 758135 6 55 236073 12-03 993462 •37 242610 12-40 757390 5 56 236795 12-01 993440 •37 243354 12-38 756646 4 57 237515 11-99 993418 •37 244097 12^36 755903 3 1 58 238235 11-97 1 993396 \ ^37 1 244839 12^34 765161 2 59 238953 11-95 993374 : -37 245579 12^32 754421 ; 1 60 239670 11-93 993351 i -37 246319 12-30 753681 1 1 Cosine. D. Sine. 1 Cotang. D. Tang. 1 m. (80 DEGBEES.) 28 (10 DEGREES.) A TABLE OF LOGARITHMIC JH. Sine. D. Cosine. D. Tan?. D. Cotang. 1 9-239670 11-93 9-993351 -37 9-246319 12-30 10-753681 60 1 240386 11-91 993329 •37 247057 12-28 762943 59 2 241101 11-89 993307 ■37 247794 ; 12-26 752206 68 3 241814 11-87 993285 •37 248530 12-24 751470 57 4 242526 11-85 993262 •37 249264 12-22 750736 66 5 243237 11-83 993240 -37 249998 12-20 750002 65 6 243947 11-81 993217 -38 250730 12-18 749270 64 7 244656 11-79 993195 -38 251461 12^17 748539 53 8 245363 11-77 993172 -38 252191 12^15 747809 62 9 246069 11-75 993149 •38 252920 12 •IS 747080 51 10 246775 11-73 993127 -38 253648 12^11 746352 60 11 9-247478 11-71 9-993104 •38 9^254374 12^09 10-745626 49 12 248181 11-69 993081 •38 255100 12^07 744900 48 13 248883 11-67 993059 •38 255824 12^05 744176 47 14 249583 11-65 993036 •38 256547 12 •OS 743453 46 1 15 250282 11-63 993013 •38 257269 12^01 742731 45 16 250980 11-61 992990 -38 257990 12^00 742010 44 17 251677 11-59 992967 -38 258710 11^98 741290 43 18 252373 11-58 992944 -38 259429 11-96 740571 42 19 253067 11-56 992921 -38 260146 11-94 739854 41 20 253761 11-54 992898 -38 260863 11-92 739137 40 21 9-254453 11-52 9-992875 -38 9-261578 11-90 10-738422 39 22 255144 11-50 992852 -38 262292 11-89 737708 38 23 255834 11-48 992829 •39 263005 11-87 736995 37 24 256523 11-46 992806 •39 263717 11-85 736283 36 25 257211 11-44 992783 •39 264428 11-83 735572 35 26 257898 11-42 992759 •39 265138 11-81 734862 34 27 258583 11-41 992736 •39 265847 11-79 734153 33 28 259268 11-39 992718 •89 266555 11-78 733445 82 29 259G51 11-37 992690 •39 267261 11-76 732739 31 30 260633 11-35 992666 •39 267967 11-74 732088 30 81 9-261314 11-33 9-992643 •39 9-268671 11-72 10-731329 29 32 261994 11-31 992619 •39 269375 11-70 730625 28 33 262673 11-80 992596 •39 270077 11-69 729923 27 84 263351 11-28 1 992572 •39 270779 11-67 729221 26 35 264027 ra-ae 992549 •39 271479 11-65 728521 26 36 264703 1J.-24 992525 •39 272178 11-64 727822 24 37 265377 11-22 992501 •39 272876 11-62 727124 23 38 266051 11-20 992478 -40 273573 11-60 726427 22 1 39 266723 11-19 992454 -40 274269 11^58 726731 21 40 267395 11-17 992430 -40 274964 11^57 725036 20 41 9-268065 11-15 9-992406 -40 9-275658 11^55 10-724342 19 42 268734 11-13 992382 •40 276351 11^53 723649 18 43 269402 11-11 992359 •40 277043 11-51 722957 17 4i 270069 11-10 992335 •40 277734 11^50 722266 16 45 270735 11-08 992311 •40 278424 11^48 721576 16 46 271400 11-06 992287 •40 279113 11^47 720887 14 47 272064 11-05 992263 •40 279801 11^45 720199 18 48 272726 11-03 992239 •40 280488 11^43 719512 12 49 273388 11-01 992214 •40 281174 11^41 718826 11 50 274049 10-99 992190 •40 281858 11^40 718142 10 51 9-274708 10-98 9-992166 •40 9-282542 11^38 10-717458 9 52 275367 10-96 992142 •40 283225 11^36 716775 8 53 276024 10-94 992117 •41 283907 11^35 716093 7 54 276681 10-92 992093 •41 284588 11-33 715412 6 55 277337 10-91 992069 •41 285268 11-31 714732 6 56 277991 10-89 992044 •41 285947 11^30 714053 4 57 278644 10-87 992020 •41 i 286624 11^28 713376 3 58 279297 10-86 991996 •41 287301 11^26 712699 2 59 279948 10-84 991971 •41 287977 11^25 712023 1 60 280599 10-82 991947 •41 288652 11-23 711348 -. Cosine. D. Sine. 1 Cotang. D. Tang. n. (79 DEGREES.) SINES AND TANGENTS. (11 DEGREES.) 29 \M. Sine. D. Cosine. D. Tang. D. 1 Cotang. 60 9-280599 10-82 9-991947 -41 9-288652 11^23 10-711348 1 281248 10-81 991922 -41 289326 11-22 710674 59 2 281897 10-79 991897 -41 289999 11-20 710001 58 8 282544 10-77 991873 -41 290671 11^18 709329 67 4: 283190 10-76 991848 -41 291342 11^17 708658 56 5 283836 10-74 991823 -41 292013 11-15 707987 55 6 284480 10-72 991799 -41 292682 11-14 707318 54 7 28fel24 10-71 991774 -42 293350 11-12 706650 58 8 285766 10-69 991749 -42 294017 11-11 705983 52 9 286408 10-67 991724 -42 294684 11-09 705316 51 10 287048 10-66 991699 -42 295349 11-07 704651 50 11 9-287687 10-64 9-991674 -42 9-296013 11-06 10-703987 49 12 288326 10-63 991649 -42 296677 11-04 708828 48 13 288964 10-61 991624 -42 297339 11-08 702661 47 14 289600 10-59 991599 -42 298001 11-01 701999 46 15 290236 10-58 991574 -42 298662 11-00 701838 45 16 290870 10-56 991549 -42 299322 10-98 700678 44 17 291504 10-54 991524 -42 299980 10-96 700020 48 18 292137 10-58 991498 -42 300638 10-95 699362 42 19 292768 10-51 991478 -42 301295 10-93 698705 41 20 293899 10-50 991448 -42 301951 10-92 698049 40 21 9-294029 10-48 9-991422 -42 9-302607 10-90 10-697398 39 22 294658 10-46 991397 -42 308261 10-89 696739 38 23 295286 10-45 991372 -43 803914 10-87 696086 37 24 295913 10-48 991346 -43 304567 10-86 695438 86 25 296539 10-42 991321 -43 305218 10-84 694782 35 26 297164 10-40 991295 -48 305869 10 ■83 694181 34 27 297788 10-39 991270 ■43 306519 10-81 693481 83 28 298412 10-37 991244 -48 807168 10-80 692832 32 29 299084 10-36 991218 •43 807815 10-78 692185 81 80 299655 10-34 991193 •43 308463 10-77 691537 30 81 9-300276 10-82 9-991167 -43 9-309109 10-75 10-690891 29 82 800895 10-31 991141 -43 309754 10-74 690246 28 33 301514 10-29 991115 ■43 310398 10-73 689602 27 34 302132 10-28 991090 •43 Slices 10-71 688958 26 35 302748 10-26 991064 •48 311685 10- '."-n 688315 25 86 803864 10-25 991088 -48 312327 10- '68 687673 24 87 308979 10-23 991012 •43 312967 10- W 687033 28 88 304593 10-22 990986 •48 813GO& 10-65 686392 22 39 305207 10-20 990960 •48 314247 10-64 685753 21 40 305819 10-19 990934 •44 814885 10-62 685115 20 41 9-806430 10-17 9-990908 •44 9-315528 10-61 10-684477 19 42 307041 10-16 990882 •44 316159 10-60 683841 18 48 307650 10-14 990855 •44 316795 10-58 688205 17 44 308259 10-13 990829 •44 317430 10-57 682570 16 45 308867 10-11 990808 •44 318064 10-55 681986 15 46 309474 10-10 990777 •44 318697 10-54 681803 14 47 810080 10-08 990750 •44 319329 10-53 680671 13 48 310685 10-07 990724 •44 319961 10-51 680039 12 49 311289 10-05 990697 •44 320592 10-50 679408 11 50 311893 10-04 990671 •44 821222 10-48 678778 10 51 9-312495 10-03 9-990644 •44 9-321851 10-47 10-678149 9 52 313097 10-01 990618 •44 322479 10-45 677521 8 53 313698 10-00 990591 •44 323106 10-44 676894 7 64 314297 9-98 990565 -44 828733 10-43 676267 6 55 314897 9-97 990538 ■44 824358 10-41 675642 5 56 815495 9-96 990511 -45 324983 10-40 675017 4 57 316092 9-94 990485 -45 325607 10-39 674893 3 58 316689 9-93 990458 -45 326231 10-37 673769 2 59 317284 9-91 990481 -45 326853 10-36 673147 1 60 317879 9-90 990404 •45 327475 10-35 672525 Cosine. D. Sine. Cotang. D. Tang. M. (78 DEGBEES.i 30 (12 DEGREES.) A TABLE OF LOGARITHMIC JH. Sine. D. Cosine. D. Tang. 1 D. Cotang. i 1 9-317879 9-90 9-990404 -45 9-327474 j 10^ 35 10 ■672526 60 1 318473 9-88 990378 -45 328095 10 33 671905 59 2 319066 9-87 990351 -45 328715 10 32 671285 58 3 319658 9-86 990324 -45 329334 10 30 670666 57 4 820249 9-84 990297 •45 329953 10^ 29 670047 56 5 320840 9-83 990270 -45 880570 10 28 669430 55 6 321480 9-82 990243 -45 381187 10^ 26 668813 54 7 322019 9-80 990215 •45 331803 10^ 25 668197 53 8 822607 9-79 990188 -45 832418 10^ 24 667582 52 g 828194 9-77 990161 -45 888033 10 ■ 23 666967 51 10 828780 9-76 990134 -45 338646 10 ■ 21 666354 50 11 9-324866 9-75 9-990107 -46 9 ■834259 10^ 20 10^665741 49 12 324950 9-73 990079 -46 334871 10^ 19 665129 48 13 825534 9-72 990052 ■46 335482 10 17 664518 47 14 826117 9-70 990025 -46 886098 10^ 16 663907 46 15 826700 9-69 989997 •46 336702 10^ 15 663298 45 16 327281 9-68 989970 •46 337311 10 18 662689 44 17 327862 9-66 989942 -46 837919 10 12 662081 43 18 328442 9-65 989915 -46 888527 10^ 11 661473 42 19 329021 9-64 989887 -46 339133 10 10 660867 41 20 329599 9-62 989860 -46 339789 10 08 660261 40 21 9-330176 9-61 9-989832 -46 9^340344 10 07 10^ 659656 89 22 880758 9-60 989804 ■46 340948 10 06 659052 38 28 381329 9-58 989T77 •46 841552 10 04 658448 37 24 381908 9-57 989749 •47 342155 10 03 657845 36 25 332478 9-56 989721 ■4.- 342757 10 02 657243 35 26 888051 9-54 989693 -47 343358 10 00 656642 34 27 333624 9-53 989665 -47 343958 9 99 656042 88 28 884195 9-52 9890.37 •47 344558 9 98 655442 32 29 334766 9-50 989609 •47 345157 9 97 654843 31 30 385337 9-49 98P58; •47 345755 9 96 654245 30 31 9-335906 9-48 9-989553 •47 9^846858 9 94 10^ 653647 29 82 836475 9-46 989525 •47 846949 9 93 658051 28 33 337043 9-45 989497 ■47 847545 9 92 652455 27 34 337610 9-4J. 989469 •47 348141 9 91 651859 26 85 388176 '9-43 989441 •47 348735 9 90 651265 25 36 888742 ... 'J-il 989413 •47 349329 9 88 650671 24 37 339306 gi-40 989384 •47 349922 9 87 650078 23 38 339871 9-39 939356 ■47 850514 9 86 649486 22 39 340434 9-37 989328 ■47 351106 9 85 648894 21 40 340996 9-36 989300 ■47 351697 9 83 648803 20 41 9-341558 9-35 9-989271 ■47 9^352287 9 82 10^647713 19 42 342119 9-34 989243 ■47 352876 9 81 647124 18 43 342679 9-82 989214 ■47 358465 9 80 646535 17 44 343289 9-31 989186 ■47 854053 9 79 645947 16 45 343797 9-80 989157 ■47 354640 9 77 645360 15 46 344355 9-29 989128 ■48 355227 9 76 644773 14 j 47 844912 9-27 989100 ■48 355813 9 75 644187 18 1 48 345469 9-26 989071 ■48 356398 9 74 643602 12 1 49 346024 9-25 989042 ■48 356982 9 73 648018 11 ' 50 346579 9-24 989014 ■48 357566 9 71 642484 10 51 9-847134 9-22 9-988985 ■48 9 ■358149 9 70 10-641851 9 52 347687 9-21 988956 ■48 358781 9 69 641269 8 53 848240 9-20 988927 ■48 359313 9 68 640687 7 54 348792 9-19 988898 -48 859893 9 67 640107 6 55 349343 9-17 988869 -48 360474 9 66 639526 5 56 349893 9-16 988840 -48 361053 9 65 638947 4 57 350443 9-15 988811 -49 361682 9 63 638368 8 58 350992 9-14 988782 -49 362210 9 ■62 637790 2 69 351540 9-13 988753 -49 362787 9 ■61 637213 1 60 852088 9-11 988724 -49 363364 9-60 636686 Cosine. D. Sine. , Cotanp;. D. 1 Tang. «. (77 DEGREES.) SINES AND TANGENTS. (13 DEGREES.) 31 M. Sine. D. Cosine. D. Tang. D. Cotang. 9-352088 9 11 9-988724 • 49 9-868364 9-60 10-636636 60 1 352685 9 10 988695 49 368940 9-59 636060 59 2 358181 9 09 988666 49 364515 9-58 635485 58 8 353726 9 08 988636 49 365090 9-57 634910 57 4 354271 9 07 988607 49 365664 9-55 634336 56 6 354815 9 05 988578 49 866287 9-54 638763 55 6 355358 9 04 988548 49 366810 9-53 633190 54 • 7 355901 9 03 988519 49 867882 9-52 632618 53 8 356443 9 02 988489 49 867953 9-31 682047 52 9 356984 9 01 988460 49 368524 9-50 631476 51 10 357524 8 99 988430 49 869094 9-49 680906 50 11 9-358064 8 98 9-988401 49 9-869668 9-48 10-680337 49 12 358603 8 97 988371 49 870232 9-46 629768 48 13 359141 8 96 988842 49 370799 9-45 629201 47 14 359678 8 95 988812 50 371867 9-44 628633 46 15 360215 8 98 988282 50 871933 9-43 628067 45 16 360752 8 92 988252 50 372499 9-42 627501 44 17 361287 8 91 988223 50 373064 9-41 626936 43 18 361822 8 90 988193 50 373629 9-40 626371 42 19 362356 8 89 988163 50 374193 9-39 625807 41 20 362889 8 88 988133 50 374756 9-38 625244 40 21 9-363422 8 87 9-988103 50 9-875319 9-37 10-624681 39 22 363954 8 85 988073 50 375881 9-35 624119 38 23 364485 8 84 988043 50 376442 9-34 623558 37 24 865016 8 83 988013 50 377003 9-33 622997 36 25 865546 8 82 987983 50 377568 9-32 622437 35 26 366075 8 81 987953 50 378122 9-31 621878 34 27 366604 8 80 987922 50 378681 9-30 621319 33 28 867131 8 79 987892 50 379239 9-29 620761 32 29 367659 8 77 987862 50 379797 9-28 620203 31 30 368185 8 76 987832 51 380354 9-27 619646 30 81 9-368711 8 75 9-987801 51 9-380910 9-26 10-619090 29 82 369236 8 74 987771 51 381466 9-25 618534 28 83 869761 8 73 987740 51 382020 9-24 617980 27 84 370285 8 72 987710 51 882575 9-23 617425 26 35 370808 8 71 987679 51 383129 9-22 616871 25 36 371380 8 70 987649 51 883682 9-21 616318 24 37 871852 8 69 987618 51 384284 9-20 615766 23 88 372378 8 67 987588 51 884786 9-19 615214 22 39 372894 8 66 987557 51 385337 9-18 614663 21 40 873414 8 65 987526 51 385888 9-17 614112 20 41 9-373933 8 64 9-987496 51 9-386438 9-15 10-618562 19 42 374452 8 63 987465 51 386987 9-14 618013 18 48 374970 8 62 987434 51 387536 9-13 612464 17 44 375487 8 61 987403 52 388084 9-12 611916 16 45 876003 8 60 987372 52 888631 9-11 611369 15 46 876519 8 59 987341 52 889178 9-10 610822 14 47 877035 8 58 987310 52 389724 9-09 61027G 13 48 377549 8 57 987279 52 390270 9-08 609730 12 49 378063 8 56 987248 52 890815 9-07 609185 11 50 378577 8 54 987217 52 391360 9-06 608640 10 51 9-379089 8 58 9-987186 52 9-391903 9-05 10-608097 9 52 379601 8 52 987155 52 892447 9-04 607553 8 53 880113 8 51 987124 52 892989 903 607011 7 54 380624 8 60 987092 52 893531 9-02 606469 6 55 381134 8 49 987061 52 894073 9-01 605927 5 56 381648 8 48 987080 52 894614 9-00 605386 4 57 382152 8 47 986998 52 895154 8-99 604846 8 58 88L661 8 46 986967 52 895694 8-98 604306 2 69 383168 8 45 986936 52 896233 8-97 603767 ! 1 1 60 383675 8-44 986904 •52 396771 8-96 608229 Cosine. D. Sine. Cotang. D. Tang. M. (76 DEGREES.") 32 (14 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tanff. D. Cotang. 9-383675 8 44 9-986904 52 9-396771 8 96 10-603229 60 1 384182 8 43 986873 53 397309 8 96 602691 59 2 384687 8 42 986841 53 397846 8 95 602154 68 3 385192 8 41 986809 53 398383 8 94 601617 67 4 385697 8 40 986778 53 398919 8 93 601081 66 5 386201 8 39 986746 53 399455 8 92 600545 65 6 386704 8 38 986714 53 399990 8 91 600010 54 7 387207 8 37 986683 53 400524 8 90 599476 53 8 387709 8 36 986651 53 401058 8 89 598942 52 g 388210 8 35 986619 53 401591 8 88 598409 51 10 388711 8 34 986587 53 402124 8 87 597876 50 11 9-389211 8 33 9-986555 53 9-402656 8 86 10-597344 49 12 889711 8 32 986523 53 403187 8 85 696813 48 13 390210 8 31 986491 53 403718 8 84 696282 47 14 890708 8 30 986459 53 404249 8 83 595751 46 15 391206 8 28 986427 53 404778 8 82 695222 45 16 391703 8 27 986395 53 405308 8 81 594692 44 17 892199 8 26 986363 54 405836 8 80 594164 48 18 392695 8 25 986331 54 406364 8 79 593636 42 19 393191 8 24 986299 54 406892 8 78 593108 41 20 398685 8 23 986266 54 407419 8 77 592581 40 21 9-394179 8 22 9-986234 54 9-407945 8 76 10-592055 39 22 394673 8 21 986202 54 408471 8 75 591529 38 28 395166 8 20 986169 54 408997 8 74 591003 37 24 395658 8 19 986137 54 409521 8 74 690479 36 25 396150 8 18 986104 54 410045 8 73 589955 35 26 396641 8 17 986072 54 410569 8 72 689431 34 27 397132 8 17 986039 54 411092 8 71 588908 88 28 397621 8 16 986007 54 411615 8 70 588385 32 29 398111 8 15 985974 54 412137 8 69 587863 31 30 398600 8 14 985942 54 412658 8 68 687342 30 31 9-399088 8 13 9-985909 55 9-413179 8 67 10-586821 29 32 399575 8 12 985876 56 413699 8 66 586301 28 83 400062 8 11 985843 55 414219 8 65 686781 27 34 400549 8 10 985811 55 414738 8 64 585262 26 35 401035 8 09 985778 55 415257 8 64 684743 25 86 401520 8 08 985745 55 415775 8 63 684225 24 87 402005 8 07 985712 55 416293 8 62 683707 23 88 402489 8 06 985679 55 416810 8 61 683190 22 39 402972 8 05 985646 55 417326 8 60 582674 21 40 403455 8 04 985613 55 417842 8 59 582158 20 41 9-403938 8 03 9-985580 55 9-418358 8 58 10-681642 19 42 404420 8 02 985547 55 418873 8 57 681127 18 43 404901 8 01 985514 55 419887 8 56 680613 17 44 405382 8 00 985480 55 419901 8 55 680099 16 45 405862 7 99 985447 55 420415 8 55 579586 15 46 406841 7 98 985414 56 420927 8 54 579073 14 47 406820 7 97 985380 56 421440 8 53 678660 13 48 407299 7 96 985347 56 421952 8 52 578048 12 49 407777 7 95 985314 56 422463 8 51 577537 11 1 50 408254 7 94 985280 56 422974 8 50 577026 10 1 51 9-408731 7 94 9-985247 56 9-423484 8 49 10-676616 9 52 409207 7 93 985213 56 423993 8 48 57,6007 8 58 409682 7 92 985180 56 424503 8 48 575497 7 54 410157 7 91 985146 56 425011 8 47 674989 6 55 410632 7 90 985113 56 425519 8 46 574481 5 56 411106 7 89 985079 56 426027 8 45 673973 4 57 411579 7 88 985045 56 426534 8 44 573466 3 58 412052 7 87 985011 56 427041 8 43 572959 2 59 412524 7 86 984978 56 427547 8 43 672453 1 60 412996 7-85 981944 •56 428052 8-42 571948 Cosine. D. Sine. Cotang. D. Tang. M. f75 DEGREES.") SINES AND TANGENTS. (15 DEGREES.) 38 m. Sine. D. Cosine. D. Tang. D. Cotang. 60 9-412996 7-85 9-984944 -57 9-428052 8-42 10-571948 1 413467 7-84 984910 -37 428557 8-41 571443 59 2 413938 7-83 984876 -57 429062 8-40 570988 58 3 414408 7-83 984842 -57 429566 8-39 570434 67 4 414878 7-82 984808 -57 430070 8-38 569930 56 5 415347 7-81 984774 -57 430573 8-38 569427 55 6 415815 7-80 984740 -57 431075 8-37 568925 54 7 416283 7-79 984706 •57 431577 8-36 568423 53 8 416751 7-78 984672 •57 432079 8^35 567921 62 9 417217 7-77 984637 •57 432580 8^34 567420 51 10 417684 7-76 984608 •57 433080 8^33 566920 50 11 9'418150 7-75 9-984569 -57 9-433580 8^32 10-566420 49 12 418615 7-74 984535 -57 434080 8-32 565920 48 13 419079 7-73 984500 •57 434579 8-31 565421 47 14 419544 7-73 984466 •57 435078 8-30 564922 46 15 420007 7-72 984432 -58 435576 8-29 564424 46 16 420470 7-71 984397 -58 436073 8-28 563927 44 IT 420933 7-70 984363 -58 '436570 8-28 563430 43 18 421395 7-69 984328 -58 437067 8-27 562933 42 19 421857 7-68 984294 -58 437563 8-26 562437 41 20 422318 7-67 984259 •58 438059 8-25 561941 40 21 9-422778 7-67 9-984224 •58 9-438554 8-24 10-561446 39 22 423238 7-66 984190 •58 439048 8-23 560952 38 23 423697 7-65 984155 •58 439543 8^23 560457 37 24 424156 7-64 984120 •58 440036 8^22 559964 36 25 424615 7-63 984085 •58 440529 8-21 659471 35 26 425073 7-62 984050 •58 441022 8-20 558978 34 27 425530 7-61 984015 •58 441514 8^19 568486 33 28 425987 7-60 983981 •58 442006 8^19 557994 32 29 426443 7-60 983946 •58 442497 8^18 557503 31 30 426899 7-59 983911 •58 442988 8-17 657012 30 31 9-427354 7-58 9-983875 •58 9-443479 8-16 10-556521 29 32 427809 7-57 983840 •59 443968 8-16 556032 28 33 428263 7-56 983805 •59 444458 8-15 555542 27 34 428717 7-55 983770 •59 444947 8^14 655053 26 35 429170 7-54 983735 •59 445435 8^13 554566 25 36 429623 7-53 983700 •59 445923 8^12 554077 24 37 430075 7-52 983664 -59 446411 8^12 553589 23 38 430527 7-52 983629 -59 446898 8^11 553102 22 39 430978 7-51 983594 ■59 447384 8-10 552616 21 40 431429 7-50 983558 -59 447870 8^09 552130 20 41 9-431879 7-49 9-983523 -59 9-448356 8-09 10-651644 19 42 432329 7-49 983487 ■59 448841 8^08 551159 18 43 432778 7-48 983452 •59 449326 8-07 550674 17 44 433226 7-47 983416 •59 449810 8^06 550190 16 45 433675 7-46 983381 •59 450294 8^06 549706 15 46 434122 7-45 983345 •59 450777 8^05 549223 ! 14 47 434569 7-44 983309 •59 451260 8^04 548740 ■ 13 48 435016 7-44 983273 •60 451743 8^03 548257 ■ 12 49 435462 7-43 983238 •60 452225 8^02 547775 11 50 435908 7-42 983202 •60 452706 8^02 547294 10 51 9-436353 7-41 9-983166 •60 9-453187 8^01 10-546813 9 52 436798 7-40 983130 •60 453668 8-00 546332 8 58 437242 7-40 983094 •60 454148 7-99 545852 7 64 437686 7-39 983058 •60 454628 7-99 545372 6 55 438129 7-38 983022 •60 455107 7-98 544893 5 56 438572 7-37 982986 •60 455586 7-97 544414 4 57 439014 7-36 982950 •60 456064 7-96 543936 3 58 439456 7-36 982914 -60 456542 ■ 7-96 543458 2 59 439897 7-35 982878 -60 457019 7-95 i 542981 1 60 440338 7-34 983842 -60 457496 7-94 542504 IH. Cosine. D. Sine. Cotang. D. Tang. (74 DEGR EES.) 34 (16 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. 1 Cotang. 9-440338 7-34 9-982842 • 60 9^457496 7-94 10-542504 60 1 440778 7-33 982805 • • 60 457973 7-93 542027 69 2 441218 7-32 982769 • 61 458449 7-93 541551 58 3 441658 7-31 982733 61 458925 7^92 541075 57 i 442096 7-31 982696 • 61 459400 7-91 540600 66 5 442535 7-30 982660 61 459875 7-90 540l25 55 6 442973 7-29 982624 ■ 61 460349 7^90 539651 54 7 443410 7-28 982587 • 61 460823 7^89 539177 53 8 443847 7-27 982551 61 461297 7^88 538703 52 9 444284 7-27 982514 61 461770 7-88 538230 51 10 444720 7-26 982477 • 61 462242 7-87 537758 50 11 9-445155 7-25 9-982441 • 61 9^462714 7-86 10-537286 49 12 445590 7-24 982404 61 463186 7^85 536814 48 13 446025 7-23 982367 • 61 463658 7^85 536342 47 14 ■446459 7-23 982331 61 464129 7-84 535871 46 15 446898 7-22 982294 - 61 464599 7^83 535401 45 16 447326 7-21 982257 61 465069 7-83 534931 44 17 447759 7-20 982220 62 465539 7-82 534461 43 18 448191 7-20 982183 62 466008 7-81 533992 42 19 448628 7-19 982146 62 466476 7-80 538524 41 20 449054 7-18 982109 62 466945 7^80 533055 40 21 9-449485 7-17 9-982072 62 9^467413 7^79 10-532587 39 22 449915 7-16 982035 62 467880 7-78 532120 38 28 450345 7-16 981998 62 468347 7^78 531653 37 24 450775 7-15 981961 62 468814 7-77 531186 36 25 451204 7-14 981924 62 469280 7-76 530720 35 26 451632 7-13 981886 62 469746 7-75 580254 34 27 452060 7-13 981849 62 470211 7-75 529789 33 28 452488 7-12 981812 62 470676 7-7-4 529324 82 29 452915 711 981774 62 471141 7-73 528859 31 80 453342 7-10 981737 62 471605 7-73 528395 30 31 9-453768 7-10 9-981699 63 9-472068 7-72 10^527932 29 32 454194 7-09 981662 63 472532 7-71 527468 28 33 454619 7-08 981625 63 472995 7-71 527005 27 34 455044 7-07 981587 63 473457 7-70 526543 26 35 455469 7-07 981549 63 473919 7-69 526081 25 86 455893 7-06 981512 63 474381 7-69 525619 24 37 456316 7-05 981474 63 474842 7-68 525158 23 38 456739 7-04 981436 63 475303 7-67 524697 22 39 457162 7-04 981399 63 475763 7-67 524237 21 40 457584 7-03 981361 63 476223 7-66 523777 20 41 9-458006 7-02 9-981323 63 9-476683 7-65 10-523817 19 42 458427 7-01 981285 63 477142 7-65 522858 18 43 458848 7-01 981247 63 477601 7-64 522399 17 44 459268 7-00 981209 63 478059 7-63 521941 16 i 45 459688 6-99 981171 63 478517 7-63 521483 15 1 46 460108 6-98 981133 64 478975 7-62 521025 14 1 47 460527 6-98 981095 64 479432 7-61 520568 13 ' 48 460946 6-97 981057 64 479889 7-61 520111 12 i 49 461364 6-96 981019 64 480345 7-60 519655 11 : 50 461782 6-95 980981 64 480801 7-59 519199 10 51 9-462199 6-95 9-980942 64 9-481257 7-59 10-518743 9 i 52 462616 6-94 980904 -64 481712 7-58 518288 8 53 463032 6-93 980866 •64 482167 7^57 517833 7 64 463448 6-93 980827 •64 482621 7-57 517379 6 55 463864 6-92 980789 •64 483075 7-56 516925 5 56 464279 6-91 980750 •64 483529 7-55 516471 4 57 464694 6-90 980712 •64 4S3982 7-55 616018 3 68 465108 6-90 980673 •64 484435 7-54 515565 2 69 465522 6-89 980635 •64 484887 7-53 515113 1 60 465935 6-88 980596 ■64 485339 7-53 514661 1 Cosine. D. Sine. j Cotang. D. i Tang. M. (73 DEGREES.) SINES AND TANGENTS. (17 DEGREES.) 35 Jtl. Sine. D. Cosine. D. Tang. D. Cotang. 1 9-465935 6 88 9-980596 64 9-485339 7-55 10-514661 i 60 1 466848 6 88 980558 64 485791 7-52 614209 1 59 2 466761 6 87 980519 65 486242 7-51 518758 ; 58 S 467178 6 86 980480 65 486693 7-51 513307 67 4 467585 6 85 980442 65 487143 7-50 612857 66 5 467996 6 85 980408 65 487598 7-49 512407 55 6 468407 6 84 980864 65 488048 7-49 511957 54 7 466817 6 88 980325 65 488492 7-48 511508 63 8 469227 6 88 980286 65 488941 7-47 • 511059 52 9 46963T 6 82 980247 65 489390 7-47 510610 51 10 470046 6 81 980208 65 489888 7-46 510162 50 11 9-470455 6 80 9-980169 65 9-490286 7-46 10-509714 49 12 470863 6 80 980130 65 490788 7-45 509267 48 18 471271 6 79 980091 65 491180 7-44 608820 47 14 471679 6 78 980052 65 491627 7-44 508378 46 15 472086 6 78 980012 65 492078 7-43 507927 45 16 472492 6 77 979973 65 492519 7-43 507481 44 17 472898 6 76 979934 66 492965 7-42 507035 48 18 473304 6 76 979895 66 493410 7-41 506590 42 19 473710 6 75 979855 66 498854 7-40 506146 41 20 474115 6 74 979816 66 494299 7-40 505701 40 21 9-474519 6 74 9-979776 66 9-494743 7-40 10-505257 39 22 474923 6 73 979737 66 495186 7-39 504814 88 28 475327 6 72 979697 66 495680 7-88 504370 37 24 475730 6 72 979658 66 496073 7-37 503927 86 25 476133 6 71 979618 ■ 66 496515 7-37 603485 85 26 476536 6 70 979579 66 496957 7-36 503043 84 27 476988 6 69 979539 66 497399 7-86 502601 38 28 477340 6 69 979499 66 497841 7-35 602159 32 29 477741 6 68 979459 66 498282 7-34 601718 31 30 478142 6 67 979420 66 498722 7-34 501278 30 31 9-478542 6 67 9-979380 66 9-499163 7-38 10-600837 29 32 478942 6 66 979340 66 499603 7-38 500397 28 33 479342 6 65 979300 1 67 500042 7-32 499968 27 34 479741 6 65 979260 1 67 500481 7-31 499519 26 35 480140 6 64 979220 67 500920 7-31 499080 25 36 480589 6 63 979180 67 501359 7-30 498641 24 37 480937 6 63 979140 67 501797 7-30 498203 28 38 481834 6 62 979100 67 502235 7-29 497765 22 39 481731 6 61 979059 67 502672 7-28 497328 21 40 482128 6 61 979019 67 603109 7-28 496891 20 41 9-482625 6 60 9-978979 67 9-603546 7-27 10-496454 19 42 482921 6 59 978989 67 503982 7-27 496018 18 43 483316 6 59 978898 67 504418 7-26 495682 17 44 483712 6 58 978858 67 504864 7-25 495146 16 45 484107 6 57 978817 67 505289 7-25 494711 15 46 484501 6 57 978777 67 505724 7-24 494276 14 47 484895 6 56 978736 67 506159 7-24 493841 13 48 485289 6 55 978696 68 506598 7-23 493407 12 49 485682 6 65 978655 68 507027 7-22 492973 11 50 486075 6 54 978615 68 507460 7-22 492540 10 51 9-486467 6 53 9-978574 68 9-507893 7-21 10-492107 9 52 486860 6 53 978588 68 608826 7-21 491674 8 53 487251 6 52 978493 68 508759 7-20 491241 7 64 487643 6 51 978452 68 509191 7-19 490809 6 65 488034 6 51 978411 68 509622 7-19 490378 5 66 488424 6 50 978870 68 510054 7-18 489946 4 67 488814 6 50 978829 68 510485 7-18 489515 8 68 489204 6 49- 978288 68 510916 7-17 489084 2 59 489598 6 48 978247 68 511346 7-16 488654 1 60 489982 6-48 978206 68 511776 7-16 488224 Cosine. D. Sine. 1 D. Cotang. D. 1 Tang. m. (72 DEGREES.) 36 (18 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. | D. Tang. D. Cotang. 9-489982 6-48 9-978206 68 9-511776 7-16 10-488224 60 1 490371 6-48 978165 68 512206 7-16 487794 59 2 490759 6-47 978124 68 612685 7-15 487365 58 3 491147 6-46 978083 69 613064 7-14 486936 57 4 491535 6-46 978042 69 613493 7-14 486507 56 5 491922 6-45 978001 69 513921 7-13 486079 55 6 492308 6-44 977959 69 514349 7-13 485651 54 7 492695 6-44 977918 69 514777 7-12 485223 53 8 493081 6-43 977877 69 515204 7-12 484796 52 9 493466 6-42 977835 69 515631 7-11 484369 51 10 493851 6-42 977794 69 616057 7-10 483943 50 11 9-494236 6-41 9-977752 • 69 9-516484 7-10 10-483616 49 12 494621 6-41 977711 69 516910 7-09 483090 48 13 495005 6-40 977669 69 617335 7-09 482665 47 14 495388 6-39 977628 69 517761 7-08 482239 46 15 495772 6-39 977586 69 518185 7-08 481815 45 16 496154 6-38 977544 70 518610 7-07 481390 44 17 496537 6-37 977503 70 519034 . 7-06 480966 43 18 496919 6-37 977461 70 519458 7-06 480542 42 19 497301 6-86 977419 70 519882 7-05 480118 41 20 497682 6-36 977377 70 620305 7-05 479695 40 21 9-498064 6-35 9-977335 70 9-520728 7-04 10-479272 39 i 22 498444 6-34 977293 70 621151 7-03 478849 38 i 23 498825 6-34 977251 70 521573 7-03 478427 37 i 24 499204 6-38 977209 70 621995 7-03 478005 36 i 25 499584 6-32 977167 70 522417 7-02 477583 35 : 26 499963 6-32 977125 70 522838 7-02 477162 34 27 500342 6-31 977083 70 623259 701 476741 33 , 28 500721 6-31 977041 70 523680 7-01 476320 32 i 29 501099 6-30 976999 70 524100 7-00 475900 81 1 80 501476 6-29 976957 70 524520 6-99 475480 30 31 9-501854 6-29 9-976914 70 9-524939 6-99 10-475061 1 29 ; 32 502231 6-28 976872 71 625359 6-98 474641 28 33 502607 6-28 976830 71 525778 6-98 474222 27 34 502984 6-27 976787 71 626197 6-97 473803 26 35 503360 6-26 976745 71 526615 6-97 473385 25 36 503735 6-26 976702 71 527033 6-96 472967 24 37 504110 6-25 976660 71 527451 6-96 472549 23 38 504485 6-25 976617 71 527868 6-95 472132 22 89 504860 6-24 976574 71 528285 6-95 471715 21 40 505234 6-23 976532 71 628702 6-94 471298 20 41 9-505608 6-28 9-976489 71 9-629119 6-93 10-470881 19 42 505981 6-22 976446 71 529535 6-93 470465 18 43 506354 6-22 976404 71 529950 6-93 470050 17 44 506727 6-21 976361 71 530366 6-92 469634 16 45 507099 6-20 976318 71 530781 6-91 469219 15 46 507471 .6-20 976275 71 631196 6-91 468804 14 47 507843 6-19 976232 72 531611 6-90 468389 13 48 508214 6-19 976189 72 532025 6-90 467975 12 49 508585 6-18 976146 72 632439 6-89 467561 11 BO 608956 6-18 976103 72 532853 6-89 467147 10 51 9-509326 6-17 9-976060 72 9-633266 6-88 10-466734 9 52 509696 616 976017 72 533679 6-88 466321 8 58 510065 e-16 975974 72 534092 6-87 466908 7 54 510434 6-15 975930 72 634504 6-87 465496 6 55 510808 6-15 975887 •72 534916 6-86 465084 5 56 511172 6-14 975844 -72 535328 6-86 464672 4 57 511540 6-13 975800 -72 686739 6-85 464261 3 58 511907 6-13 975757 -72 536150 6-85 463850 2 ; 59 512275 6-12 975714 -72 636561 6-84 463439 1 '■ 60 512642 6-12 975670 -72 536972 6-84 463028 Cosine. D. Sine. D. Cotang. D. Tang. (71 DEGREES.) SINES AND TANGENTS. (19 DEGREES.) 37 JH. Sine. D. Cosine. D. Tang. D, Cotang. 60 9-512642 6-12 9-975670 -73 9-636972 6-84 10-463028 1 513009 6-11 976627 -73 537382 6-83 462618 59 2 513375 6-11 975583 -78 537792 6-83 462208 68 3 513741 6-10 975539 -73 538202 6-82 461798 57 4: 614107 6-09 975496 -73 538611 6-82 461389 56 5 514472 6-09 975462 •73 539020 6'81 460980 55 6 614837 6-08 976408 -73 539429 6-81 460571 54 7 515^02 6-08 975366 -78 539837 6-80 460163 53 8 515566 6-07 975321 -73 540245 6-80 469755 52 9 515930 6-07 976277 -73 540653 6-79 459347 61 10 516294 6-06 975233 -73 641061 6-79 458939 50 11 9-516657 6-05 9-975189 -73 9-541468 6-78 10^458532 49 12 517020 6-05 975145 -73 541875 6-78 458125 48 13 517382 6-04 975101 ■73 542281 6-77 467719 47 14 517746 6-04 976057 -73 642688 6-77 457312 46 15 518107 6-03 975013 -73 643094 6-76 456906 45 16 518468 6-03 974969 -74 543499 6-76 456501 44 17 518829 6-02 974925 •74 643906 6-75 456096 43 18 519190 6-01 974880 -74 644310 6-75 456690 42 19 519551 6-01 974836 -74 544715 6-74 456285 41 20 519911 6-00 974792 -74 545119 6-74 454881 40 21 9-520271 6-00 9-974748 -74 9-546624 6-73 10-464476 39 22 520631 5-99 974703 -74 545928 6-73 454072 38 23 520990 5-99 974659 -74 546331 6-72 453669 37 24 521849 5-98 974614 -74 546735 6-72 453265 36 25 521707 6-98 974570 -74 647138 6^71 462862 35 26 522066 6-97 974625 -74 547540 6^71 462460 34 27 522424 5-96 974481 -74 647943 6-70 452057 33 28 522781 5-96 974436 -74 548345 6-70 451655 32 29 523138 5-95 974391 -74 548747 6-69 461253 31 30 523495 5-95 974347 -75 549149 6-69 450861 30 31 9-523852 5-94 9-974302 -75 9 649560 6-68 10-450450 29 32 524208 5-94 974257 -75 549951 6-68 450049 28 33 524664 5-93 974212 -75 550362 6-67 449648 27 34 524920 5-93 974167 ■75 560752 6-67 449248 26 35 525275 5-92 974122 •75 551152 6-66 448848 25 36 525630 5-91 974077 -75 561562 6-66 448448 24 37 525984 5-91 974032 ■76 551952 6-65 448048 23 38 526339 5-90 973987 -75 562351 6-65 447649 22 39 526693 5-90 973942 -75 552750 6-65 447260 21 40 527046 5-89 973897 -75 563149 6-64 446851 20 41 9-527400 5-89 9-973852 -75 9-563548 6-64 10-446452 19 42 527753 5-88 978807 -75 653946 6-63 446054 18 43 528105 5-88 973761 ■75 554344 6-63 446666 17 44 528458 5-87 973716 •76 554741 6^62 445259 16 45 528810 5-87 973671 •76 556139 6-62 444861 15 46 529161 5-86 973626 •76 555536 6-61 444464 14 47 529513 5-86 973580 •76 555933 6-61 444067 13 48 529864 5-85 973536 •76 556329 6-60 443671 12 49 530215 5-85 973489 •76 556725 6-60 443275 11 50 530565 5-84 973444 •76 567121 659 442879 10 51 9-530915 5-84 9-973398 •76 9-557517 6-59 10-442483 9 52 531265 5-83 973352 •76 567913 6-59 442087 8 53 631614 5-82 973307 •76 558308 6-58 441692 7 54 531963 6-82 973261 •76 558702 6-68 441298 6 55 532312 5-81 973215 •76 569097 6-57 440903 5 56 532661 6-81 973169 •76 559491 6-67 440509 4 57 533009 5-80 973124 •76 559885 6-66 440115 3 58 533357 5-80 973078 •76 560279 6-56 439721 2 59 633704 5-79 973032 -77 560673 6-56 439327 1 60 534052 5-78 972986 -77 561066 6-55 438934 M. I Cosine* D. Sine. D. Cotang. D. Tang. (70 DEGB .EES.) 38 (20 DEGREES.) A TABLE OF LOGABITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-534052 5-78 9-972986 77 9-561066 6 65 10-438934 60 1 534399 5-77 972940 77 661459 6 64 438641 69 2 534745 5-77 972894 77 661851 6 54 438149 68 3 535092 5-77 972848 77 662244 6 53 437766 67 4 535438 5-76 972802 77 662636 6 53 437364 56 6 635783 5-76 972755 77 663028 6 53 436972 66 6 536129 5-75 972709 77 663419 6 52 436681 64 7 536474 5-74 972663 77 663811 6 52 436189 63 8 536818 5-74 972617 77 564202 6 51 435798 52 9 587168 5-73 972570 77 564592 6 61 435408 51 10 537507 5-73 972524 77 564983 6 60 435017 50 11 9-537851 5-72 9-972478 77 9-565373 6 50 10-434627 49 12 538194 5-72 972431 78 565763 6 49 434237 48 13 538538 5-71 972385 78 566153 6 49 433847 47 14 588880 5-71 972338 78 566542 6 49 433458 46 15 539223 5-70 972291 78 566932 6 48 433068 45 16 539565 5-70 972245 78 567320 6 48 432680 44 17 539907 5-69 972198 78 567709 6 47 432291 43 18 540249 5-69 972151 78 568098 6 47 431902 42 19 540590 5-68 972105 78 568486 6 46 431614 41 20 540931 5-68 972058 78 568873 6 46 431127 40 21 9-541272 5-67 9-972011 78 9-569261 6 46 10-430739 39 22 541613 5-67 971964 78 569648 6 45 430352 38 23 541953 5-66 971917 78 570035 6 45 429965 37 24 542293 5-66 971870 78 570422 6 44 429678 86 25 542632 5-65 971823 78 570809 6 44 429191 35 26 542971 5-65 971776 78 571195 6 43 428805 84 27 543310 5-64 971729 79 571581 6 43 428419 33 28 543649 5-64 971682 79 571967 6 42 428033 32 29 543987 5-63 971635 79 572352 6 42 427648 31 80 544325 5-63 971588 79 572738 6 42 427262 30 31 9-544663 5-62 9-971540 79 9-573123 6 41 10-426877 29 32 545000 5-62 971493 79 573507 6 41 426493 28 33 545838 5-61 971446 79 573892 6 40 426108 27 34 545674 5-61 971398 79 674276 6 40 425724 26 35 546011 5-60 971351 79 574660 6 39 426340 25 86 546347 5-60 971303 79 675044 6 39 424956 24 37 546683 5-59 971256 79 675427 6 39 424573 23 38 547019 5-59 971208 79 575810 6 38 424190 22 39 547354 5-58 971181 79 676193 6 38 423807 21 40 547689 5-58 971118 79 676576 6 37 423424 20 41 9-548024 5-57 9-971066 80 9-676968 6 37 10-423041 19 42 548359 5-57 971018 80 677341 6 36 422659 18 48 548693 5-56 970970 80 677728 6 36 422277 17 44 549027 5-56 970922 80 678104 6 36 421896 16 45 549360 5-55 970874 80 578486 6 35 421614 16 46 549693 5-55 970827 80 578867 6 35 421133 14 47 550026 5-54 970779 80 579248 6 34 420762 13 48 550359 6-54 970731 80 579629 6 34 420371 12 49 550692 5-53 970683 80 580009 6 34 419991 11 50 551024 5-53 970635 80 680389 6 33 419611 10 51 9-551356 5-52 9-970586 80 9-680769 6 33 10-419231 9 52 551687 5-52 970538 80 581149 6 32 418851 8 53 552018 5-52 970490 80 581628 6 32 418472 7 54 552349 5-51 970442 80 581907 6 32 418093 6 55 552680 5-51 970394 80 582286 6 31 417714 6 56 553010 5-50 970345 81 582665 6 31 417335 4 57 553341 6-50 970297 81 583043 6 30 416957 3 58 553670 5-49 970249 81 583422 6 30 416578 2 59 554000 5-49 970200 81 583800 6 29 416200 1 60 554329 5-48 970152 -81 584177 6-29 416823 1 Cosine. D. Sine. D. Cotang. D. Tang. M. (69 DEGREES.1 SINES AND TANGENTS. (21 DEGREES.) 39 JH. Sine, j D. Cosine. D. Tang. D. Cotang. 9-554329 5-48 9-970152 -81 9-584177 6-29 10 •415823 60 1 554658 5-48 970108 •81 584655 6-29 415445 59 2 554987 5-47 970056 -81 584932 6-28 415068 58 3 555315 5-47 970006 -81 585309 6^28 414691 57 4 555643 5-46 969967 -81 585686 6^27 414314 56 5 555971 5-46 969909 •81 586062 6^27 413988 55 6 556.299 5-45 969860 -81 586439 6-27 413561 54 7 656626 5-45 969811 -81 686815 6^26 413186 53 8 556953 5-44 969762 -81 587190 6^26 412810 52 9 657280 6-44 969714 -81 687566 6^25 412434 51 10 557606 5-48 969666 -81 587941 6^25 412059 50 11 9-567932 5-48 9-969616 -82 9-688316 6^25 10-411684 49 12 558258 5-48 969567 •82 588691 6^24 411309 48 13 558583 5-42 969518 -82 589066 6^24 410934 47 14 658909 5-42 969469 ■ -82 589440 6^23 410560 46 15 559234 5-41 969420 -82 589814 623 410186 45 16 559558 5-41 969370 -82 590188 6^23 409812 44 17 559883 5-40 969321 •82 690562 6^22 409438 43 18 560207 5-40 969272 -82 590935 6^22 409065 42 19 560631 5-39 969228 -82 591308 6^22 408692 41 20 560855 5-89 969178 •82 691681 6-21 408319 40 21 9-561178 5-38 9-969124 •82 9-592064 6^21 10^407946 39 22 561501 5-38 969075 ■82 592426 6^20 407674 38 23 661824 5-37 969025 •82 592798 6-20 407202 37 24 562146 5-87 968976 •82 593170 6^19 406829 36 25 562468 5-86 968926 •83 598542 6^19 406468 35 26 562790 5-36 968877 •83 593914 6^18 406086 34 27 563112 5-36 968827 •88 694285 6^18 405716 38 28 563483 5-35 968777 •83 694656 6^18 405344 32 29 663755 5-35 968728 •83 595027 6^17 404973 81 80 564075 5-34 968678 •83 595398 6^17 404602 30 81 0-564396 5-34 9-968628 •83 9-595768 6^17 10 • 404232 29 82 664716 5-33 968578 •83 696138 6^16 403862 28 38 665036 5-33 968528 •83 596508 6^16 403492 27 84 565366 5-32 968479 •83 596878 6^16 408122 26 85 666676 5-32 968429 •83 597247 6^16 402753 25 36 565995 5-31 968379 ■83 697616 6^15 402384 24 37 566314 5-81 968329 ■83 597986 6-16 402015 28 38 566632 5-81 968278 •83 698364 6^14 401646 22 89 566951 5-30 968228 •84 598722 6^14 401278 21 40 667269 5-30 968178 -84 599091 6^13 400909 20 41 9-567587 5-29 9-968128 -84 9-599469 6^13 10-400641 19 42 567904 5-29 968078 -84 699827 6^18 400178 18 43 668222 5-28 968027 -84 600194 6-12 899806 17 44 668539 5-28 967977 •84 600562 6^12 899438 16 45 668856 5-28 967927 •84 600929 6^11 399071 15 46 560172 5-27 967876 •84 601296 6^11 398704 14 47 569488 5-27 967826 -84 601662 6^11 398338 13 48 669804 5-26 967775 -84 602029 6^10 397971 12 49 570120 5-26 967725 -84 602395 6^10 397605 11 50 570485 5-26 967674 -84 602761 6-10 397239 10 51 9-570751 5-26 9-967624 -84 9-608127 6^09 10^396878 9 52 671066 6-24 967573 •84 603493 6^09 896507 8 53 571380 5-24 967522 -85 603858 6^09 896142 7 54 571695 5-23 967471 -85 604223 6^08 396777 6 55 572009 5-23 967421 -85 604588 6^08 895412 5 56 572323 5-23 967370 -85 604953 6^07 395047 4 57 572636 5-22 967319 •85 606317 6^07 394683 3 68 572950 5-22 967268 •86 605682 6^07 394318 2 59 573263 6-21 967217 -85 606046 6^06 393954 1 60 573575 5-21 967166 -85 606410 6^06 393590 |M.j Cosine. D. Sine. D. Cotang. D. Tang. (68 DEGEEES.) 40 (22 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tans. D. Cotang. 9-573575 5-21 9-967166 86 9-606410 6-06 10-893690 60 1 578888 5-20 967116 85 606773 6-06 393227 69 2 574200 5-20 967064 85 607137 6-06 392863 58 3 574512 5-19 967013 85 607500 6-05 392500 57 4 574824 5-19 966961 85 607863 6-04 392137 56 6 675136 5-19 966910 86 608225 6-04 391775 55 6 675447 5-18 966859 86 608688 6-04 391412 54 7 576768 5-18 966808 86 608950 6-03 391060 53 8 576069 5-17 966756 86 609312 6-03 390688 52 9 576379 5-17 966705 86 609674 6-03 390326 51 10 576689 516 966663 86 610036 6-02 389964 60 11 9-576999 6-16 9-966602 86 9-610397 6-02 10-389603 49 12 577809 5-16 966560 86 610769 6-02 389241 48 13 577618 6-15 966499 86 611120 6-01 388880 47 14 677927 6-16 966447 86 611480 6-01 388520 46 15 578236 6-14 966396 86 611841 6-01 888159 45 16 578545 5-14 966344 86 612201 6-00 387799 44 17 678853 5-13 966292 86 612561 6-00 887439 43 18 579162 5-13 966240 86 612921 6-00 387079 42 19 579470 6-13 966188 86 613281 5-99 386719 41 20 579777 5-12 966136 86 613641 5-99 886359 40 21 9-580085 5-12 9-966085 87 9-614000 5-98 10-386000 39 22 580392 511 966033 87 614359 5-98 385641 38 23 580699 5-11 965981 87 614718 5-98 885282 37 24 581005 6-11 965928 87 616077 6-97 384923 36 25 581312 5-10 965876 87 615435 5-97 384565 35 26 581618 6-10 965824 87 615793 5-97 384207 34 27 681924 5-09 965772 87 616151 5-96 383849 33 28 582229 6-09 965720 87 616509 6-96 383491 32 29 582535 5-09 965668 87 616867 5-96 383133 31 80 582840 5-08 965615 87 617224 5-95 882776 30 31 9-583145 6-08 9-965563 87 9-617582 5-95 10-382418 29 32 583449 5 07 965611 87 617989 5-96 882061 28 83 583754 5-07 965458 87 618295 5-94 881705 27 84 584058 6-06 965406 87 618652 5-94 881348 26 85 584361 5-06 965353 88 619008 5-94 880992 25 86 584665 5-06 965301 88 619364 5-93 880636 24 37 584968 5-05 966248 88 619721 5-93 880279 23 88 585272 5-05 965195 88 620076 5-93 879924 22 89 585574 5-04 965143 88 620432 5-92 379568 21 40 685877 5-04 966090 88 620787 5-92 379213 20 41 9-586179 5-03 9-965037 88 9-621142 6-92 10-378858 19 42 686482 5-03 964984 88 621497 6-91 378503 IS 48 586783 5-03 964931 88 621852 5-91 878148 17 44 587085 5-02 964879 88 622207 5-90 877793 16 45 587886 . 6-02 964826 88 622561 5-90 877439 15 46 587688 5-01 964773 88 622915 5-90 377086 14 47 587989 6-01 964719 88 628269 5-89 376731 13 48 588289 5-01 964666 89 623623 5-89 376377 12 49 588590 6-00 964613 89 623976 6-89 376024 11 50 588890 5-00 964660 89 624330 5-88 375670 10 51 9-589190 4-99 9-964507 89 9-624683 5-88 10-375317 9 62 689489 4-99 964454 89 625036 5-88 3X4964 8 53 589789 4-99 964400 89 625388 5-87 374612 7 54 . 590088 4-98 964347 89 626741 5-87 374259 6 55 590387 4-98 964294 89 626093 5-87 878907 5 56 590686 4-97 964240 89 626445 5-86 878655 4 57 590984 4-97 964187 89 626797 5-86 373203 8 58 591282 4-97 964133 89 627149 5-86 872851 2 59 591580 4-96 964080 89 627501 5-85 372499 1 60 591878 4-96 964026 89 627862 5-85 372148 Cosine. D. Sine. D. Cotang. D. Tang. M. (67 DEOREES.1 SINES AND TANGENTS. (23 DEGREES.) 41 m. Sine. D. Cosine. D. Tang. D. Cotang. 60 9-591878 4-96 9-964026 -89 9^627862 5^85 10-372148 1 592176 4-95 963972 -89 628208 5^86 871797 59 2 592478 4-96 963919 -89 628664 5^85 371446 58 8 592770 4-95 963865 -90 628905 5^84 871095 57 i- 593067 4-94 963811 ■90 629255 6^84 870746 56 5 593363 4-94 963767 •90 629606 6^83 370394 55 6 593659 4-93 968704 •90 629956 5^83 370044 64 T 593955 4-98 963650 -90 630806 5^88 369694 53 8 594261 4-93 963696 -90 630656 5^88 369344 52 9 594547 4-92 963542 -90 681005 5^82 868996 51 10 594842 4-92 963488 -90 631356 5^82 368645 50 11 9-595137 4-91 9-963434 -90 9 •631704 6^82 W368296 49 12 696432 4-91 963379 •90 6320DJ 5^81 367947 48 13 695727 4-91 963825 •90 682401 5^81 367699 47 14 696021 4-90 963271 -90 632750 5^81 367250 46 15 596315 4-90 963217 -90 633098 6^80 366902 45 16 596609 4-89 963163 -90 633447 5-80 866563 44 17 596903 4-89 968108 •91 633796 5^80 366205 43 18 597196 4-89 963054 •91 634143 5^79 365857 42 19 697490 4-88 962999 •91 634490 5^79 365510 41 20 697783 4-88 962945 •91 634838 5^79 865162 40 21 9-598075 4-87 9-962890 •91 9 ■635186 6^78 10 •364816 39 22 598368 4-87 962836 •91 635532 5-78 864468 38 23 598660 4-87 962781 •91 635879 5^78 364121 37 24 598952 4-86 962727 •91 636226 5^77 363774 36 25 699244 4-86 962672 •91 636672 6^77 863428 35 26 599536 4-85 962617 •91 686919 5^77 363081 34 27 699827 4-85 962562 •91 637265 5^77 862735 33 28 600118 4-85 962508 •91 687611 5^76 362389 32 29 600409 4-84 962453 •91 637956 5^76 362044 31 30 600700 4-84 962398 •92 638802 5^76 861698 30 31 9-600990 4-84 9-962343 •92 9 ■638647 5^75 10^ 361353 29 82 601280 4-83 962288 •92 638992 6^75 361008 28 33 601570 4-83 962283 •92 639337 5-75 360663 27 34 601860 4-82 962178 •92 639682 5^74 360318 26 85 602160 4-82 962123 •92 640027 5-74 359973 26 36 602439 4-82 962067 •92 640371 5-74 859629 24 37 602728 4-81 962012 •92 640716 5-73 359284 23 38 603017 4-81 961957 •92 641060 5-78 358940 22 39 603305 4-81 961902 •92 641404 5-73 358596 21 40 603594 4-80 961846 •92 641747 6-72 368258 20 41 9-603882 4-80 9-961791 •92 9^642091 5-72 10^367909 19 42 604170 4-79 961785 •92 642484 5^72 857566 18 43 604457 4-79 961680 ■92 642777 5-72 857223 17 44 -> 604745 4-79 961624 •98 643120 5^71 366880 16 45 606032 4-78 961669 •98 643468 5-71 366637 16 46 605319 4-78 961613 •93 643806 5-71 366194 14 47 605606 4-78 961468 •98 644148 5-70 355852 13 48 605892 4-77 961402 •98 644490 6-70 855610 12 49 606179 4-77 961846 •93 644832 5-70 355168 11 50 606465 4-76 961290 •98 645174 6-69 854826 10 51 9-606751 4-76 9-961235 •93 9 •646616 5-69 10^854484 9 52 607036 4-76 961179 •93 645867 569 364148 8 53 607322 4-75 961123 •93 646199 6-69 353801 7 64 607607 4-75 961067 •93 646640 5-68 363460 6 55 607892 4-74 961011 •93 646881 5-68 353119 5 66 608177 4-74 960956 •93 647222 5-68 352778 4 67 608461 4-74 960899 •93 647562 5-67 352438 3 58 608745 4-73 960843 •94 647903 6-67 362097 2 59 609029 4-73 960786' •94 648243 6-67 351757 1 60 609313 4-78 960730 •94 648583 5^66 351417 I Cosine< D. Slue. D. Cotaug. D. Tang. M. C66 DEGI lEES.) 42 (24' DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-609313 4-73 9-960730 94 9-648583 5-66 10-351417 60 1 609597 4-72 960674 94 648923 5-66 351077 59 2 609880 4-72 960618 94 649263 5-66 350737 58 3 610164 4-72 960561 94 649602 5-66 350398 57 i 610447 4-71 960505 94 649942 5-65 350058 56 5 610729 4-71 960448 94 650281 5-65 349719 55 1 6 611012 4-70 960392 94 650620 5-65 349380 54 i ■7 611294 4-70 960335 94 650959 5-64 349041 53 8 611576 4-70 960279 94 651297 5-64 348703 52 9 611858 4-69 960222 94 651636 5-64 348364 51 10 612140 4-69 960165 94 651974 5-63 348026 50 11 9-612421 4-69 9-960109 95 9-652312 5-63 10-347688 49 12 612702 4-68 960052 95 652650 5-63 347350 48 13 612983 4-68 959995 95 652988 5-63 347012 47 14 613264 4-67 959938 95 653326 5-62 346674 46 15 613545 4-67 959882 95 653663 5-62 346337 45 16 613825 4-67 959825 95 654000 5-62 846000 44 17 614105 4-66 959768 95 654337 5-61 345663 43 18 614385 614665' 4-66 959711 95 654674 5-61 345326 42 19 4-66 959654 95 655011 5-61 344989 41 20 614944 4-65 959596 95 655348 5-61 344652 40 21 9-615223 4-65 9-959539 95 9-655684 5-60 10-344316 39 22 615502 4-65 959482 95 656020 5-60 343980 38 23 615781 4-64 959425 95 656356 5-60 343644 37 21 616060 4-64 959368 95 656692 5-59 343308 36 25 616338 4-64 959310 96 657028 5-59 342972 35 26 616616 4-63 959253 96 657364 5-59 342636 34 27 616894 4-63 959195 96 657699 5-59 342301 33 28 617172 4-62 959138 96 658034 5-58 341966 82 29 617450 4-62 959081 96 658369 5-58 341631 31 80 617727 4-62 959023 96 658704 5-58 341296 30 31 9-618004 4-61 9-958965 96 9-659039 6-58 10-340961 29 32 618281 4-61 958908 96 659373 5-57 340627 28 33 618558 4-61 958850 96 659708 5-57 340292 27 34 618834 4-60 958792 96 660042 5-57 339958 26 35 619110 4-60 958734 96 660376 5-57 339624 25 36 619386 4-60 958677 96 660710 5-56 339290 24 37 619662 4-59 958619 96 661043 5-56 338957 23 • 38 619938 4-59 958561 96 661377 5-56 338623 22 : 39 620213 4-59 958503 97 661710 5-55 338290 21 40 620488 4-58 958445 97 662043 5-55 337957 20 41 9-620763 4-58 9-958387 97 9-662376 5-55 10-337624 19 i 42 621038 4-57 958329 97 662709 5-54 337291 18 i 48 621313 4-57 958271 97 663042 5-54 336958 17 1 44 621587 4-57 958213 97 663375 5-54 336625 16 ; 45 621861 4-56 958154 97 663707 5-54 386293 15 1 46 622135 4-56 958096 97 664039 5-53 335961 14 ! 47 622409 4-56 958038 97 664371 5-53 335629 13 48 622682 4-55 957979 97 664703 5-53 335297 12 49 622956 4-55 957921 97 665035 5-58 334965 11 50 623229 4-55 957863 97 665366 5-52 834634 10 51 9-623502 4-54 9-957804 97 9-665697 5-52 10-334303 9 i 52 623774 4-54 957746 98 666029 5-52 333971 8 ! 53 624047 4-54 957687 98 666360 5-51 833640 7 ! 54 624319 4-53 957628 98 666691 5-51 333309 6 1 55 624591 4-53 957570 98 667021 5-51 332979 5 56 624863 4-58 957511 98 667352 5-51 832648 4 57 625135 4-52 957452 98 667682 5-50 332318 3 58 625406 4-52 957393 98 668013 5-50 331987 2 59 625677 4-52 957335 98 668343 5-50 331657 1 60 625948 4-51 957276 98 668672 5-50 331328 Tang. I Cosine. D. Sine. D, Cotang. D. SI. (65 DEGREES.) SINES AND TANGENTS. (25 DE6EEES.) 43 JH. Sine. D. Cosine. D. Tang, i D. Cotang. J 9-625948 4-51 9-957276 -98 9-668673 5-50 10 ■331327 60 1 626219 4-51 1 957217 ■98 669002 5-49 330998 59 2 626490 4-51 957158 -98 669332 5-49 330668 58 3 626760 4-50 957099 -98 669661 5-49 330389 57 4 627030 4-50 ! 957040 -98 669991 5-48 330009 56 5 627300 4-50 956981 •98 670320 5-48 329680 55 6 627.570 4-49 956921 •99 670649 5-48 329351 54 7 627840 4-49 956862 -99 670977 5-48 329023 53 8 628109 4-49 956803 -99 671306 5 '47 328694 52 9 628378 4-48 956744 -99 671634 5-47 328366 51 10 628647 4-48 956684 -99 671963 5-47 328037 50 11 9-628916 4-47 9-956625 -99 9-672291 5-47 10-327709 49 12 629185 4-47 956566 -99 672619 5-46 327381 48 13 629453 4-47 956506 -99 672947 5-46 327053 47 14 629721 4-46 956447 -99 673274 5-46 326726 46 15 629989 4-46 956387 -99 673602 5-46 326398 45 16 630257 4-46 956327 •99 673929 5-45 326071 44 17 630524 4-46 956268 -99 674257 5-45 ! 325743 43 18 630702 4-45 956208 1-00 674584 5-45 ! 325416 42 19 631059 4-45 956148 1-00 674910 5-44 i 325090 41 20 631326 4-45 956089 1-00 675237 5-44 ; 324763 40 21 0-631598 4-44 9-956029 1-00 9-675564 5-44 ! 10-324436 39 22 631859 4-44 955969 1-00 675890 5-44 : 324110 38 23 632125 4-44 955909 1-00 676216 5-43 323784 37 24 632392 4-43 955849 1-00 676543 5-43 323457 36 25 632658 4-43 955789 1-00 676869 5-43 323131 35 26 682923 4-43 955729 1-00 677194 5-43 322806 34 27 638189 4-42 955669 1-00 677520 5-42 322480 33 28 633454 4-42 955609 1-00 677846 5-42 322154 32 29 633719 4-42 955548 1-00 678171 5-42 321829 31 30 633984 4-41 955488 1-00 678496 5-42 321504 30 31 9-634249 4-41 9-955428 1-01 9-678821 - 5-41 10-321179 29 32 634514 4-40 955368 1-01 679146 5-41 320854 28 33 634778 4-40 955307 1-01 679471 5-41 320529 27 84 635042 4-40 955247 1^01 679795 6-41 320205 26 85 635306 4-39 955186 101 680120 5-40 319880 25 36 635570 4-39 955126 1^01 680444 5-40 319556 24 37 635834 4-39 955065 I'Ol 680768 5-40 319232 23 38 636097 4-38 955005 1^01 681092 5-40 318908 22 39 686360 4-88 954944 101 681416 5-39 318584 21 40 636623 4-38 954883 1-01 681740 5-39 318260 20 41 9-636886 4-37 9-954823 1-01 9-682063 5-39 10-317937 19 42 637148 4-37 954762 1^01 682387 5-39 317613 18 43 637411 4-37 -954701 1^01 682710 5'38 317290 17 44 637673 4-37 954640 1^01 683033 5-38 316967 16 45 637935 4-36 954579 1^01 683356 5-38 316644 15 46 638197 4-36 954518 1-02 683679 5-38 316321 14 47 638458 4-36 954457 1-02 684001 5-37 315999 13 48 638720 4-85 954396 1-02 684324 5-37 315676 12 49 638981 4-35 954335 1-02 684646 5-87 315354 11 50 639242 4-35 954274 1-02 684968 5-37 315032 10 51 9-639503 4-34 9-954213 1-02 9-685290 5-36 10-314710 9 52 639764 4-34 954152 1-02 685612 5-36 314388 8 53 640024 4-34 954090 1-02 685934 5-36 314066 7 54 640284 4-33 954029 1-02 686255 5-36 313745 6 65 640544 4-83 953968 1-02 686577 5-35 313423 5 66 640804 4-83 953906 1-02 686898 5-35 313102 4 57 641064 . 4-32 953845 1^02 687219 5-35 312781 3 58 641324 4-82 953783 1'02 687540 5-35 i 312460 2 69 641584 4-32 953722 1-03 687861 5-34 312139 1 60 641842 4-31 953660 1-03 688182 5-34 311818 Cosine. D. Sine. D. Cotang. D. Tang. M. (64 DEGREES.) 44 (26 DEGREES. ) A TABLE OF LOGARITHMIC JH. Sine. D. Cosine. D. Tang. D. . Cotang. 10-811818 60 9-641842 4-31 9-953660 1-03 9-688182 5-34 1 642101 4-31 953599 1-03 688602 5-34 311498 69 2 642360 4-31 963687 1-03 688823 6-34 311177 58 3 642618 4-30 953476 1-03 689143 6-33 310867 57 4 642877 4-30 953413 1-03 689463 5-33 810537 56 6 643135 4-30 963362 1-03 689783 5-33 810217 66 6 643393 4-30 953290 1-03 690103 5-33 809897 54 T 643650 4-29 953228 1-03 690423 5-33 309577 53 8 643908 . 4-29 963166 1-03 690742 6-32 309258 62 9 644165 4-29 963104 1-03 691062 6-32 808938 51 10 644423 4-28 963042 1-03 691381 5-32 308619 50 11 9-644680 4-28 9-962980 1-04 9-691700 5-31 10-808300 49 12 644936 4-28 962918 1-04 692019 6-31 307981 48 13 645193 4-27 952855 1-04 692338 5-31 307662 47 14 645450 4-27 962793 1-04 692656 5-31 307344 46 16 645706 4-27 962731 1-04 692975 5-31 307025 45 16 646962 4-26 962669 1-04 693293 5-30 306707 44 17 646218 4-26 952606 1-04 693612 5-30 306388 43 18 646474 4-26 962544 1-04 693930 6-30 306070 42 19 646729 4-26 952481 1-04 694248 6-30 305752 41 20 646984 4-25 952419 1-04 694666 5-29 305434 40 21 9-647240 4-26 9-952356 1-04 9-694883 5-29 10- 305117 39 22 647494 4-24 952294 1-04 695201 5-29 804799 38 23 647749 4-24 952231 1-04 695518 6-29 304482 37 24 648004 4-24 952168 1-05 695836 5-29 804164 36 25 648258 4-24 952106 1-05 696158 6-28 303847 35 26 648512 4-23 952043 1-05 696470 5-28 303530 34 27 648766 4-23 951980 1-05 696787 6-28 803213 33 28 649020 4-23 951917 1-05 697103 5-28 302897 32 29 649274 4-22 961854 1-06 697420 5-27 302580 31 80 649527 4-22 951791 1-05 697736 5-27 302264 30 81 9-649781 4-22 C- 951728 1-05 9-698053 6-27 10-301947 29 82 650034 4-22 951666 1-05 698369 6-27 301631 28 33 650287 4-21 961602 1-05 698685 5-26 301315 27 84 650539 4-21 951539 1-05 699001 5-26 300999 26 35 650792 4-21 961476 1-06 699316 5-26 300684 25 36 651044 4-20 961412 1-06 699632 5-26 300368 24 37 651297 4-20 951349 1-06 699947 5-26 300053 23 38 661649 4-20 951286 1-06 700263 5-25 299737 22 89 651800 4-19 961222 1-06 700578 5-25 299422 21 40 652062 4-19 961169 1-06 700893 5-25 299107 20 41 9-652304 4-19 9-951096 1-06 9-701208 5-24 10-298792 19 42 652555 4-18 951032 1-06 701623 5-24 298477 18 43 652806 4-18 960968 1-06 701837 5-24 298163 17 44 653057 4-18 950905 1-06 702152 5-24 297848 16 45 663308 4-18 950841 1-06 702466 5-24 297534 15 46 653558 4-17 950778 106 702780 5-23 297220 14 47 653808 4-17 950714 1-06 703096 5-23 296906 13 48 654059 4-17 960650 1-06 703409 5-23 296691 12 49 664309 4-16 950586 1-06 703723 5-23 296277 11 50 654558 4-16 960622 1-07 704036 5-22 295964 10 51 9-664808 4-16 9-950458 1-07 9-704350 5-22 10-295660 9 52 655058 4-16 950394 1-07 704663 5-22 295337 8 53 655307 4-15 960880 1-07 704977 5-22 296023 7 54 655566 4-15 950266 1-07 705290 6-22 294710 6 55 655805 4-15 950202 1-07 705603 5-21 294397 6 56 666054 4-14 960188 1-07 705916 5-21 294084 4. 57 656302 4-14 950074 1-07 706228 5-21 293772 3 58 656651 4-14 950010 1-07 706541 6-21 293459 2 69 656799 4-13 949945 1-07 706854 6-21 298146 1 60 657047 4-13 949881 1-07 707166 5-20 292834 Cosine. D. Sine. D. Cotang. D. Tang. n. (63 DEGREES.) SINES AND TANGENTS. (27 DEGREES.) 45 m. Sine. D. Cosine. D. Tang. ». Cotang. 60 9-657047 4-13 9-949881 1-07 9-707166 5-20 LO-292834 1 657295 4-13 949816 1-07 707478 6-20 292622 59 2 657542 4-12 949752 1-07 707790 5-20 292210 58 3 657790 4-12 949688 1-08 708102 5-20 291898 57 4 658037 4-12 949623 1-08 708414 5-19 291686 56 5 658284 4-12 949558 1-08 708726 6-19 291274 65 6 658531 4-11 949494 1-08 709037 5-19 290963 54 7 658778 4-11 949429 1-08 709349 5-19 290651 63 8 659025 4-11 949364 1-08 709660 6-19 290340 62 9 659271 4-10 949300 1-08 709971 5-18 290029 51 10 659517 4-10 949235 1-08 710282 5-18 289718 60 11 9-659763 4-10 9-949170 1-08 9-710593 5-18 10-289407 49 12 660009 4-09 949105 1-08 710904 5-18 289096 48 13 660255 4-09 949040 1-08 711216 6-18 288785 47 14 660501 409 948975 1-08 711625 6-17 288475 46 15 660746 4-09 948910 1-08 711836 5-17 288164 45 16 660991 4-08 948845 1-08 712146 5-17 287864 44 17 661236 4-08 948780 1-09 712466 6-17 287644 43 18 661481 4-08 948715 1-09 712766 6-16 287284 42 19 661726 4-07 948660 1-09 713076 6-16 286924 41 20 661970 4-07 948584 1-09 713386 6-16 286614 40 21 9-662214 4-07 9-948519 1-09 9-713696 5-16 10-286304 39 22 662459 4-07 948454 1-09 714005 5-16 285995 38 23 662703 4-06 948388 1-09 714314 6-15 285686 37 24 662946 4-06 948323 1-09 714624 5-15 285376 36 25 663190 4-06 948257 1-09 714933 6-16 286067 35 26 663433 4-05 948192 1-09 715242 6-15 284758 34 27 668677 4-05 948126 1-09 715661 5-14 284449 33 28 663920 4-05 948060 109 715860 6-14 284140 32 29 664163 4-05 947996 1-10 716168 5-14 283832 31 30 664406 4-04 947929 1-10 716477 5-14 283523 30 31 9-664648 4-04 9-947863 1-10 9-716785 5-14 10-283215 29 82 664891 4-04 947797 1-10 717098 5-13 282907 28 -33 665133 4-OS 947731 1-10 717401 6-13 282599 27 84 665375 4-03 947665 1-10 717709 5-13 282291 26 35 665617 4-03 947600 1-10 718017 6-13 281983 25 86 665869 4-02 947533 1-10 718325 5-13 281670 24 37 666100 4-02 947467 1-10 718633 5-12 281367 23 38 666342 4-02 947401 1-10 718940 5-12 281060 22 39 666583 4-02 947335 1-10 719248 6-12 280752 21 40 666824 4-01 947269 1-10 719565 5-12 280445 20 41 9-667065 4-01 9-947203 1-10 9-719862 6-12 10-280138 19 42 667305 4-01 947136 1-11 720169 5-11 279831 18 48 667546 4-01 947070 1-11 720476 5-11 279624 17 44 667786 4-00 947004 1-11 720783 6-11 279217 16 46 668027 4-00 946937 1-11 721089 5-11 278911 16 46 668267 4-00 946871 1-11 721396 6-11 278604 14 47 668506 3-99 946804 1-11 721702 6-10 278298 13 48 668746 3-99 946738 1-11 722009 5-10 277991 12 49 668986 8-99 946671 1-11 722315 5-10 277686 11 50 669225 3-99 946604 1-11 722621 5-10 277879 10 51 9-669464 3-98 9-946538 1-11 9-722927 6-10 10-277073 9 52 669703 3-98 946471 1-11 723232 5-09 276768 8 53 669942 8-98 946404 1-11 723638 5-09 276462 7 54 670181 3-97 946337 1-11 723844 5-09 276156 6 55 670419 3-97 946270 1-12 724149 6-09 275851 6 56 670658 8-97 946203 1-12 724464 5-09 275546 4 57 670896 3-97 946136 1-12 724759 5-08 276241 3 58 671134 8-96 946069 1-12 725065 5-08 274935 2 69 671372 3-96 946002 1-12 725869 5-08 274631 1 60 671609 3-96 945935 1-12 725674 6-08 274826 Cosine. D. Sine. D. Cotang. 1 »• Tang. M. (62 DEGREES.) 46 (28 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-671609 3 96 9-946935 1-12 9-725674 5-08 10-274326 60 1 671847 3 95 945868 1-12 725979 5-08 274021 69 2 672084 3 95 945800 1-12 726284 5-07 273716 68 3 672321 3 95 945733 1-12 726588 5-07 273412 57 4 672558 3 95 945666 1-12 726892 5-07 273108 56 6 672795 3 94 945598 1-12 727197 5-07 272803 65 6 673032 3 94 945531 1-12 727501 5-07 272499 54 7 673268 3 94 945464 1-13 727805 5-06 272195 53 8 673505 3 94 945396 1-13 728109 5-06 271891 52 g 673741 3 93 945328 113 728412 5-06 271588 51 10 673977 3 93 945261 1-18 728716 506 271284 50 11 9-674213 3 93 9-945198 1-13 9-729020 5-06 10-270980 49 12 674448 3 92 945125 1-13 729323 5-05 270677 48 13 674684 3 92 945058 1-13 729626 5-05 270374 47 14 674919 3 92 944990 1-13 729929 5-05 270071 46 15 675155 8 92 944922 1-13 780233 5-05 269767 45 16 675390 8 91 944854 1-13 780585 5-05 269465 44 17 675624 3 91 944786 1-13 730888 5-04 269162 48 18 675859 3 91 944718 1-13 731141 5-04 268859 42 19 676094 8 91 944650 1-13 781444 5-04 268556 41 20 676328 3 90 944582 1-14 731746 5-04 268254 40 21 0-676562 8 90 9-944514 1-14 9-732048 5-04 10-267952 89 22 676796 3 90 944446 1-14 732.851 5-08 267649 38 23 677030 8 90 944377 1-14 732653 503 267347 87 24 677264 3 89 944309 1-14 732955 5-08 267045 36 25 677498 8 89 944241 1-14 733257 5-08 266748 85 26 677731 3 89 944172 1-14 738558 5-03 266442 34 27 677964 3 88 944104 1-14 733860 502 266140 33 28 678197 3 88 944036 1-14 734162 5-02 265838 32 29 678430 3 88 943967 1-14 734463 5-02 265587 31 80 678663 8 88 948899 1-14 734764 5-02 265236 30 31 9-678895 3 87 9-943830 1-14 9-735066 5-02 10-264934 29 32 679128 3 87 943761 1-14 735367 5-02 264683 28 33 679360 3 87 943693 1-15 735668 5-01 264332 27 34 679592 3 87 943624 1-15 735969 «-01 264031 26 85 679824 3 86 948555 1-15 736269 5-01 263731 26 36 680056 3 86 943486 1-15 736570 5-01 263430 24 37 680288 3 86 943417 1-15 736871 5-01 263129 23 38 680519 3 85 943348 1-15 737171 5-00 262829 22 39 680750 3 85 948279 1-15 787471 5-00 262529 21 40 680982 8 85 943210 1-15 737771 5-00 262229 20 41 9-681213 8 85 9-943141 1-15 9-738071 5-00 10-261929 19 42 681443 8 84 943072 1-15 *73837l 5-00 261629 18 43 681674 8 84 943003 1-15 738671 4-99 261329 17 44 681905 3 84 942984 1-15 738971 4-99 261029 16 45 682135 3 84 942864 1-15 789271 4-99 260729 15 ; 46 682365 8 83 942795 1-16 739570 4-99 260430 14 ' 47 682595 8 88 942726 1-16 739870 4-99 260130 13 48 682825 3 88 942656 1-16 740169 4-99 259881 12 49 683055 3 83 942587 1-16 740468 4-98 259532 11 50 683284 8 82 942517 1-16 740767 4-98 ■ 259238 10 1 51 9-683514 3 82 9-942448 1-16 9-741066 4-98 10-258984 9 52 683743 3 82 942378 1-16 741365 4-98 258635 8 53 683972 3 82 942808 1-16 741664 4-98 258336 7 54 684201 8 81 942239 1-16 741962 4-97 258038 6 55 684430 3 81 942169 1-16 742261 4-97 257789 5 56 684658 3 81 942099 1-16 742559 4-97 257441 4 57 684887 8 80 942029 1-16 742858 4-97 257142 3 58 685115 3 80 941959 1-16 743156 4-97 256844 2 59 685343 3 80 941889 1-17 743454 4-97 256546 1 60 685571 8-80 941819 1-17 743752 4-96 256248 M. Cosine. D. Sine. -». Cotang. D. Tang. (61 DEGREES.) SINES AND TANGENTS. (29 DEGREES.) 47 IH. Sine. D. Cosine. 1 D. Tang. 1 D. Cotang. ~o" 9-685571 3-80 9-941819 1-17 9-743752 4-96 10-266248 60 1 685799 8-79 941749 1-17 744060 4-96 255950 59 2 686027 3-79 941679 1-17 744348 4-96 256652 58 3 686254 3-79 941609 1-17 744645 4-96 255355 57 4 686482 3-79 941639 1-17 744943 4-96 266057 56 5 686709 3-78 941469 1-17 745240 4-96 254760 55 6 686936 3-78 941398 1-17 746638 4-95 254462 54 7 687163 3-78 941328 1-17 746885 4-95 254165 58 8 687389 3-78 941258 1-17 746132 4-95 258868 62 g 687616 3-77 941187 1-lT 746429 4-95 253571 51 10 687843 3-77 941117 1-17 746726 4-95 253274 50 11 9-688069 3-77 9-941046 1-18 9-747028 4-94 10-25'2977 49 12 688295 3-77 940976 1-18 747319 4-94 252681 48 13 688521 3-76 940905 1-18 747616 4-94 252384 47 14 688747 3-76 940834 1-18 747913 4-94 252087 46 15 688972 8-76 940763 1-18 748209 4-94 251791 45 16 689198 3-76 940693 1-18 748605 4-93 251496 44 17 689423 3-75 940622 1-18 748801 4-93 261199 43 18 689648 3-75 940551 1-18 749097 4-93 250903 42 19 689873 3-75 940480 1-18 749393 4-93 250607 41 20 690098 3-75 940409 1-18 749689 4-93 260311 40 21 9-690323 3-74 9-940338 1-18 9-749986 4-93 10-250015 39 22 690548 3-74 940267 1-18 750281 4-92 249719 38 23 690772 3-74 940196 1-18 750576 4-92 249424 37 24 690996 8-74 940125 1-19 750872 4-92 249128 86 25 691220 3-73 940054 1-19 751167 4-92 248883 35 26 691444 3-73 939982 1-19 761462 4-92 248538 84 27 691668 3-73 939911 1-19 751757 4-92 248248 33 28 691892 8-73 939840 1-19 752062 4-91 247948 32 29 692115 8-72 989768 1-19 762347 4-91 247653 31 30 692339 3-72 939697 1-19 752642 4-91 247358 30 31 9-692562 3-72 9-939625 1-19 9-752937 4-91 10-247063 29 32 692785 3-71 939654 1-19 753231 4-91 246769 28 33 693008 3-71 939482 1-19 763526 4-91 246474 27 34 693231 3-71 939410 119 753820 4-90 246180 26 35 693458 3-71 939339 1-19 754115 4-90 245885 26 36 693676 3-70 939267 1-20 764409 4-90 246691 24 37 693898 3-70 939196 1-20 754703 4-90 245297 23 38 694120 3-70 939128 1-20 754997 4-90 246003 22 39 694342 3-70 989052 1-20 755291 4-90 244709 21 40 694564 3-69 938980 1-20 755686 4-89 244416 20 41 9-694786 8-69 9-938908 1-20 9-755878 4-89 10-244122 19 42 695007 3-69 988836 1-20 756172 4-89 248828 18 43 695229 3-69 988763 1-20 756465 4-89 243535 17 44 695450 3-68 938691 1-20 766759 4-89 243241 16 45 695671 8-68 988619 1-20 757062 4-89 242948 15 46 695892 8-68 938547 1-20 757346 4-88 242666 14 47 696113 3-68 938475 1-20 767638 4-88 242382 13 48 696834 3-67 938402 1-21 767931 4-88 242069 12 49 696554 3-67 938330 1-21 768224 4-88 241776 11 50 696775 8-67 938258 1-21 768617 4-88 241483 10 61 9-696995 3-67 9-988185 1-21 9-758810 4-88 10-241190 9 52 697215 3-66 988113 1-21 759102 4-87 240898 8 53 697435 8-66 938040 1-21 759396 4-87 240605 7 54 697654 8-66 937967 1-21 759687 4-87 i 240318 6 55 697874 3-66 937895 1-21 759979 4-87 240021 6 56 698094 3-65 937822 1-21 760272 4-87 239728 4 57 698313 3-66 937749 1-21 760564 4-87 1 239436 8 58 698532 3-66 937676 1-21 760866 4-86 239144 2 69 698751 3-65 987604 1-21 761148 4-86 1 238852 1 60 698970 8-64 937581 1-21 761439 4-86 238661 Cosine. D. Sine. D. Cotang. D. Tang. SI. (60 DEGREES.) 48 (30 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-698970 3-64 9-937581 1-21 9-761439 4-86 10-238561 60 1 699189 8-64 987458 1-22 761731 4-86 238269 69 2 699407 8-64 937885 1-22 762023 4-86 237977 58 8 699626 3-64 937312 1-22 762814 4-86 287686 57 4 699844 3-63 937238 1-22 762606 4-85 287894 56 5 700062 8-63 937165 1-22 762897 4-85 237103 55 6 700280 3-63 937092 1-22 763188 4-85 236812 54 7 700498 8-68 987019 1-22 763479 4-85 286521 53 8 700716 3-63 986946 1-22 768770 4-85 286280 52 9 700933 3-62 936872 1-22 764061 4-85 235989 51 10 701151 8-62 936799 1-22 764852 4-84 235648 50 11 9-701868 3-62 9-936725 1-22 9-764643 ' 4-84 10-235357 49 12 701585 3-62 986652 1-23 764933 4-84 235067 48 18 701802 3-61 986578 1-23 765224 4-84 234776 47 14 702019 3-61 936505 1-23 765514 4-84 234486 46 15 702236 3-61 986431 1-23 765805 4-84 234195 45 16 702452 8-61 986857 1-23 766095 4-84 233905 44 17 702669 8-60 936284 1-23 766885 4-83 233615 43 18 702885 3-60 936210 1-23 766675 4-83 233825 42 19 703101 3-60 936136 1-23 766965 4-83 233035 41 20 703317 8-60 936062 1-23 767255 4-83 232745 40 21 9-703533 3-59 9-935988 1-23 9-767545 4-83 10-232455 39 22 708749 3-59 -. 935914 1-23 767834 4-83 232166 38 28 703964 3-59 935840 1-23 768124 4-82 231876 87 24 704179 3-59 935766 1-24 768413 4-82 231587 36 25 704895 3-59 935692 1-24 768703 4-82 231297 36 26 704610 3-58 985618 1-24 768992 4-82 231008 34 27 704825 3-58 935543 1-24 769281 4-82 230719 33 28 705040 3-58 935469 1-24 769570 4-82 230430 32 29 705254 3-58 935395 1-24 769860 4-81 230140 31 80 705469 3-57 935320 1-24 770148 4-81 229852 80 81 9-705683 3-57 9-935246 1-24 9-770437 4-81 10-229568 29 82 705898 3-57 935171 1-24 770726 4-81 229274 28 88 706112 8-57 935097 1-24 771015 4-81 228985 27 34 706326 3-56 935022 1-24 771303 4-81 228697 26 35 706539 3-56 984948 1-24 771592 4-81 228408 25 36 706753 8-56 934873 1-24 771880 4-80 228120 24 37 706967 3-56 934798 1-25 772168 4-80 227832 23 38 707180 3-55 934723 1-25 772457 4-80 227548 22 39 707398 3-55 984649 1-25 772745 4-80 227255 21 40 707606 3-55 934574 1-25 773033 4-80 226967 20 41 9-707819 3-55 9-934499 1-25 9-773821 4-80 10-226679 19 42 708032 8-54 934424' 1-25 773608 4-79 226392 18 43 708245 8-54 934849 1-25 773896 4-79 226104 17 44 708458 3-54 984274 1-25 774184 4-79 225816 16 45 708670 8-54 934199 1-25 774471 4-79 225529 16 46 708882 8-58 934123 1-25 774759 4-79 225241 14 47 709094 8-53 984048 1-25 775046 4-79 224954 13 48 709806 3-53 983973 1-25 775333 4-79 224667 12 49 709518 8-53 988898 1-26 775621 4-78 224379 11 50 709730 8-53 983822 1-26 775908 4-78 224092 10 51 9-709941 3-52 9-938747 1-26 9-776195 4-78 10-223805 9 52 710158 8-52 933671 1-26 776482 4-78 228518 8 53 710364 3-52 933596 1-26 776769 4-78 223231 7 54 710575 3-52 983520 1-26 777055 4-78 222945 6 55 710786 3-51 933445 1-26 777842 4-78 222658 6 56 710997 3-51 988869 1-26 777628 4-77 222872 4 57 711208 3-51 938298 1-26 777915 4-77 222085 8 58 711419 3-51 933217 1-26 778201 4-77 221799 2 59 711629 3-50 933141 1-26 778487 4-77 221512 1 60 711839 3-50 983066 1-26 778774 4-77 221226 Cosine. D. Sine. ». Cotang. D. Tang. «. (59 DEGBEESj SINES AND TANGENTS. (31 DEGREES.) 49 M. Sine. D. Cosine. D. Tang. D. Cotang. 60 9-711839 3-50 9-938066 1-26 9-778774 4-77 10-221226 1 712060 3-50 932990 1-27 779060 4-77 220940 59 2 712260 3-60 932914 1-27 779346 4-76 220664 58 3 712469 8-49 932838 1-27 779632 4-76 220368 57 4 712679 3-49 932762 1-27 779918 4-76 220082 56 6 712889 3-49 932685 1-27 780203 4-76 219797 65 6 713098 3-49 982609 1-27 780489 4-76 219611 64 7 713308 3-49 932533 1-27 780775 4-76 219226 53 8 713617 3-48 932457 1-27 781060 4-76 218940 52 9 713726 8-48 932380 1-27 781346 4-75 218654 61 10 713935 8-48 932304 1-27 781631 4-75 218369 50 11 9-714144 8-48 9-982228 1-27 9-781916 4-76 10-218084 49 12 714352 3-47 932161 1-27 782201 4-75 217799 48 13 714661 8-47 932075 1-28 782486 4-75 217514 47 14 714769 3-47 931998 1-28 782771 4-75 217229 46 15 714978 8-47 981921 1-28 788066 4-75 216944 45 16 715186 3-47 931846 1-28 788341 4-75 216659 44 17 715894 8-46 931768 1-28 783626 4-74 216874 43 18 715602 8-46 931691 1-28 783910 4-74 216090 42 19 715809 3-46 931614 1-28 784195 4-74 215806 41 20 716017 3-46 981537 1-28 784479 4-74 215521 40 21 9-716224 3-45 9-931460 1-28 9-784764 4-74 10-216286 89 22 716432 3-45 931383 1-28 786048 4-74 214952 38 23 716639 3 -45 931S06 1-28 785332 4-73 214668 37 24 716846 3-45 931229 1-29 785616 4-73 214384 86 25 717053 8-45 931152 1-29 785900 4-73 214100 35 26 717269 3-44 981076 1-29 786184 4-73 213816 34 27 717466 3-44 980998 1-29 786468 4-73 213532 38 28 717673 3-44 930921 1-29 786762 4-73 218248 32 29 717879 3-44 930843 1-29 787036 4-73 212964 31 80 718086 8-43 930766 1-29 787319 4-72 212681 30 31 9-718291 3-43 9-930688 1-29 9-787603 4-72 10-212397 29 32 718497 3-43 980611 1-29 787886 4-72 212114 28 83 718703 3-43 930633 1-29 788170 4-72 211830 27 84 718909 8-43 930456 1-29 788463 4-72 211547 26 85 719114 3-42 930378 1-29 788736 4-72 211264 25 86 719820 3-42 930300 1-80 789019 4-72 210981 24 87 719525 8-42 930228 1-30 789302 4-71 210698 23 88 719730 8-42 930145 1-30 789586 4-71 210415 22 39 719935 3-41 930067 1-30 789868 4-71 210182 21 40 720140 8-41 929989 1-80 790151 4-71 209849 20 41 9-720346 3-41 9-929911 1-30 9-790433 4-71 10-209567 19 42 720549 3-41 929883 1-80 790716 4-71 209284 18 43 720754 3-40 929765 1-30 790999 4-71 209001 17 44 720968 8-40 929677 1-30 791281 4-71 208719 16 46 721162 3-40 929699 1-80 791563 4-70 208437 16 46 721366 3-40 929621 1-30 791846 4-70 208154 14 47 721670 3-40 929442 1-30 792128 4-70 207872 18 48 721774 3-39 929364 1-31 792410 4-70 207590 12 49 721978 3-39 929286 1-81 792692 4-70 207308 11 50 722181 3-39 929207 1-31 792974 4-70 207026 10 51 9-722385 8-89 9-929129 1-31 9-793256 4-70 10-206744 9 52 722688 3-89 929060 1-31 793538 4-69 206462 8 63 722791 3-38 928972 1-31 793819 4-69 206181 7 64 722994 3-38 928893 1-31 794101 4-69 205899 6 55 723197 3-88 928815 1-81 794383 4-69 206617 5 66 728400 8-38 928736 1-31 794664 >4-69 205336 4 57 723603 8-37 928657 1-31 794945 4-69 206065 3 58 723805 3-37 928578 1-31 795227 4-69 204773 2 59 724007 3-87 928499 1-31 796508 4-68 204492 1 60 724210 3-37 928420 1-31 795789 4-68 204211 Cosine. D. Sine. D. Cotang. D. Tang. M. (58 DEGI tEES.) 50 (32 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-724210 8 87 9-928420 1-82 9-795789 4-68 10-204211 60 1 724412 8 37 928342 1-32 796070 4-68 203930 59 2 724614 3 36 928263 1-32 796851 4-68 203649 58 3 724816 8 86 928183 1-32 796632 4-68 203368 57 4 725017 3 86 928104 1-32 796913 4-68 203087 56 5 725219 3 36 928025 1-32 797194 4-68 202806 55 6 725420 3 35 927946 1-32 797475 4-68 202525 54 7 725622 8 85 927867 1-32 797755 4-68 202245 53 8 725823 3 85 927787 1-32 798036 4-67 201964 52 9 726024 3 85 927708 1-32 798316 4-67 201684 51 10 726225 3 35 927629 1-32 798596 4-67 201404 50 11 9-726426 3 34 9-927549 1-32 9-798877 4-67 10-201123 49 12 726626 3 34 927470 1-33 799157 4-67 200843 48 13 726827 8 84 927390 1-83 799437 4-67 200568 47 14 727027 3 34 927310 1-83 799717 4-67 200283 46 15 727228 3 34 927231 1-88 799997 4-66 200008 45 16 727428 3 33 927151 1-83 800277 4-66 199723 44 17 727628 8 33 927071 1-83 800557 4-66 199443 43 18 727828 8 88 926991 1-33 800886 4-66 199164 42 19 728027 3 33 926911 1-33 801116 4-66 198884 41 20 728227 3 33 926831 1-88 801396 4-66 198604 40 21 9-728427 3 32 9-926751 1-33 9-801675 4-66 10-198325 39 22 728626 3 32 026671 1-33 801955 4-66 198045 38 23 728825 3 32 926591 1-33 802234 4-65 197766 37 24 729024 3 32 926511 1-34 802513 4-65 197487 36 25 729223 3 31 926431 1-34 802792 4-65 197208 35 26 729422 8 31 926351 1-84 803072 4-65 196928 34 27 729621 3 31 926270 1-34 803351 4-65 196649 83 28 729820 3 81 926190 1-34 803630 4-65 196370 32 29 730018 3 30 926110 1-84 803908 4-65 196092 31 80 730216 3 80 926029 1-34 804187 4-65 195813 80 31 9-730415 3 30 9-925949 1-34 9-804466 4-64 10-195534 29 32 730613 8 30 925868 1-34 804745 4-64 195255 28 33 730811 3 30 925788 1-34 805023 4-64 194977 27 34 731009 8 29 925707 1-34 805802 4-64 194698 26 35 731206 8 29 925626 1-34 805580 4-64 194420 25 36 781404 3 29 925545 1-35 805859 4-64 194141 24 37 731602 3 29 925465 1-35 806137 4-64 193863 23 38 731799 3 29 925384 1-35 806415 4-63 193585 22 39 731996 3 28 925303 1-35 806693 4-63 193307 21 40 732193 8 28 925222 1-35 806971 4-63 193029 20 41 9-732390 3 28 9-925141 1-35 9-807249 4-68 10-192751 19 42 732587 3 28 925060 1-35 807527 4-63 192473 18 43 732784 3 28 924979 1-35 807805 1 4-68 192195 17- 44 732980 3 27 924897 1-85 808088^ 4-68 191917 16 45 733177 8 27 924816 1-35 808361 4-68 191639 15 46 733373 3 27 924735 1-36 808638 4-62 191362 14 47 733569 3 27 924654 1-86 808916 4-62 191084 13 48 788765 3 27 924572 1-36 809193 4-62 190807 12 49 733961 3 26 924491 1-36 809471 4-62 190529 11 50 784157 3 26 924409 1-36 809748 4-62 190252 10 51 9-734353 3 26 ,9'924328 1-36 9-810025 4-62 10-189975 9 52 734549 8 26 924246 1-36 810302 4-62 18d698 8 53 734744 3 25 924164 1-36 810580 4-62 189420 7 54 734939 8 25 924083 1-36 810857 4-62 189143 6 1 1 55 735135 3 25 924001 1-36 811134 4-61 188866 6 : : 56 735330 8 25 928919 1-86 811410 4-61 188590 4 1 : 57 735525 3 25 923837 1-36 811687 4-61 188313 3 1 ; 58 735719 8 24 923755 1-37 811964 4-61 188036 2 ; 59 735914 3 24 923673 1-87 812241 4-61 187759 1 i 60 736109 8-24 923591 1-87 812517 4-61 187483 Cosine. D. Sine. D. Cotang. D. Tang. IH. (57 DEGREES.) SINES AND TANGENTS. (33 DEGREES.) 51 M. Sine. D. Cosine. D. Tang. D. Cotang. 9-786109 8-24 9-928691 1-37 9-812517 4-61 10-187482 60 1 786303 3-24 923509 1-87 812794 4-61 187206 59 - 2 736498 3-24 928427 1-87 818070 4-61 186930 58 3 786692 3-23 923845 1-37 818347 4-60 186658 57 i 736886 8-23 923268 1-37 813628 4-60 186377 66 6 787080 3-23 928181 1-37 813899 4-60 186101 55 6 787.274 8-23 928098 1-87 814175 4-60 185825 54 7 787467 8-23 923016 1-37 814452 4-60 185548 63 8 787661 3-22 922933 1-37 814728 4-60 185272 52 9 787855 8-22 922851 1-37 816004 4-60 184996 51 10 788048 8-22 922768 1-38 815279 4-60 184721 50 H 9-788241 3-22 9-922686 1-88 9-815555 4-59 10-184445 49 12 738434 8-22 922603 1-38 815881 4-59 184169 48 13 788627 8-21 922520 1-38 816107 4-59 188893 47 14 738820 3-21 922488 1-88 816382 4-59 183618 46 15 739013 3-21 922355 1-38 816658 4-59 183342 45 16 739206 3-21 922272 1-38 816938 4-59 183067 44 17 739398 8-21 922189 1-38 817209 4-59 182791 48 18 739590 3-20 922106 1-38 817484 4-59 182516 42 19 739783 3-20 922023 1-38 817759 4-59 182241 41 20 739975 3-20 921940 1-38 818085 4-58 181965 40 21 9-740167 8-20 9-921857 1-39 9-818810 4-58 10-181690 39 22 740359 8-20 921774 1-39 818585 4-68 181415 38 28 740550 819 921691 1-39 818860 4-58 181140 87 24 740742 3-19 921607 1-39 819185 4-68 180865 38 25 740934 3-19 921524 1-39 819410 4-58 180590 35 26 741125 8-19 921441 1-89 819684 4-58 180316 34 27 741316 3-19 921857 1-39 819959 4-68 180041 33 28 741508 3-18 921274 1-39 820234 4-58 179766 32 29 741699 8-18 921190 1-39 820508 4-57 179492 81 30 741889 3-18 921107 1-89 820783 4-57 179217 80 31 9-742080 8-18 9-921023 1-39 9-821057 4-57 10-178948 29 32 742271 3-18 920939 1-40 821332 4-57 178668 28 33 742462 3-17 920856 1-40 821606 4-57 178894 27 34 742662 8-17 920772 1-40 821880 4-57 178120 26 85 -742842 8-17 920688 1-40 822154 4-57 177846 25 36 743033 3-17 920604 1'40 822429 4-57 177571 24 37 743223 3-17 920520 1-40 822703 4-57 177297 23 38 743413 8-16 920436 1-40 822977 4-56 177023 22 39 748602 3-16 920352 1-40 823250 4-56 176750 21 40 743792 8-16 920268 1-40 823524 4-56 176476 20 41 9-743982 8-16 9-920184 1-40 9-823798 4-56 10-176202 19 42 744171 8-16 920099 1-40 824072 4-56 176928 18 43 744361 8-15 920015 1-40 824845 4-56 175655 17 44 744660 3-15 919931 1-41 824619 4-56 175381 16 45 744739 8-15 919846 1-41 824898 4-56 175107 16 46 744928 3-15 919762 1-41 825166 4-56 174884 14 47 745117 8-15 919677 1-41 826489 4-55 174661 13 48 745306 3-14 919593 1-41 825718 4-65 174287 12 49 745494 3-14 919508 1-41 826986 4-55 174014 11 50 745683 3-14 919424 1-41 826259 4-65 173741 10 51 9-745871 314 9-919339 1-41 9-826682 4-55 10-173468 9 52 746059 3-14 919264 1-41 826805 4-65 173195 8 53 746248 8-18 919169 1-41 827078 4-55 172922 7 54 746436 3-13 919085 1-41 827351 4-55 172649 6 66 746624 3-18 919000 1-41 827624 4-55 172376 5 66 746812 3-13 918916 1-42 827897 4-54 172103 4 57 746999 3-13 .918830' 1-42 828170 4-54 171880 3 58 747187 3-12 918745 1-42 828442 4-64 171568 2 59 747874 8-12 91866ft 1-42 828716 4-54 171286 1 60 747562 3-12 918574 1-42 828987 4-54 171018 Cosine. D. Sine. D. Cotang. D. Tang. M. (56 DEGB EES.) 52 (34 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-747562 3-12 9-918574 1-42 9-828987 4-54 10-171013 60 1 747749 3-12 918489 1-42 829260 4-54 170740 59 2 747936 3-12 918404 1-42 829532 4-54 170468 58 8 748123 3-11 918318 1-42 829805 4-54 170195 57 i 748810 3-11 918233 1-42 830077 4-54 169923 56 6 748497 8-11 918147 1-42 830349 4-53 169651 55 6 748683 3-11 918062 1-42 830621 4-53 169379 54 7 748870 3-11 917976 1-43 830893 4-53 169107 53 8 749056 3-10 917891 1-43 831165 4-53 168835 52 9 749243 3-10 917805 1-43 831437 4-53 lt>8563 51 10 749429 310 917719 1-43 831709 4-53 168291 50 11 9-749615 3-10 9-917634 1-43 9-831981 4-53 10-168019 49 12 749801 3-10 917548 1-43 832253 4-53 167747 48 13 749987 3-09 917462 1-43 832525 4-53 167475 47 14 750172 3-09 917376 1-43 832796 4-53 167204 46 15 750358 3-09 917290 1-43 833068 4-62 166932 45 16 750543 3-09 917204 1-43 833339 4-52 166661 44 1 17 750729 8-09 917118 1-44 833611 4-52 166389 43 18 750914 3-08 917032 1-44 833882 4-52 166118 42 19 751099 3-08 916946 1-44 834164 4-52 165846 41 20 7.51284 8-08 916859 1-44 834425 4-52 165575 40 21 9-761469 3-08 9-916773 1-44 9-834696 4-52 10-165304 39 22 751654 3-08 916687 1-44 834967 4-52 165033 88 23 761889 8-08 916600 1-44 835238. 4-52 164762 37 24 752023 3-07 916514 1-44 835509 4-52 164491 gB 25 752208 307 916427 1-44 835780 4-51 164220 35 1 26 752392 3-07 916341 1-44 836051 4-51 163949 84 27 752576 3-07 916254 1-44 836322 4-51 163678 33 28 752760 3-07 916167 1-45 836593 4-51 163407 82 29 762944 3-06 916081 1-45 836864 4-51 163136 31 80 753128 3-06 915994 1-45 837134 4-51 162866 30 81 9-753812 3-06 9-915907 1-45 9-837405 4-51 10-162595 29 32 753495 8-06 915820 1-45 837675 4-51 162325 28 83 753679 3-06 915733 1-4S 837946 4-51 162054 27 84 753862 3-05 915646 1-45 838216 4-51 161784 26 35 754046 8-05 915559 1-45 838487 4-50 161513 25 36 754229 8-05 915472 1-45 838757 4-50 161243 24 37 754412 8-05 915385 1-45 839027 4-50 160973 23 88 754595 3-05 915297 1-45 839297 4-50 160703 22 89 754778 3-04 915210 1-45 839568 4-50 160432 21 40 754960 3-04 916123 1-46 839838 4-50 160162 20 1 41 9-755143 3-04 9-915035 1-46 9-840108 4-50 10-159892 19 i 42 755326 8-04 914948 1-46 840378 4-50 169622 18 48 755506 8-04 914860 1-46 840647 4-50 159353 17 ; 44 755690 8-04 914773 1-46 840917 4-49 159083 16 1 45 .755872 8-03 914685 1-46 841187 4-49 158813 15 46 756054 8-03 914598 1-46 841457 4-49 158548 14 • 47 756236 3-03 914510 1-46 841726 4-49 158274 13 48 756418 3-03 914422 1-46 841996 4-49 158004 12 49 756600 8-03 914334 1-46 842266 4-49 157734 11 60 756782 3-02 914246 1-47 842585 4-49 157465 10 61 9-756963 3-02 9-914158 1-47 9-842805 4-49 10-157195 9 52 757144 3-02 914070 1-47 843074 4-49 15^926 8 58 757326 3-02 913982 1-47 843343 4-49 156657 7 54 757507 3-02 913894 1-47 843612 4-49 156388 ■ 6 66 757688 8-01 913806 1-47 843882 4-48 156118 5 56 757869 3-01 913718 1-47 844151 4-48 155849 4 57 758050 3-01 918630 1-47 844420 4-48 155580 3 68 758230 8-01 913541 1-47 844689 4-48 155311 2 69 758411 3-01 913453 1-47 844958 4-48 155042 1 60 758591 8-01 918365 1-47 845227 4-48 154773 Cosine. D. Sine. D. Cotang. D. Tang. m. (55 DEGREES.) SINES AND TANGENTS. (35 DEGREES.) 53 M. Sine. D. Cosine. D. Tang. D. Cotang. 9-758591 8-01 9-918366 1-47 9-846227 4-48 10-154778 60 1 758772 3-00 913276 1-47 845496 4-48 154504 59 2 768952 8-00 913187 1-48 845764 4-48 154236 58 8 759132 3-00 918099 1-48 846038 4-48 153967 57 4 759812 8-00 913010 1-48 846302 4-48 163698 56 6 759492 8-00 912922 1-48 846570 4-47 153430 56 6 759672 2-99 912833 1-48 846839 4-47 168161 64 7 759852 2-99 912744 1-48 847107 4-47 152893 58 8 760081 2-99 912655 1-48 847376 4-47 152624 62 9 760211 2-99 912666 1-48 847644 4-47 152356 51 10 760890 2-99 912477 1-48 847918 4-47 152087 50 11 9-760569 2-98 9-912888 1-48 9-848181 4-47 10-151810 49 12 760748 2-98 912299 1-49 848449 4-47 151551 48 13 760927 2-98- 912210 1-49 848717 4-47 151283 47 14 761106 2-98 912121 1-49 848986 4-47 151014 46 IB 761285 2-98 912031 1-49 849254 4-47 150746 45 16 761464 2-98 911942 1-49 849522 4-47 150478 44 17 761642 2-97 911858 1-49 849790 4-46 150210 43 18 761821 2-97 911768 1-49 860068 4-46 149942 42 19 761999 2-97 911674 1-49 850325 4-46 149675 41 20 762177 2-97 911584 1-49 850598 4-46 149407 40 21 9-762356 2-97 9-911495 1-49 9-850861 4-46 10-149189 89 22 762584 2-96 911405 1-49 851129 4-46 148871 38 23 762712 2-96 911815 1-50 851896 4-46 148604 37 24 762889 2-96 911226 1-50 851664 4-46 148836 36 25 763067 2-96 911186 1-50 851981 4-46 148069 35 26 768245 2-96 911046 1-50 852199 4-46 147801 34 27 763422 2-96 910956 1-50 852466 4-46 147534 38 28 768600 2-95 910866 1-50 852738 4-45 147267 32 29 763777 2-95 910776 1-50 853001 4-45 146999 31 30 768954 2-95 910686 1-50 863268 4-45 146732 80 81 9-764181 2-95 9-910596 1-60 9-868586 4-45 10-146465 29 32 764308 2-95 910506 1-50 853802 4-45 146198 28 83 764486 2-94 910415 1-50 854069 4-45 146931 27 84 764662 2-94 910325 1-51 854386 4-45 145664 26 86 764888 2-94 910235 1-51 854603 4-45 145397 25 36 765015 2-94 910144 1-51 864870 4-45 146130 24 87 765191 2-94 910054 1-51 855137 4-45 144868 23 38 765367 2-94 909963 1-51 855404 4-45 144596 22 39 765544 2-93 909873 1-51 855671 4-44 144329 21 40 765720 2-98 909782 1-61 866938 4-44 144062 20 41 9-765896 2-98 9-909691 1-51 9-856204 4-44 10-143796 19 42 766072 2-98 909601 1-61 856471 4-44 143529 18 48 766247 2-93 909510 1-51 856787 4-44 143263 17 44 766428 2-93 909419 1-51 867004 4-44 142996 16 45 766598 2-92 909328 1-52 857270 4-44 142730 15 46 766774 2-92 909237 1-52 857537 4-44 142463 14 47 766949 2-92 909146 1-52 857803 4-44 142197 13 48 767124 2-92 909055 1-52 858069 4-44 141931 12 49 767800 2-92 908964 1-52 858336 4-44 141664 11 50 767475 2-91 908878 1-52 858602 4-48 141398 10 51 9-767649 2-91 9-908781 1-52 9-858868 4-43 10-141132 9 52 767824 767999 2-91 908690 1-52 859184 4-48 140866 8 53 2-91 908599 1-52 859400 4-48 140600 7 54 768178 2-91 908507 1-52 859666 4-43 140834 6 65 768348 76852fe 2-90 908416 1-58 859932 4-43 140068 5 56 2-90 908824 1-58 860198 4-43 189802 4 67 768697 2-90 908288 1-53 860464 4-43 189536 3 68 768871 2-90 908141 1-53 860780 4-43 139270 2 69 769045 2-90 908049 1-58 860995 4-43 139005 1 60 769219 2-90 907958 1-58 861261 4-48 138739 Cosine. D. Sine. D. Cotang. D. Tang. m. (54 DEGREES.) 64 (86 DEGREES.) A TABLE OF LOUARITHMIC M. Sine. U. Cosine. D. Tang. D. Cotang. 9-769219 2-90 9-907958 1-53 9-861261 4-43 10-188739 60 1 769398 2-89 907866 1-58 861527 4-43 188473 69 2 769566 2-89 907774 1-53 861792 4-42 138208 58 3 769740 2-89 907682 1-53 862058 4-42 137942 57 4 769918 2-89 907590 1-53 862323 4-42 187677 66 5 770087 2-89 907498 1-53 862589 4-42 137411 55 6 770260 2-88 907406 1-53 862854 4-42 187146 54 7 770433 2-88 907314 1-54 863119 4-42 136881 53 8 770606 2-88 907222 1-54 863385 4-42 136615 52 9 770779 2-88 907129 1-54 863650 4-42 136350 61 10 770952 2-88 907037 1-54 863915 4-42 136085 50 11 9-771125 2-88 9-906945 1-54 9-864180 4-42 10 135820 49 12 771298 2-87 906852 1-54 864445 4-42 135665 48 13 771470 2-87 906760 1-54 864710 4-42 135290 47 14 771643 2-87 906667 1-54 864975 4-41 185025 46 15 771815 2-87 906575 1-54 865240 4-41 184760 46 16 771987 2-87 906482 1-54 865505 4-41 134495 44 17 772159 2-87 906389 1-55 865770 4-41 134230 43 18 772331 2-80 906296 1-55 866035 4-41 138965 42 19 772503 2-86 906204 1-55 866300 4-41 133700 41 20 772675 2-86 906111 1-55 866564 4-41 133436 40 21 9-772847 2-86 9-906018 1-55 9-866829 4-41 10-133171 39 22 773018 2-86 905926 1-55 867094 4-41 132906 38 23 773190 2-86 905832 1-55 867358 4-41 132642 37 24 773361 2-85 905739 1-55 867623 4-41 132377 36 25 773533 2-83 905645 1-55 867887 4-41 132113 35 26 773704 2-85 905552 1-55 868152 4-40 131848 34 27 773875 2-85 905459 1-55 868416 4-40 131584 33 28 774046 2-85 905366 1-56 868680 4-40 131320 32 29 774217 2-85 905272 1-56 868945 4-40 181066 31 30 774388 2-84 905179 1-56 869209 4-40 180794 30 31 9-774558 2-84 9-905085 1-56 9-869473 4-40 10-130527 29 32 774729 2-84 904992 1-56 869737 4-40 130263 28 33 774899 2-84 904898 1-56 870001 4-40 129999 27 34 775070 2-84 904804 1-56 870265 4-40 129735 26 35 775240 2-84 904711 1-56 870529 4-40 129471 25 36 775410 2-83 P04617 1-56 870793 4-40 129207 24 37 775580 2-83 904523 1-56 871057 4-40 128943 23 33 775750 2-83 904429 1-57 871821 4-40 128679 22 39 775920 2-83 904335 1-57 871585 4-40 128415 21 40 776090 2-83 904241 1-57 871849 4-39 128151 20 41 9-776259 2-83 9-904147 1-57 9-872112 4-39 10-127888 19 42 776429 2-82 904058 1-57 872376 4-39 127624 18 43 776598 2-82 908959 1-57 872640 4-39 127360 17 44 776768 2-82 903864 1-57 872903 4-39 127097 16 45 776937 2-82 903770 1-57 873167 4-39 126883 15 46 777106 2-82 903676 1-57 873430 4-39 126570 14 47 777275 2-81 903581 1-57 873694 4-39 126306 13 48 777444 2-81 903487 1-57 873957 4-39 126043 1? 49 777613 2-81 903392 1-58 874220 4-39 125780 11 BO 777781 2-81 903298 1-58 874484 4-39 125516 10 El 9-777950 2-81 0-903203 1-58 9-874747 4-39 10-125253 9 52 778119 2-81 903108 1-58 875010 4-39 124990 8 53 778287 2-80 903014 1-58 875273 4-38 124727 7 54 778455 2-80 902919 1-58 875536 4-38 124464 6 55 778624 2-80 902824 1-58 875800 4-38 124200 5 56 778792 2-80 902729 1-58 876063 4-38 123937 4 57 778960 2-80 902634 1-58 876326 4-38 123674 3 SB 779128 2-80 902689 1-59 876589 4-38 123411 2 59 779295 2-79 902444 1-59 876851 4-38 123149 1 60 779468 2-79 902349 1-59 877114 4-38 122886 Cosine. D. Sine. D. Cotang. D. Tang. M. (53 DEGREES.) SINES AKD TANGENTS. (37 DEGREES.) 65 jh. Sine. D. Cosine. D. Tang. D. Cotang. 60 9-779468 2 79 9-902349 1-59 9-877114 4-38 10-122886 1 779631 2 79 902258 1-59 877377 4-88 122623 69 2 779798 2 79 902168 1-59 877640 4-38 122360 68 8 779966 2 79 902063 1-59 877903 4-38 122097 57 4 780133 2 79 901967 1-69 878165 4-88 121885 56 5 780300 2 78 901872 1-69 878428 4-38 121572 55 6 780467 2 78 901776 1-59 878691 4-38 121309 54 7 780634 2 78 901681 1-69 878953 4-87 121047 63 8 780801 2 78 901686 1-59 879216 4-37 120784 52 g 780968 2 78 901490 1-59 879478 4-87 120522 51 10 781134 2 78 901394 1-60 879741 4-37 120259 50 n 9- 781301 2 77 9-901298 1-60 9 '880003 4-87 10-119997 49 12 781468 2 77 901202 1-60 880265 4-37 119735 48 13 781684 2 77 901106 1-60 880528 4-37 119472 47 14 781800 2 77 901010 1-60 880790 4-87 119210 46 15 781966 2 77 900914 1-60 881052 4-87 118948 45 16 782132 2 77 900818 1-60 881314 4-37 118686 44 17 782298 2 76 900722 1-60 881576 4-37 118424 43 18 782464 2 76 900626 1-60 881839 4-87 118161 42 19 782630 2 76 900529 1-60 882101 4-37 117899 41 20 782796 2 76 900488 1-61 882363 4-36 117637 40 21 9-782961 2 76 9-900387 1-61 9 '882625 4-36 10-117875 39 22 783127 2 76 900240 1-61 882887 4-36 117113 88 28 783292 2 75 900144 1-61 883148 4-36 116862 37 24 783458 2 75 900047 1-61 883410 4-36 116590 86 25 788623 2 75 899951 1-61 888672 4-36 116828 35 26 783788 2 75 899854 1-61 883934 4-86 116066 34 27 783958 2 75 899757 1-61 884196 4-86 116804 33 28 784118 2 75 899660 1-61 884457 4-86 115543 32 29 784282 2 74 899564 1-61 884719 4-36 116281 31 80 784447 2 74 899467 1-62 884980 4-86 115020 30 31 9-784612 2 74 9-899870 1-62 9-885242 4-36 10-114758 29 32 784776 2 74 899273 1-62 885503 4-86 114497 28 33 784941 2 74 899176 1-62 885765 4-36 114235 27 84 785105 2 74 899078 1-62 886026 4-36 113974 26 35 786269 2 73 898981 1-62 886288 4-36 118712 26 86 785483 2 73 898884 1-62 886549 4-35 113451 24 87 785597 2 73 898787 1-62 886810 4-35 118190 23 38 7857S1 2 73 898689 1-62 887072 4-35 112928 22 39 785925 2 78 898592 1-62 887333 4-35 112667 21 40 786089 2 73 898494 1-68 887594 4-35 112406 20 41 9-786252 2 72 9-898397 1-63 9-887855 4-35 10-112145 19 42 786416 2 72 898299 1-63 888116 4-35 111884 18 48 786679 2 72 898202 1-68 888877 4-85 111623 17 44 786742 2 72 898104 1-63 888639 4-36 111361 16 45 786906 2 72 898006 1-63 888900 4-35 111100 16 46 787069 2 72 897908 1-63 889160 4-85 110840 14 47 787282 2 71 897810 1-63 889421 4-85 110679 13 48 787395 2 71 897712 1-63 889682 4-85 110818 12 49 787657 2 71 897614 1-63 889943 4-86 110067 11 50 787720 2 71 897516 1-68 890204 4-34 109796 10 51 9-787888 2 71 9-897418 1-64 9-890466 4-34 10-109535 9 52 788046 2 71 897320 1-64 890726 4-84 109275 8 68 788208 2 71 897222 1-64 890986 4-34 109014 7 54 788870 2 70 897123 1-64 891247 4-34 . 108763 6 65 788532 2 70 897025 1-64 891507 4-34 108493 5 66 788694 2 70 896926 1-64 891768 4-34 108232 4 67 788856 2 70 , 896828 1-64 892028 4-34 107972 3 68 789018 2 70 896729 1-64 892289 4-34 107711 2 69 789180 2 70 896681 1-64 892549 4-84 107461 1 60 789842 2-69 896632 1-64 892810 4-34 107190 Cosine. D. Sine. D. Cotang. D. Tang. jn. '52 DEGREES. 1 56 (38 DEGEEES.) A TABLE OF LOGARITHMIC IH. Sine. D. Cosine. D. Tan?. D. Cotang. 9-789342 2-69 9-896532 1-64 9-892810 4-34 10-107190 60 ' 1 789504 2-69 896433 1-65 893070 4-34 106930 59 I 2 789665 2-69 896335 1-65 893831 4-34 106669 58 8 789827 2-69 896236 1-65 893591 4-34 106409 57 4 789988 2-69 896187 1-65 898851 4-34 106149 56 5 790149 2-69 896038 1-65 894111 4-84 105889 55 6 790310 2-68 895939 1-65 894371 4-34 105629 54 7 790471 2-68 895840 1-65 894632 4-33 105368 53 8 790632 2-68 895741 1-65 894892 4-33 105108 52 9 790793 2-68 895641 1-65 895152 4-33 104848 51 10 790954 2-68 895542 1-65 895412 4-33 104588 50 11 9 '791115 2-68 9-895443 1-66 9-895672 4-33 10-104328 49 12 791275 2-67 895343 1-66 895982 4-33 104068 48 13 791436 2-67 895244 1-66 896192 4-33 108808 47 14 791596 2-67 895145 1-66 896452 4-38 108548 46 15 791757 2-67 895045 1-66 896712 4-33 103288 45 16 791917 2-67 894945 1-66 896971 4-33 103029 44 17 792077 2-67 894846 1-66 897231 4-33 102769 43 18 792237 2-66 894746 1-66 897491 4-33 102509 42 19 792397 2-66 894646 1-66 897751 4-33 102249 41 20 792557 2-66 894546 1-66 898010 4-33 191990 40 21 9-792716 2-66 9-894446 1-67 9-898270 4-33 10-101780 39 22 792876 2-66 894346 1-67 898530 4-33 101470 88 28 793035 2-66 894246 1-67 898789 4-33 101211 37 24 793195 2-65 894146 1-67 899049 4-32 100951 86 25 798354 2-65 894046 1-67 899308 4-32 100692 35 26 793514 2-65 893946 1-67 899568 4-32 100432 34 27 793673 2-65 893846 1-67 899827 4-82 100173 33 28 793832 2-65 893745 1-67 900086 4-32 099914 82 29 r98991 2-65 893645 1-67 900346 4-32 099654 31 SO 794150 2-64 893544 1-67 900605 4-32 099395 30 81 9-794308 2-64 9-893444 1-68 9-900864 4-32 10-099136 29 32 794467 2-64 893343 1-68 901124 4-82 098876 28 88 794626 2-64 898243 1-68 901383 4-82 098617 27 34 794784 2-64 893142 1-68 901642 4-32 098358 26 85 794942 2-64 898041 1-68 901901 4-82 098099 25 86 795101 2-64 892940 1-68 902160 4-82 097840 24 87 795259 2-63 892889 1-68 902419 4-32 097581 28 88 795417 2-68 892789 1-68 902679 4-32 097821 22 89 795575 2-63 892638 1-68 902938 4-82 097062 21 40 795783 2-63 892536 1-68 908197 4-31 096803 20 41 «^-795891 2-63 9-892435 1-69 9-903455 4-31 10-096545 19 42 796049 2-63 892334 1-69 908714 4-31 096286 18 48 796206 2-63 892283 1-69 903973 4-81 096027 17 44 796364 2-62 892132 1-69 904232 4-31 095768 16 45 796521 2-62 892030 1-69 904491 4-31 095509 15 46 796679 2-62 891929 1-69 904750 4-31 095250 14 47 796836 2-62 891827 1-69 905008 4-31 094992 18 48 796993 2-62 891726 1-69 905267 4-31 094788 12 49 797150 2-61 891624 1-69 905526 4-31 094474 11 50 797307 2-61 891523 1-70 905784 4-31 094216 10 61 9-797464 2-61 9-891421 1-70 9-906043 4-31 10-093957 9 52 797621 2-61 891319 1-70 906302 4-31 . 098698 8 53 797777 2-61 891217 1-70 906560 4-31 093440 7 54 797984 2-61 891115 1-70 906819 ^4-31 093181 6 55 798091 2-61 891013 1-70 907077 4-31 092928 5 56 798247 2-61 890911 1-70 907336 4-31 092684 4 57 798403 2-60 890809 1-70 907594 4-31 092406 3 58 798560 2-60 890707 1-70 907852 4-31 092148 2 59 798716 2-60 890605 1-70 908111 4-30 091889 1 60 798872 2-60 890503 1-70 908869 4-80 091631 Cosine. D. Sine. D, Cotang. D. Tang. M. (51 DEGREES.' SINES AND TANGENTS. (39 DEaREES.) 57 IH. Sine. D. Cosine. D. Tang. D. Cotang. 9-798872 2-60 9-890503 1-70 9-908369 4-30 10091631 60 1 799028 2-60 890400 1-71 908628 4-80 091372 59 2 799184 2-60 890298 1-71 908886 4-30 091114 58 3 799389 2-59 890195 1-71 909144 4-30 090856 57 4 799495 2-59 890093 1-71 909402 4-30 090598 56 5 799651 2-59 889990 1-71 909660 4-30 090340 55 6 799806 2-59 889888 1-71 909918 4-30 090082 54 7 799962 2-59 889785 1-71 910177 4-30 089823 53 8 800117 2-59 889682 1-71 910435 4-30 089565 52 9 800272 2-58 889579 1-71 910693 4-30 089307 51 10 800427 2-58 889477 1-71 910951 4-30 089049 50 11 9-800582 2-58 9-889374 1-72 9-911209 4-30 10-088791 49 12 800737 2-58 889271 1-72 911467 4-30 088533 48 13 800892 2-58 889168 1-72 911724 4-80 088276 47 14 801047 2-58 889064 1-72 911982 4-30 088018 46 15 801201 2-58 888961 1-72 912240 4-30 087760 45 16 801356 2-57 888858 1-72 912498 4-30 087502 44 17 801511 2-57 888755 1-72 912756 4-30 087244 43 18 801665 2-57 888651 1-72 913014 4-29 086986 42 19 801819 2-57 888548 1-72 913271 4-29 086729 41 20 801973 2-57 888444 1-73 913529 4-29 086471 40 21 9-802128 ' 2-57 9-888341 1-73 9-913787 4-29 10-086213 39 22 802282 2-56 888237 1-73 914044 4-29 085956 38 23 802436 2-56 888134 1-73 914302 4-29 085698 37 24 802589 2-50 888030 1-73 914560 4-29 085440 36 25 802743 2-56 887926 1-73 914817 4-29 085183 35 26 802897 2-56 887822 1-73 915075 4-29 084925 84 27 803050 2-56 887718 1-73 915332 4-29 084668 33 28 803204 2-56 887614 1-73 915590 4-29 084410 32 29 803357 2-55 887510 1-73 915847 4-29 084153 31 30 803511 2-55 887406 1-74 916104 4-29 083896 30 31 9-803664 2-55 9-887302 1-74 9-916362 4-29 10-083638 29 32 803817 2-55 887198 1-74 916619 4-29 083881 28 33 803970 2-55 887093 1-74 916877 4-29 083123 27 34 804123 2-55 886989 1-74 917134 4-29 082866 26 35 804276 2-54 886885 1-74 917391 4-29 082609 25 36 804428 2-54 886780 1-74 917648 4-29 082352 24 37 804581 2-54 886676 1-74 917905 4-29 082095 23 38 804734 2-54 886571 1-74 918163 4-28 081837 22 39 804886 2-54 886466 1-74 918420 4-28 081580 21 40 805039 2-54 886362 1-75 91867T 4-28 081323 20 41 9-805191 2-54 9-886257 1-75 9-918934 4-28 10-081066 19 42 805343 2-53 886152 1-75 919191 4-28 080809 18 43 805495 2-53 886047 1-75 919448 4-28 080552 17 44 805647 2-53 885942 1-75 919705 4-28 080295 16 45 805799 2-53 885837 1-75 919962 4-28 080038 15 46 805951 2-53 885732 1-75 920219 4-28 079781 14 i 47 806103 2-53 885627 1-75 920476 4-28 079524 13 , 48 806254 2-53 885522 1-75 920733 4-28 079267 12 49 806406 2-52 885416 1-75 920990 4-28 079010 11 50 806557 2-52 885311 1-76 921247 4-28 078753 10 51 9-806709 2-52 9-885205 1-76 9-921503 4-28 10-078497 9 52 806860 2-52 885100 1-76 921760 4-28 0782^0 8 53 807011 2-52 884994 1-76 92201T 4-28 077983 7 54 807163 2-52 884889 1-76 922274 4-28 077726 6 55 807314 2-52 884783 1-76 922530 4-28 077470 5 56 807465 2-51 884677 1-76 922787 4-28 077213 4 57 807615 2-51 884572 1-76 923044 4-28 076956 3 58 807766 2-51 884466 1-76 928300 4-28 076700 2 59 807917 2-51 884360 1-76 928557 4-27 076443 1 60 808067 2-51 884254 1-77 i 923813 4-27 076187 Cosine. D. Sine. D. 1 Cotang. D. Tang. JH. ^50 DEGI lEES.) 58 (40 DEGREES.) A TABLE OF LOGAKITHMIC M. Sine. D. Cosine. D. Tang. ! D. Cotang. 60 9-808067 2-51 9-884264 1-77 9-928813 4-27 10- 076187 1 808218 2-51 884148 1-77 924070 4-27 075930 59 2 808368 2-51 884042 1-77 924327 4-27 075673 58 8 808519 2-50 883936 1-77 924583 4-27 075417 57 i 808669 2-50 883829 1-77 924840 4-27 075160 56 B 808819 2-50 883723 1-77 925096 4-27 074904 55 6 808969 2-50 888617 1-77 925352 4-27 074648 54 7 809119 2-50 883510 1-77 925609 4-27 074391 58 8 809269 2-50 883404 1-77 925865 4-27 074135 52 9 809419 2-49 883297 1-78 926122 4-27 078878 51 10 809569 2-49 883191 1-78 926378 4-27 073622 50 11 9-809718 2-49 9-883084 1-78 9-926634 4-27 10- 078866 49 12 809868 2-49 882977 1-78 926890 4-27 073110 48 13 810017 2-49 882871 1-78 927147 4-27 072863 47 14 810167 2-49 882764 1-78 927403 4-27 072597 46 15 810316 2-48 882657 1-78 927659 4-27 072841 46 16 810465 2-48 882550 1-78 927915 4-27 072085 44 17 810614 2-48 882448 1-78 928171 4-27 071829 48 18 810763 2-48 882836 1-79 928427 4-27 071673 42 19 810912 2-48 882229 1-79 928683 4-27 071317 41 20 811061 2-48 882121 1-79 928940 4-27 071060 40 21 9-811210 2-48 9-882014 1-79 9-929196 4-27 10-070804 89 22 811358 2-47 881907 1-79 929462 4-27 070548 38 23 811507 2-47 881799 1-79 929708 4-27 070292 37 24 811665 2-47 881692 1-79 929964 4-26 070036 36 25 811804 2-47 881684 1-79 930220 4-26 069780 85 26 811952 2-47 ,881477 1-79 980475 4-26 069525 34 27 812100 2-47 881869 1-79 930731 4-26 069269 33 28 812248 2-47 881261 1-80 980987 4-26 069013 32 29 812896 2-46 881153 1-80 931243 4-26 068767 31 80 812544 2-46 881046 1-80 931499 4-26 068501 80 31 9-812692 2-46 9-880988 1-80 9-931755 4-26 10-068246 29 82 812840 2-46 880830 1-80 932010 4-26 067990 28 33 812988 2-46 880722 1-80 932266 4-26 067734 27 34 813185 2-46 880613 1-80 932522 4-26 067478 26 35 813283 2-46 880505 1-80 932778 4-26 067222 25 36 813430 2-45 880397 1-80 933083 4-26 066967 24 37 813578 2-45 880289 1-81 938289 4-26 066711 28 38 813725 2-45 880180 1-81 933545 4-26 066455 22 39 818872 2-45 880072 1-81 933800 4-26 066200 21 40 814019 2-45 879963 1-81 934056 4-26 065944 20 41 9-814166 2-45 9-879856 1-81 9-984811 4-26 10-065689 19 42 814313 2-45 879746 1-81 934567 4-26 065433 18 43 814460 2-44 879637 1-81 934823 4-26 066177 17 44 814607 2-44 879529 1-81 935078 4-26 064922 16 45 814768 2-44 879420 1-81 985338 4-26 064667 15 46 814900 2-44 879311 1-81 985589 4-26 064411 14 " 47 815046 2-44 879202 1-82 935844 4-26 064156 13 48 815193 2-44 879093 1-82 936100 4-26 063900 12 49 815339 2-44 878984 1-82 936855 4-26 068645 11 60 815485 2-43 878876 1-82 936610 4-26 063390 10 51 9-815631 2-48 9-878766 1-82 9-936866 4-25 10-063134 9 52 816778 2-43 878656 1-82 937121 4-25 062879 8 63 815924 2-43 878547 1-82 937376 4-25 062624 7 64 816069 2-43 878438 1-82 987632 4-26 062368 6 55 816215 2-43 878828 1-82 937887 4-25 062118 6 56 816361 2-43 878219 1-83 938142 4-25 061858 4 i 57 816507 2-42 878109 1-88 938398 4-25 061602 3 ' 68 816652 2-42 877999 1-83 988653 4-26 061347 2 69 816798 2-42 877890 1-83 938908 4-25 061092 1 60 816943 2-42 877780 1-88 989163 4-25 060837 Cosine. D. Sine. D. Cotang. D. Tang. m. (49 DEGI lEES.) SINES AND TANGENTS. (4l DEGREES.) 59 M. Sine. D. Cosine. D. Tang. D. Cotang. ld-06083f i 9-816948 i 2-42 9-877780 1-83 9-939163 4-25 60 i 1 817088 ! 2-42 877670 1-83 989418 4-25 060582 59 2 817238 1 2-42 877560 1-83 989673 4-25 060327 58 8 817379 1 2-42 877450 1-83 939928 4-25 060072 57 4 817524 i 2-41 877340 1-83 940183 4-25 059817 56 5 817668 j 2-41 877230 1-84 940438 4-26 059562 55 6 817813 ! 2-41 877120 1-84 940694 4-25 059306 54 7 817958 2-41 877010 1-84 940949 4-25 069061 58 8 818108 2-41 876899 1-84 941204 4-25 068796 52 9 818247 2-41 876789 1-84 941458 4-25 058542 51 10 818392 2-41 876678 1-84 941714 4-25 058286 50 11 9-818536 2-40 9-876568 1-84 9-941968 4-25 10-058032 49 12 818681 2-40 876457 1-84 942223 4-25 057777 48 13 818825 2-40 876347 1-84 942478 4-25 067522 47 14 818969 2-40 876286 ■ 1-85 942733 4-25 067267 46 15 819113 2-40 876125 1-8D 942988 4-25 057012 45 16 819257 2-40 876014 1-85 948243 4-25 056767 44 17 819401 2-40 875904 1-85 943498 4-26 056502 43 18 819545 2-39 875793 1-85 943752 4-25 056248 42 19 819689 2-39 875682 1-85 944007 4-25 065993 41 20 819832 2-89 876571 1-85 944262 4-25 055738 40 21 9-819976 2-39 9-875459 1-85 9-944517 4-26 10-055483 39 22 820120 2-39 875348 1-85 944771 4-24 055229 38 23 820263 2-39 875287 1-85 945026 4-24 054974 37 24 820406 2-39 875126 1-86 945281 4-24 054719 36 25 820550 2-38 875014 1-86 945535 4-24 054465 35 26 820698 2-88 874903 1-86 945790 4-24 054210 34 27 820836 2-38 874791 1-86 946045 4-24 053955 38 28 820979 2-88 874680 1-86 946299 4-24 053701 82 29 821122 2-38 874568 1-86 946554 4-24 053446 31 30 821266 2-38 874456 1-86 946808 4-24 063192 30 31 9-821407 2-38 9-874344 1-86 9-947063 4-24 10-052987 29 32 821550 2-88 874232 1-87 947818 4-24 052682 28 33 821693 2-37 874121 1-87 947572 4-24 062428 27 34 821835 2-37 874009 1-87 947826 4-24 052174 26 35 821977 2-37 878896 1-87 948081 4-24 051919 25 36 822120 2-37 873784 1-87 948336 4-24 061664 24 37 822262 2-87 873672 1-87 948590 4-24 051410 23 38 822404 2-37 873560 1-87 948844 4-24 051156 22 39 822546 2-37 873448 1-87 949099 4-24 060901 21 40 822688 2-36 873385 1-87 949353 4-24 050647 20 41 9-822830 2-36 9-873228 1-87 9-949607 4-24 10-050393 19 42 822972 2-36 873110 1-88 949862 4-24 060138 18 43 823114 2-36 872998 1-88 950116 4-24 049884 17 44 823256 2-36 872885 1-88 960370 4-24 049630 16 46 823397 2-36 872772 1-88 950625 4-24 049375 15 46 823539 2-86 872659 1-88 950879 4-24 049121 14 47 823680 2-85 872547 1-88 951183 4-24 048867 IS 48 823821 2-35 872484 1-88 951388 4-24 048612 12 49 823963 2-35 872321 1-88 951642 4-24 048358 11 60 824104 2-35 872208 1-88 951896 4-24 048104 10 51 9-824245 2-35 9-872095 1-89 9-952150 4-24 10-047860 9 52 824886 2-35 871981 1-89 952405 4-24 047595 8 63 824527 2-35 871868 1-89 952659 4-24 047341 7 64 824668 2-84 871755 1-89 952913 4-24 047087 6 55 824808 2-34 871641 1-89 953167 4-23 046833 5 56 824949 2-34 871528 1-89 953421 4-23 046579 4 57 825090 2-84 871414 1-89 953675 4-23 046325 3 68 825230 2-84 871801 1-89 958929 4-23 046071 2 69 825371 2-84 871187 1-89 954183 4-28 046817 1 60 825511 2-34 871073 1-90 954437 4-23 045563 JH. Cosine* D. 1 Sine. 1 D, Cotang. D. Tang. (48 DEGREES.) 60 (42f DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine. D. Tang. D. Cotang. 9-825511 2-34 9-871078 1-90 9-954437 4-23 10-045563 60 1 825651 2-33 870960 1-90 954691 4-23 046309 69 2 825791 2-33 870846 1-90 954945 4-28 046065 68 8 825931» 2-38 870782 1-90 955200 4-23 044800 67 4: 826071 2-33 870618 1-90 955454 4-23 044646 66 : B 826211 2-33 870504 1-90 955707 4-28 044298 66 1 6 826351 2-33 870390 1-90 955961 4-28 044039 64 7 826491 2-33 870276 1-90 956215 4-23 043785 53 ; 8 826631 2-33 870161 1-90 966469 4-23 048531 62 1 9 826770 2-32 870047 1-91 966723 4-28 043277 61 i 10 826910 2-82 869933 1-91 956977 4-23 048023 50 1 11 9-827049 2-32 9-869818 1-91 9-957281 4-28 10-042769 49 12 827189 2-32 869704 1-91 957485 4-28 042615 48 13 827828 2-32 869589 1-91 967739 4-28 042261 47 14 827467 2-32 869474 1-91 957993 4-23 042007 46 15 827606 2-32 869360 1-91 958246 4-23 041754 45 16 827745 2-32 869245 1-91 958500 4-23 041500 44 17 827884 2-31 869130 1-91 968754 4-23 041246 43 18 828023 2-31 869015 1-92 959008 4-23 040992 42 19 828162 2-81 868900 1-92 959262 4-23 040738 41 20 828301 2-81 868785 1-92 959516 4-23 040484 40 21 9-828439 2-81 9-868670 1-92 9-959769 4-23 10-040231 39 22 828578 2-31 868555 1-92 960023 4-23 089977 88 23 828716 2-81 868440 1-92 960277 4-23 039723 37 24 828855 2-30 868324 1-92 960531 4-23 039469 86 25 828993 2-30 868209 1-92 960784 4-23 039216 85 26 829131 2-30 868093 1-92 961038 4-28 038962 34 ' 27 829269 2-30 867978 1-93 961291 4-23 038709 33 28 829407 2-30 867862 1-93 961546 4-23 038455 32 29 829545 2-30 867747 1-93 961799 4-28 038201 31 30 829683 2-30 867681 1-93 962052 4-23 037948 30 31 9-829821 2-29 9-867515 1-93 9-962306 4-23 10-037694 29 32 829959 2-29 867899 1-93 962560 4-23 037440 28 33 830097 2-29 867283 1-93 962813 4-23 037187 27 34 830234 2-29 867167 1-98 968067 4-28 086933 26 35 830372 2-29 867051 1-93 968320 4-23 036680 25 36 830509 2-29 866935 1-94 963574 4-23 036426 24 37 830646 2-29 866819 1-94 963827 4-28 036173 23 ! 38 830784 2-29 866703 l'-94 964081 4-23 085919 22 39 830921 2-28 866586 1-94 964835 4-23 085665 21 ! 40 831058 2-28 866470 1-94 964588 4-22 036412 20 41 9-831195 2-28 9-866358 1-94 9-964842 4-22 10-086158 19 42 831332 2-28 866287 1-94 966095 4-22 084905 18 43 831469 2-28 866120 1-94 965349 4-22 034661 17 ; 44 831606- 2-28 866004 1-95 965602 4-22 034398 16 ' 46 831742 2-28 865887 1-95 966855 4-22 034145 15 46 831879 2-28 865770 1-95 966105 4-22 033891 14 47 832015 2-27 865653 1-95 966362 4-22 033638 13 48 832162 2-27 865536 1-95 966616 4-22 033384 12 49 832288 2-27 865419 1-95 966869 4-22 033131 11 , 50 832425 2-27 865302 1-95 967123 4-22 032877 10 . 51 9-832561 2-27 9-865185 1-95 9-967376 4-22 10-082624 9 ■ 52 832697 2-27 865068 1-95 967629 4-22 032371 8 < 53 832833 2-27 864950 1-95 967883 4-22 032117 7 1 54 832969 2-26 864833 1-96 968186 4-22 081864 6 1 55 833106 2-26 864716 1-96 968389 4-22 031611 5 ■ 56 833241 2-26 864598 1-96 968643 4-22 031357 4 57 833377 2-26 864481 1-96 968896 4-22 081104 3 58 883512 2-26 864363 1-96 969149 4-22 030851 2 59 833648 2-26 864245 1-96 969408 4-22 080597 1 60 838783 2-26 864127 1-96 969656 4-22 030344 Cosine. D. Sine. D. Cotang. D. Tang. M. (47 DEGREES.) SINES AND TANGENTS. (43 DEGREES.) JH. Sine. D. Cosine. D. Tang. D. Cotang. 1 9 •833783 2-26 9-864127 1-96 9-969656 4-22 10 030844 60 1 833919 2-25 864010 1-96 969909 4-22 030091 59 2 834054 2-25 868892 1-97 970162 4-22 029888 58 3 834189 2-25 863774 1-97 970416 4-22 029584 57 4 834325 2-25 863656 1-97 970669 4-22 029381 56 5 884460 2-25 863588 1-97 970922 4-22 029078 56 6 834595 2-25 863419 1-97 971175 4-22 028825 54 7 8347dO 2-25 868801 1-97 971429 4-22 028571 53 8 884865 2-25 863183 1-97 971682 4-22 028318 52 9 834999 2-24 868064 1-97 971935 4-22 028065 51 10 835134 2-24 862946 1-98 972188 4-22 027812 50 11 9-835269 2-24 9-862827 1-98 9-972441 4-22 10-027559 49 12 885403 2-24 862709 1-98 972694 4-22 027306 48 13 885538 2-24 862590 1-98 972948 4-22 027052 47 14 835672 2-24 862471 1-98 973201 4-22 026799 46 15 835807 2-24 862353 1-98 973454 4-22 026546 45 16 835941 2-24 862234 1-98 973707 4-22 026298 44 17 836075 2-28 862115 1-98 973960 4-22 026040 43 18 836209 2-23 861996 1-98 974218 4-22 025787 42 19 836343 2-23 861877 1-98 974466 4-22 025584 41 20 836477 2-23 861758 1-99 974719 4-22 025281 40 21 9-836611 2-23 9-861638 1-99 9-974978 4-22 10-025027 89 22 836745 2-23 861519 1-99 975226 4-22 024774 38 28 836878 2-23 861400 1-99 975479 4-22 024521 87 24 837012 2-22 861280 1-99 975782 4-23 024268 36 25 837146 2-22 861161 1-99 975985 4-22 034015 35 26 837279 2-22 861041 1-99 976288 4-22 023762 84 27 887412 2-22 860922 1-99 976491 4-22 023509 83 28 887546 2-22 860802 1-99 .976744 4-22 023256 32 29 837679 2-22 860682 2-00 976997 4-22 028008 81 30 887812 2-22 860562 2-00 977250 4-22 033750 30 31 9-837945 2-22 9-860442 2-00 9-977503 4-22 10-023497 29 32 838078 2-21 860322 2-00 977756 4-22 032244 28 33 838211 2-21 860202 200 978009 4-22 021991 27 84 838344 2-21 860082 2-00 978262 4-23 021738 26 35 838477 2-21 859963 2-00 978515 4-22 021485 26 86 888610 2-21 859842 2-00 978768 4-22 021282 24 37 888742 2-21 859721 2-01 979021 4-22 020979 23 38 838875 2-21 859601 2-01 979274 4-22 020726 23 89 839007 2-21 859480 2-01 979527 4-22 020473 31 40 889140 2-20 859360 2-01 979780 4-22 020220 20 41 9-839272 2-20 9-859239 2-01 9-980088 4-22 10-019967 19 42 839404 2-20 859119 2-01 980286 4-22 019714 18 43 889536 2-20 858998 2-01 980538 4-22 019462 17 44 839668 2-20 858877 2-01 980791 4-21 019209 16 45 839800 2-20 858756 2-02 981044 4-21 018956 15 46 839932 2-20 858635 2-02 981297 4-21 018703 14 47 840064 2-19 858514 2-02 981550 4-21 018450 13 48 840196 2-19 858893 2-02 981808 4-21 018197 12 49 840328 2-19 858272 2-02 982056 4-21 017944 11 60 840459 2-19 858151 2-02 982309 4-21 017691 10 51 9-840591 2-19 9-858029 2-02 9-982562 4-21 10-017438 9 52 840722 2-19 857908 2-02 982814 4-21 017186 8 58 840854 2-19 857786 2-02 983067 4-21 016983 7 54 840985 2-19 857665 203 988820 4-21 016680 6 55 841116 2-18 857543 203 983573 4-21 016427 5 56 841247 2-18 857422 2-03 988826 4-21 016174 4 57 841378 2-18 857300 2-03 984079 4-21 015921 8 1 58 841509 2-18 857178 ■2-03 984831 4-21 015669 2 59 841640 2-18 857056 2-08 984584 4-21 015416 1 60 841771 2-18 856934 2-03 ,984837 4-21 i 015163 Cosiue. D. Sine. D. Cotang. D. ! Tang. 01. (46 DEGI lEES.") 62 (44 DEGREES.) A TABLE OF LOGARITHMIC M. Sine. D. Cosine* D. Tang. D. Cotang. 9-841771 2-18 9-866934 2-03 9-984837 4-21 10-015163 60 1 841902 2-18 866812 2-03 985090 4-21 014910 59 2 842033 2-18 856690 2-04 985343 4-21 014657 58 3 842163 2-17 856668 2-04 986596 4-21 014404 57 4: 842294 2-17 856446 2-04 985848 4-21 014162 56 1 5 842424 2-17 856323 2-04 986101 4-21 013899 65 1 6 842655 2-17 856201 2-04 986354 4-21 013646 54 i 7 842685 2-17 856078 2-04 986607 4-21 013393 53 8 842816 2-17 856956 2-04 986860 4-21 013140 52 g 842946 2-17 856833 2-04 987112 4-21 012888 51 1 10 843076 2-17 855711 2-05 987366 4-21 012635 50 11 9-843206 2-16 9-855588 2-05 9-987618 4-21 10-012382 49 12 843336 2-16 855466 2-06 987871 4-21 012129 48 13 843466 2-16 856342 2-05 988123 4-21 011877 47 14 843595 2-16 856219 2-05 988376 4-21 011624 46 15 843725 216 865096 2-05 988629 4-21 011371 45 16 843855 2-16 854973 2-05 988882 4-21 011118 44 17 843984 216 854860 2-05 989134 4-21 010866 43 18 844114 2-15 854727 2-06 989387 4-21 010613 42 19 844243 2-16 854603 2-06 989640 4-21 010360 41 20 844372 2-15 864480 2-06 989898 4-21 010107 40 21 9-844502 2-15 9-854356 206 9-990145 4-21 10-009855 39 22 844681 2-15 864233 2-06 990398 4-21 009602 38 23 844760 2-16 854109 2-06 990651 4-21 009349 37 ! 24 844889 2-15 853986 2-06 990903 4-21 009097 36 i 25 845018 2-15 853862 2-06 991156 4-21 008844 35 i 26 845147 2-15 853788 2-06 991409 4-21 008591 34 1 27 845276 2-14 863614 2-07 991662 4-21 008338 33 1 28 845405 2-14 853490 2-07 991914 4-21 008086 32 1 29 846633 2-14 863366 - 2-07 992167 4-21 007833 31 i 80 845662 2-14 853242 2-07 992420 4-21 007580 30 1 81 9-845790 2-14 9-853118 2-07 9-992672 4-21 10-007328 29 i 82 845919 2-14 852994 2-07 992925 4-21 007075 28 83 846047 2-14 852869 2-07 993178 4-21 006822 27 34 846175 2-14 862745 2-07 993430 4-21 006670 26 i 85 846304 2-14 852620 2-07 993683 4-21 006317 25 ■■ 36 846432 2-13 862496 2-08 993936 4-21 006064 24 87 846560 2-13 852371 2-08 994189 4-21 005811 23 ! 88 846688 2-13 852247 2-08 994441 4-21 005559 22 1 89 846816 2-18 852122 2-08 994694 4-21 005306 21 1 40 846944 2-13 861997 2-08 994947 4-21 006053 20 1 41 9-847071 2-13 9-861872 2-08 9-995199 4-21 10-004801 19 1 42 847199 2-13 851747 208 996462 4-21 004548 18 j 43 847327 2-13 851622 2-08 996706 4-21 004295 17 1 44 847454 2-12 851497 2-09 995967 4-21 004043 16 ! 45 847582 2-12 851372 2-09 996210 4-21 003790 15 1 46 847709 2-12 851246 209 996468 4-21 003537 14 : 47 847836 2-12 851121 2-09 996715 4-21 003286 13 48 847964 2-12 850996 2-09 996968 4-21 003032 12 49 848091 2-12 850870 2-09 997221 4-21 002779 11 50 848218 2-12 850745 2-09 997473 4-21 002627 10 51 9-848345 2-12 9-850619 2-09 9-997726 4-21 10-002274 9 62 848472 2-11 850493 2-10 997979 4-21 002021 8 63 848599 2-11 850368 2-10 998231 4-21 001769 7 54 848726 2-11 850242 2-10 998484 4-21 001616 6 65 848852 2-11 850116 2-10 998737 4-21 001263 5 66 848979 2-11 849990 2-10 998989 4-21 001011 4 57 849106 2-11 849864 2-10 999242 4-21 000768 3 68 849282 2-11 849738 2-10 999496 4-21 000506 2 59 849359 2-11 849611 2-10 999748 4-21 000253 1 60 849486 2-11 849485 2-10 10-000000 4-21 10-000000 Cosine. D. Sine. D. Cotang. D. Tang. SI. (45 DEGREES.) A TABLE OF NATURAL SINES. 63 Deg. 1 1 Deg. 2 Deg. 8 Deg. 4 Deg. M _S. 00000 C. S. S. as. S. C.8. S. C.S. 8. C. S. Unit. 01745 99985 03490 99938 o5234 99863 06976 99755 60 I 00029 1 -0000 01774 oi8o3 99984 o35i9 03548 o5263 99861 07005 99754 U 2 00058 I -0000 99984 99937 05292 99860 07034 99753 3 00087 1 -0000 oi832 99983 03577 99936 o532i 99858 07063 99750 U i 001 [6 l-OOOO 01862 99983 o36o6 99935 o535o ^^l 07092 99748 5 00145 1.0000 01891 99982 03635 99934 o5379 05408 07121 99745 55 6 00175 I. 0000 01920 99982 03664 99933 99854 07i5o 99744 54 I 00204 1.0000 01949 01978 99981 03693 99932 05437 99853 07179 07208 99743 53 00333 I. 0000 99980 03723 99931 05466 99851 99740 53 9 003t)3 I. 0000 02007 99980 03752 99930 05495 99849 V,lU 99738 5i 10 00391 I .0000 02036 99979 03781 o38io 99929 o5524 99847 99736 5o II 00330 99999 02065 99979 99978 99927 05553 99846 07295 99734 ii 11 00349 99999 02094 03839 99926 05582 99844 07324 99731 i3 00378 99999 02123 99977 03868 99925 o56ii 99842 07353 99739 ii 14 00407 99999 021 52 99977 03897 99924 o564o 99841 07382 99737 99725 i5 00436 99999 03l8l 99976 03926 99923 05669 99839 07411 45 16 00465 99999 02311 99976 03955 99922 05698 99838 07440 99723 44 \l 00495 99999 02240 99975 03984 99921 05727 99836 07460 07498 99721 43 oo534 99999 99998 02269 02298 99974 04013 99919 99918 05706 99834 99719 43 "9 00553 99974 04042 05785 o58i4 99833 07527 99715 41 30 oo582 99998 02327 99973 04071 99917 99831 07556 997' 4 40 31 0061 1 99998 02356 99973 04100 99916 o5844 99829 07585 99712 It 33 00640 99998 02385 99972 04129 99915 05873 99827 07614 99710 33 00669 99998 02414 9997' 04159 99913 05902 99826 07643 99708 li 34 00698 99998 02443 99970 04188 99912 05931 99824 07672 99705 35 00737 99997 02473 99969 04217 99911 05960 99822 07701 99703 35 36 00756 99997 025ol '^fet 04246 99910 ll^ 99821 07730 34 11 00785 00814 99997 o353o 042751 99909 99819 07759 07788 07846 99699 33 99997 0256o 99967 o43o4 99907 06047 06076 99817 99815 99696 32 39 00844 99996 02580 02618 99966 04333 99906 99694 3i 30 00873 99996 99966 04363 99905 o6io5 99813 99692 3o 3i 00902 99996 02647 99965 04391 99904 061 34 99813 07875 99689 11 32 00931 99996 02676 99964 04430 99902 o6i63 99810 07904 Itl^ 33 00960 99995 02705 99963 04449 04478 99901 06192 9980R 07933 11 34 00989 01018 99995 02734 99963 if 06221 99806 07962 99583 35 99995 02763 99962 o45o1 o625o 99804 07991 08020 99680 35 36 01047 99995 02792 02821 99961 04536 99896 06279 o63o8 99803 99678 24 37 01076 99994 99960 04565 99801 08049 08078 99676 23 38 oiio5 99994 o285o 99959 04594 99894 06337 99799 99673 22 39 01134 99994 02879 02008 99959 04633 99893 06366 99797 08107 o8i36 99671 21 40 01 164 99993 99958 04653 99893 06395 99795 99668 20 41 01193 99993 02938 999^7 04683 99888 06424 99793 o8i65 99666 \t 43 01222 99993 02967 99956 04711 06453 99793 08194 99664 43 OI25l 99992 02996 99955 04740 06482 99790 99788 08223 99661 \l 44 01280 99093 o3o25 99954 04769 04798 99886 o65ii 08252 99559 45 oi3o9| 9999'! o3o54 99953 99885 06540 99786 08281 99557 i5 46 01 338 99991 o3o83 99952 04827 99883 06509 06598 99784 o83io 99554 ■4 47 01367 99991 01396 99990 03ll3 99952 04856 99882 99782 08339 08368 90652 i3 48 o3i4i 99951 04885 99881 066271 99780 99549 11 49 01425 99900 01454' 99989 c3i70 99950 04914 99879 99878 06656 99778 08397 08426 99647 II sS 03228 99940 99948 . 04943 06685 99776 99644 10 fl 014831 99989 04972 99876 06714 99774 08455 99643 t 52 oi5i3' 999S9 01 542 1 99988 03257 99947 o5ooi 99875 06743 99773 08484 99639 S3 03286; 99946 o5o3o 99873 06773 06802 99770 o85i3 99537 I S4 01 571] 99988 o33i6' 99945 o5o59 o5o88 99872 99768 08543 99535 55 01600 99987 o3345i 99944 99870 o683i 99766 08571 99533 5 56 01620 99987 01658 99986 03374 99943 o5ii7 99869 06860 99764 0&600 9Q6J0 4 IS o34o3 99942 o5i46 99867 06889 06918 99763 08639I 99627 08658. 99635 3 01687 99986 03433 99941 o5i75 99866 99760 3 59 01716 99985 03461: 99940 o52o5 99864 06947 C. S. 99758 08687 99621 8 I U C. 8. i S. c. s. 1 s. C. S. s. S. C. 8. _89Deg. 88 De^ . 87 Deg. 8« Deg. 85 De^. 94 A TABLE OF NATURAL SINE3. M 6Deg. 6 Deg. 7 Deg. 8 Deg. 1 9 Deg. u S. |C. s. S. as. s. 1 c. s. S. 13917 C. S S. 1 C 8. 087161 99619 10453 99452 I3I87' 99355 99027 99023 1 5643 98769 6a ■ 08745 99617 10482 99449 133l5 90351 13946 15672 98754 59 3 08774 99«>'4 lo5ii 99446 13245 99348 13975! 99019 15701 98760! 58 3 08803 99613 io54o 99443 12374 99»44 14004 99015 i573o 98755, 57 4 o883i 99609 10569 99440 12303, 99340 i4o33 99011 15758 98751 56 5 08860 99607 \°J>U 99437 , I333i! 99237 1 133501 99333 14061 99006 15787 98746 55 6 08889 08918 99604 99434 14090 99002 ^998 i58i6 98741 54 I 99602 10655 99431 13389' 99330 14119 14148 1 5845 98737 53 08947 99599 10684 99428 1341S 99225 98994 15873 9^732 53 9 08976 99596 10713 99434 12447 12470 99222 14177 l42o5 15902 98728 5i 10 09005 99594 10742 99431 99219 99215 15931 98723 5o 11 09034 ^^^^8 10771 10800 99418 12504 14234 ?;8982 15959 98718 tt 13 09063 99415 12533 99211 14263 98978 15988 98714 i3 09093 99586 10829 10858 99412 12562 99208 14292 98973 16017 98709 % 14 09121 99583 99400 99406 12591 99204 14320 $^t 16046 98704 i5 09150 99580 10887 12630 99200 14349 16074 98700 45 i6 09179 09308 99578 10916 99402 13649 12678 99197 14378 98961 i5io3 98695 44 \l 99575 10945 99395 99193 99180 99186 14407 98957 i6i32 98600 43 09337 99572 10973 12706 14435 98953 16160 98686 43 '9 09366 99570 11002 99393 12735 ■4454 98948 16189 16218 98681 41 30 09295 99567 lio3i 99390 12764 99182 14493 98944 98675 40 31 09324 99564 11060 99386 12793 12822 99178 14522 98940 16246 98671 ^2 33 09353 99562 11089 I1118 99383 99175 14551 98935 16275 98667 38 33 09382 99559 99380 12851 99171 14580 9893, i53o4 98662 ll 34 09411 99556 11 147 9937T 12880 99161 99153 14608 98927 98923 15333 98657 35 09440 99553 11176 99374 1 2908 14637 i535i 98652 35 3b 09469 09498 99551 II2o5 IZ 12937 99160 14555 98919 15390 98648 34 11 99548 1 1 234 13955 99155 14695 98914 16419 96643 33 09527 99545 1 1 263 99364 12995 99152 14723 98910 16447 16476 98638 32 39 09556 99542 1 1 291 99350 l3o24 99148 14752 98906 98633 3i 3o 09585 99540 Il320 99357 i3o53 99144 14781 98902 i65o5 98629 3o 3i 09614 99537 11349 II378 99354 l3o8i 99'4i 14810 98897 16533 98624 \t 33 09642 99534 99351 13110 •^'^l 14838 98893 98889 10552 98619 33 09671 99531 1 1407 99347 I3i39 14867 16591 98614 27 34 09700 99528 11436 99344 13 168 99129 99125 14895 98884 r662o 98609 25 35 Z^l 99526 11465 99341 13 197 14925 98880 16648 98604 25 36 99523 ii49Vi 99337 13226 99122 14954 98875 16677 16706 9K600 34 ll 09787 09816 99520 11523 99334 13254 99118 14982 98871 98.195: 23 99517 11552 99331 13283 99114 i5oii 9R867J 16734 98590I 22 985851 21 39 09845 99514 Ii58o 99327 13312 99110 i5o4o 98863 16763 46 09874 99511 11609 11638 99324 13341 99106 i5o59 98858 16792 16820 9858oi 20 41 09903 99508 99320 13370 99102 i5097 98854 985751 10 985701 18 43 09932 99506 11667 99317 13399 99098 l5i26 98849 16849 43 09961 995o3 1 1696 99314 13427 13455 99094 i5i55 98845 16878 98565 \l 4a 09990 9q5oo 11725 99310 99091 i5i84 98841 16906 98551 45 10019 99497 U754 99307 13485 99087 15212: 98835 16935 98555 i5 46 10048 99494 1 1783 I1812 993o3 i35i4 99083! 15241 1 98832 15964 98551 141 4-' 1 10077 99488 99300 13543 990791 99075 15270 98827! 16992, 9854.' I3 48 1 10106 11S40 99297 13572 15292 98823 17021 98541 12 49 1 ioi35 50 10164 99485 11869 99293 1 3600 99071 15327 98818 15355 98814 1 10501 98535 III 99482 1,18981 99290 \l^t Ztl 1707S' 98531 n>\ 5i 10192 99419 1I937' 99386 15385 98809 i-'w' 9852b 17136; 98521' 8 53 10331 99476 11956 99383 13087 99059! i54i4 9880D 53 1025o 99473 11985' 99279 13716 99055 15442 9880c ni64' 985i6l , 7 54 10279 io3oS 99470 12014 99276 13744 9905 1 15471 98796 171931 98511 6 55 99467 13043 99272 13773 99047 i55oo 9?79' 17222' 98506 5 56 10337. 99^6^ 12071 99269 1 2 100 99265 i38o2 99043 15529 98787 17250' j85oi 4 U io3b6 99461 i333i 99039 15557 15586 98782 172 9 98496 17308 98491 3 10395 9^458 1 2 1 29 99262 13158 99258 C. S. 1 s. i585o 99035 98778 2 S9 It 10424! 99455 13889, 99o3i l35i5 98773 17336 C. S. 98486 I c. s. 1 s. c. s. 1 s. C. S. S. S. MDeR. 88 Doe. 82 Deif. 81 Deg 80D 26640 283i8 f§l 29987 31648 94860 33298 Itt 33 36668 96379 28346 3001 5 31675 94851 33325 3a 29 26696 9637' 28374 30043 95380 3i7o3 94842 33353 94274 3i |3o 26724 96353 28403 30071 95372 3 1730 94832 33381 94264 3o 1 3> 267S2 96355 38439 95874 30098 95363 31758 94823 33408 94254 It 32 26780 26808 96347 38457 95853 30136 95354 31785 3i8i3 94814 33435 94245 33 96340 28485 95857 30154 95345 94805 33463 94235 11 34 26836 96332 283i3 95849 30182 95337 31841 94795 94786 33490 94225 35 26864 96334 28341 95841 30209 95328 3i858 33518 94215 35 36 26B92 96316 28569 95832 30237 95319 31896 94777 94768 33545 94206 34 U 26920 95308 28397 95824 30265 95310 31923 33573 94186 33 26948 95301 38525 95816 30292 95301 3i95i 94758 33600 33 39 26976 tt^ 28552 95807 3o32o 95293 95284 31979 94749 33627 94176 ai 40 27004 38680 95799 30348 32005 94740 33555 94167 30 41 27032 96377 38708 lt!,V. 30375 95275 32034 94730 33682 94157 \t 42 27060 96269 28735 3o4o3 95255 32061 94721 33710 94147 43 37088 96261 38764 95774 3043 1 95257 32089 94712 33737 94137 \l 44 271 16 96253 38793 95755 30459 953481 32116 94702 33764 94127 94118 45 27144 96245 38820 95757 30486 95240 32144 94693 33792 i5 46 27172 96338 28847 95749 3o5i4 95231 32171 94684 33819 33845 94108 14 92254 40195 9.566 ^'m 9085. 43 93534 37003 93903 93893 93881 ^^^T 92243 40331 9.555 4i8i3 90839 \l 44 3540a 93534 37039 38644 92231 40348 91543 41840 9o8a6 45 35439 93514 37056 38671 92230 40275 9i53i 41866 90814 i5 46 35456 935o3 37083 93870 38698 92309 4o3oi 9i5o8 4189a 9080a 14 $ 35484 93493 93483 37110 93859 5?^?^ 93198 40338 4191? 41945 90790 i3 355 1 1 37137 92849 93838 ^f^H 92 .86 40355 91496 91484 90778 la 49 35538 93472 37164 ^m 92175 4o38i 419T 90766 II So 35565 93462 37191 93837 93164 40408 9147a 41998 90753 10 Si 3559a 9345« 37318 9a8i6 38833 931 52 40434 91461 J 3034 90741 t 55 35619 93441 37345 92805 38859 93141 40461 91449 42o5i 90729 53 35647 93431 37372 92794 38886 93i3o 40488 91437 43077 90717 I 54 35674 93420 37299 92784 3891a 92119 4o5i4 91435 43104 90704 55 35701 93410 37326 92773 38939 92107 4o54l 91414 43. 3o. 90693 43.56; 90680 5 56 35718 93400 37353 93763 38966 92096 4o567 91403 4 11 35755 93389 37380; 93751 38993 930H5 40594 91390 43183; 90668 3 3578a 35810 93321 37407 1 93740 39020 92073 40621 9.378 42309' 90655 1 S9 37434 93739 39046 92063 40647 9.356 42235 C. S. 90643 I M C.8. S. c. s. s. C. S. S. C. S. S. S. M 69 D(«. II 68I>eR. 67 Deg. 1 £6 Deg. D 65 Beat. 19 A TABLE OF NATURAL SINES. 25 i)eg. 9 lu II 12 i3 14 i5 \l '9 10 21 22 23 24 25 26 423'J2 42388! 423 1 5 42341 1 4?3o7 42394 42420 42446 42473 42491 42521 42552 42578 42604 4263 1 42667 16 I 42683 ' 42709 43736 42762 42788 42815 42841 42867 42894 42920 42946 42972 42999 43o25 43o5i 43077 43104 43i3o 43 1 56 43182 43300 43:35 43261 43287 433 1 3-1 43340 43366 43392 43418 43445 43471 43497 43523 4354c 43575 43602 43638 43654 43680 43706 43733 43759 43i85 43811 C.S. 3o 3i 32 33 34 35 36 ll 39 40 41 43 43 44 45 46 u Si 53 53 D4 55 56 M )o63i 9061R 90606 qo5q4 905&2 90569 90557 90545 90533 90620 90607 90495 90483 90470 90468 90446 90433 90421 90408 90396 90383 90371 9o358 90346 90334 90321 90309 90306 90284 90271 90369 90246 90333 90321 90208 90196 90183 90171 90168 90146 90133 90120 90108 9OU96 90082 90070 90067 90046 90033 90019 00007 69994 89901 89968 89966 89943 89930 899'8 89906 89892 8. MDcg. Deg. S. 0. S. 43837 43863 43889 43916 43942 43968 43994 44020 44046 44073 44098 44124 44161 44177 442o3 44339 44255 44281 44307 44333 44359 44385 44411 44437 44464 44490 44616 44542 44668 44694 44620 44646 44673 44698' 44724. 44760! 44776 44803 44828 44864 44880 44906 44933 44968 44984 45oioj 46o36 460631 460S8; 461 14 46140! 46166! 46192 46318 46243 46269 46296 45321 45347 45373 C. S. 89879 89867 898641 8984 ll 89828' 89816 898031 89790, 89764! 89753I 89730 89736 89713 89700 89687 89674 89663 8964( 8963( 89623 89610 89607 89584 89671 89668 89645 89632 89619 89606 89493 89480 89467 89454 89441 89416 89402 89389 89376 89363 89360 89337 89334 8931 1 89398 89386 89373 89360 89246 89333 89219 89306 89193 89181 8916' 8915: 89140 89137 89114 S. 08 Deg. 27 Deg. 28 Deg. S. C. S. 46399 46435 46461 46477 46603 46639 46664 46680 466061 45633 46668, 45684, 457101 46736! 46763 45787 46813 46839 46860 46891 46917 46943 46968 46994 46030 46046 46072 46097 46123 46149 46170 46201 46226 46263 46378 46304 46330 46366 46381 46407 46433 46468 46484 46610 46536 46661 46687 46613 46639 46664 46690 46716 46743 46767 46793 46819 46844 46870 46896 46921 1 891 01 89087 89074 89061 89048 89035 89031 89008 88955 88943 88938 88916 88876 88862 88848 88835 88822 88808 88796 88782 88768 88755 88741 88738 88716 88701 88688 88674 88661 88647 88634 8S630 88607 88693 88580 88566 88663 88539 88636 88513 88499 88486 88473 88468 88446 88431 88417 88404 88390 88377 88363 88349 88336 88323 883o8 C^S. I S. 82 Dog. S. C. S. 46947 46973 46999 47024 47060 47076 47101 47127 47163 47178 47204 47339 47355 47281 47306 47333 47358 47383 47409 47434 47460 47486 475ii 4763i 47662 47688 47614 47631 4766I 47690 47716 47741 47767 47793 47818 47844 47869 47895 47920 47946 47971 47997 48032 48048 48073 48099 48134 48160 48176 48201 48226 48263 48277 483o3 48338 48364' 48379 48406 48430 48456 C. S. 88396 88381 88367 88354 88340 88226 88213 88199 88180 88172 88168 88144 88i3o 88117 88io3 88076 88063 88048 88034 88020 88006 87993 87979 87960 87961 87937 87923 87909 87896 87882 87868 87854 87840 87826 87812 87798 87784 87770 87766 87743 87729 87715 87701 87687 87673 87669 87645 87631 87617 87603 87689 87675 37661 87646 87532 67618 87604 87490 87476 S. 61 Deg. 29 Hog. C. ST S. M 48481 1 87461! 6c 48606 87448 48632 87434 48667. 87430 48583 I 87406 48608 87391 : 48634 48669 48684 48710 48735 48761 48786 48811 48837 48862 u 55 48913 48938 48964 48989 49014 49040 49066 49090 49116 49141 49166 49192 49217 49243 49268 49293 49318 49344 49369 49394 49419 49445 49470 49496 49631 49646 49671 49696, 87377 54 87363 53 87349 87335 87321 87306 87392 87278 87364 87360. 46 87235! 44 87331 43 87207 43 87193 41 37178 40 87164 39 87160 38 87i36| 37 871 21 1 36 871071 36 870931 34 87079' 33 870641 32 87060 3i 87036 3i 61 5o 87031 87007 86993 86978 86964 86949 86935 86921 86906 86893 86863 3o 3' ll 36 34 23 33 31 20 :g ]i 16 86834 49622! 86830 49647, 868o5i 14 49672' 86791! 13 49697' 86777- 13 49723 867621 II 49748' 807481 o 49773| 867331 g 49798 86719J 49824. 86704! 498491 866901 49874I 86676! 49899 86661 I 49924 86646 49960 86632 49975 86617 crs. ! s. 60 Deg . A TABLE OF NATURAL SINES. M 80 ] 3eg. C. S. 81 Deg. 82 Deg. 88 Deg. U Deg. M S. S. ! C. S. S. C. S. S. 1 c. s s. c. s. 50000 866o3 5i5o4 85717 62993 84806 54464 83867 55919 83904 5o I 50025 86588 5i529 85702 53017 84789 54488 83851 55943. 82887 5o 66968 33871 1 58 C5992 32865 57 66016. 82839 56 a 5oa5a 86S73 5i554 85687 53041 84774 53o66 84769 53091 34743 545i3 83835 3 50076 86559 5i579 85b72 54537 83819 4 5oioi 85544 51604 85657 54661 83804 ■5 5oi36 86530 51628 85642 53ii6, 84728 54586 83788 6604c 82822 5S 6 5oi5i 8651 5 5i653 85627 53 140; 84713 54610 83772 56o64 82806I 54 I 50176 86501 51678 85612 53164, 84697 531891 84681 64635 83756 66088. 82790 33 5030I 86486 51703 85597 51728 85582 546I9 83740 661121 82773 56i36i 82767 S3 9 50327 86471 53214! 84666 64583 83724 61 10 5o252 86457 51753 85567 53238 84660 64708 83708 66160 82741 60 II 50277 86442 51778 85551 5i8o3; 85536 63253 84635 54732 83692 55i84 82734 ^ 13 5o3o} 86427 86413 53288 84619 64766 83676 8366o 56208 82708 i3 5o327 5i838l 85521 63312 84604 54781 54805 56232 82692 47 i4 5o352 863^4 5i852. 855o6 53337 84588 83645 55255 82675 66280, 82669 45 i5 5o377 51877 85491 53361 84673 64829 83629 46 i6 5o4o3 86369 51902 85476 63386 84667 54864 836i3 663o5' 82643 44 \l 50438 86354 51927 86461 6341 1 84542 54878 .83597 8358 1 66329 82626 43 50453 86340 5 1 953 86446 634301 84626 64902 66353 82610 42 '9 50478 86325 51977 85431 53460I 84611 64927 83565 65377 82693 41 ao 5o5o3 86310 52002 85416 53484: 84495 53309 84480 54961 83549 83533 66401 82677 40 31 5o528 86295 86281 52026 85401 64975 66426 826611 39 66449 82644 38 32 5o553 52o5i 85385 53534 84464 64999 835 17 33 50578I 86266 52076 85370 53658 84448 66024 83501 55473 826281 37 34 5o6o3i 8625i 52ior 85355 63583 84433 66048 83485 66497 826111 35 35 50628 86237 52126, 85340 53607 84417 66072 83469 83453 66521 82496, 36 36 5o654 86222 52i5ii 85325 63632 84402 66097 55545 82478 34 11 5o679 86207 52175 85310 63655 84386 66121 83437 65559 82462 55593 82446 33 50704 86192 52200' 85294 53681 1 84370 66146 83421 33 39 50729 86178 52225 85279 53706 84356 55169 83406 66617 82429I 3i' 66641, 824i3 3a 3o 50734 86163 52250 85264 53730 84339 66194 83389 3i 50779 5o8o4 86148 52275 85249 53764 84324 66218 83373 56665 82396 29 66689 82380 28 32 86i33 52299 85234 637791 843o8 63804 84292 66242 83356 33 50829 86119 52324 85218 55265 83340 66713 82363 17 34 5o854 86104 52349 85203 53828 84277 66291 83324 66735 82347 i 26 35 50879 86089 523741 85i88 53853 84261 653 1 5, 833o8 66760 823301 26 36 50904 86074 52399 85i73 52423 85i57 53877 84245 55339 83292 66784 823141 24 66808 822971 23 55832 82281 1 22 11 5og29 86o59 53902 84230 55363 83276 50954 86043 52448 85 142 53926 84214 55388, 83260 39 50979 86o3o 52473 85i27 53931 84198 53976 84182 55412 83244 66866 822641 31 40 5ioo4 8601 5 52498 85112 55436' 83228 66880 82248I 20 41 5ioaQ 86000 52522 85oo6 52547 85o8i 54000 84167 554601 83212 66904 8223l| 19 42 5io54 85985 64024 84161 55484 83196 56928 82214 |8| 43 5 1079 85970 52572 85o66 64049 84135 64073 84120 65509 55633 83179 83i63 66962 81198 56976 82181 n. 44 5iio4 85956 525g7 85o5i '", 45 . 5ii29 83941 52621 85o35 54097 84104 55557 83i47 57001I1 83i65 l5 46 5ii54 85926 62646 85o20 54122 84088 55581 83i3i 57034 81148 57047I 82i3a 14 ii 5ii79 85011 52671' 85oo5 64145 84072 556o6 83ii5 i3 5 1204 52696 84989 54171 84067 56530 83098 83083 67071 ?21l5 12 49 5 1319 8588 1 52720 84974 54196 84041 66654 57096 82008 57110 82082 67143 82066 11 5o 51254 85866 52745 1 84959 62770' 84943 .54220 84026 66678 83o66 10 5i 51279 85851 54244 84009 55702 83o6o § 53 5i3o4 85836 52794 84928 64260 83994 64293 83978 64317 83962 66726 83o34 67167 82048 53 5 1 329 85821 52819 84913 > 52844 84897 52869 84882 52893 84866 66750 83oi7 67191 82032 I 54 5i354' 858o6 55775 83001 67216 82016 55 5i379' 85792 54342 83945 5579? 83986 67233 81999 57262 81982 5 56 5i4o4 85777 54366 83930 56833 82969 4 u 61439 85763 62918 84861 54391 839151 56847 54416 83899II 55871 82953 67286 81966 3 51454 85747 52943 84836 82935 67310 S1949 3 5q 5i479 85733 63967: 84830 54440. 83883 55896 8292a 57334, 81931 C. S. 1 8. I H C. S. 1 S. C. S. 1 s. C. S. 1 S. C. S. S. 69De(r. 68 Deg. 67 Deg. 66 Deg. 65 Eeg. 70 A TABLE OF NATURAL SINES. M 86~ .- ■5735I Deg. 86 Deg. 87 Deg. 88 Deg. 89 Deg. M 60 iC.S. S. 0. S. S. c. s. S. 61566 C. S. 78801 S. C.S. 81915 58779 58802 80902 8o885 60183 79864 6'w932 'X I 57381 81899 57405 81S83 6o2o5 79846 61 58, 78783 63,55 U 1 58826 80867 60228 79839 61613 78765 1 63,77 ;t& 3 57439 81 855 574531 81848 58849 58873 8o85o 6o25i 79811 6i635 78747 63ooo II 4 80833 6027^ 79793 6i658 78729 63023 77541 6 57477 8i832 58896 80816 60298 79776 61681 787.1 63045 77623 55 e 575011 8i8i5 58920 8^]^ 6o32i 79758 61704 786,4 53068 77605 54 7 ; 575J4' 81798 6 57548 8178J 53,43 60344 79741 61726 78676 63o,o 77586 53 58967 80755 60367 79723 61749 78658 63ii3 77558 S3 g 57572 81755 58990 80748 6o3,o 79706 61772 78640 63i35 77550 Si ic 1 575951 81748 59014 80710 60414 79688 61795 5i8i8 78622 63i58 77531 5o II 57619 81731 57643 81714 59037 80713 60437 7,671 78604 63i3o 775i3, 49 I] 59061 80696 60460 7,553 61841 78585 632o3 77494 48 l3 57667 81698 59084 80679 60483 7,535 61864 78558 53225 77475 77458 47 14 57691 8I68I 59108 8.ji2 6o5o6 79618 61887 78550 63348 45 i5 57715 81664 59131 80644 6o52, 79600 61909 78532 53371 77439 45 16 57738 81647 59154 80627 60553 79583 61932 78514 632,3 77421 ii \l 57762 8i63i 59178 80610 6->576 79565 61955 784,5 633 16 77403 43 57786 81614 59201 80593 60099 79547 61978 78478 6o338 77384 43 >9 57810 '^Ul 59225 80576 60622 79530 62001 78460 63361 77366 41 30 57833 59248 8o558 60645 79512 62024 78442 63383 77347 40 31 57857 8i563 59272 80541 60553 794,4 62045 78424 63405 77329 It 33 57881 81 546 59295 8o524 60691 79477 62069 78405 53428 77310 33 57904 8i53o 59318 8o5o7 60714 79459 62092 78387 53401 772,2 u 34 57928 8i5i3 59342 80489 60738 79441 62ii5 7835, 63473 77255 35 57952 81495 5,365 80472 80455 60761 7,424 62 1 38 78351 634,5 35 36 57976 81479 59389 60784 60807 7,4o5 62160 78333 635 18 77236 34 11 57999 81462 59412 80438 7,388 62183 783 1 5 63540 77218 33 58o2j, 81445 59435 80420 6o83o 7,371 62206 782,7 63563 77<99 77181 33 29 58047 81438 59459 80403 5o853 79353 62229 & 53585 3i 3a 58070 81413 59482 8o385 60876 79335 52231 636o8 77162 3o 3i 58094 81395 59506 8o368 608,, 7,3i8 62274 78243 63ci3o 77144 39 33 58ii8 81378 8i36i 59529 8o35i 60922 7,3oo 62297 78225 53653 77125 38 33 58141 5,552 80334 60945 7,282 62320 78206 63675 ]V^l 11 34 58i65 8i344 59575 8o3i5 60968 79264 52343 78188 63698 35 58189 81327 59099 802,9 60,, 1 79247 62365 78170 62388 78152 63720 77070 25 36 53212 8i3io 59622 80282 6ioi5 7,22, 63742 77o5i 34 11 58236 81293 5,545 80264 6io38 7,211 6241 1 78134 53765 77033 23 58260 81276 582831 81259 5^6^? 80247 61061 7,1,3 62433 78116 53^10 77014 i3 39 8o23o 61084 79176 7,i58 62456; 78098 76996 21 40 583o7| 81243 5,716 80213 61107 62479 78079 62502 78061 63832 76977 20 41 58330 81225 III 80195 6ii3o 79140 63854 76959 :§ 43 58354 81208 80178 6ii53 79122 62524 78043 63877 76940 43 58378 81191 5,80, 80160 61176 79105 62547 78025 6389, 76931 \i 44 58401 81174 8ii57| 80143 611,, 79087 62570 78007 62592 77,88 53922 76^84 45 58425 5,833 80125 61223 79069 53944 i5 46 58449 81140 5,855 80108 61345 7905 1 6261 5: 77,70 52638 77952 63966 75855 14 4J 58475 31123 59879 80091 61368 79033 53989 76847 76828 i3 4^ 58495' 8iio5 5,902 80073 8ooS5 612,1 79015 52660 77934 64011 13 «5 , 585i9 81089 5,926 6i3i4 62683 77016 62706 778,7 54033 76810 II S , 58543 81072 8io55 59949 8oo38 61337 64055 76791 10 5l 58557 i,972 80021 6i36o 78962 62738, 77879 64078 76771 4I S3 ' 58590 8io38 59995 8ooo3 6i383 78944 62751 1 77861 64100 76754 8 S3 585i4 81021 60019 79986 61406 78926 537741 77843 54123 76735 'I ii 53637 81004 60042 79968 6142, 78,08 627,6 52819 77834 54145 7.6711 55 58661 80987 6oo65 79,51 6i45i 788,1 77806 64167 766,8 5 56 58684 80970 60089 79934 61474] 78873 62842 77788 641,01 76679 4 u 38708 80953 60112 79916 614,7! 78855 52864 77769 64212 76661 3 587311 Rnn,16 60135 V^. 6i530| 78837 62887 77751 64334 76642 a 59 587551 80919 601 58 61543 78819 62909 77733 S. 64356 76623 I M c. s. i s. 0. S. S. c. s. s. C. S. C. 8. i S. "M 54 DeR. 68 Dee. 1 6S Dee. 51 Deg. 1 60 Deg. A TABLE OF NATURAL SINKS. n n 40 Dog 41 Deg. 42 Beg. 48 Deg. H 44 Deg. Jt 8. C. S. S. C.S. B. C.S. S. C. S. S. |C.8. 64379 76604 65605 75471 66913 74314 68200 73i35 69466^ 71934 to I 64301 76586 65528 7545a 55935 74295 68221 73116 69487 71914 So » 64323 75548 65650 75433 66955 ]iiit 68242 73096 69508I 7,§;4 5S 3 64346 55573 75414 55978 68264 73076 69539 71873: 5i i 64368 76530 65694 75395 55999 74237 58285, 73o56 69549 7i8i3 56 5 64390 765ii 65716 75355 67021 74217 74n8 683o5 73o35 1 69670 71833 ss 6 1 64412; 76493 1 65738 67043 68327 73016 69591 718131 Si 2 64435; 76473 65759 '4111 67064 74178 68349 72996 69612 7179a, S3 64457 76455 65781 670&5 74159 68370 72976 69633 71773: 5j 9 64479 76435 658o3 tx 67107 74139 68391 72957 69654 7175a Si 10 645oi 76417 76398 76380 65835 67129 74120 68412 72937 69675 71733 So II 64524 65847 75261 671 5i' 74100 68433 72917 72897 69696 71711 71091 4I la 64546 65859 75241 67172 74080 68455 69717 i3 64568! 76361 55891 75222 67194 74061 58476 72877 '^11 71671 7i65o ii i< 64590 76342 659i3 75203 672T5 74041 68497 685i8 72857 iS 64612 76323 55935 75184 67237 74022 73837 69779 7i63o 45 16 54635 76304 55955 75i65 67358 74002 58539 72817 69800 71610 44 \l 64657 76285 65978 75i45 67280 73983 58561 72797 69821 71590 43 64679 76267 76248 66000 75126 67301 73963 68582 72777 69842 71369 43 •9 64701 66022 7508? 67323 73944 686o3 72757 69862 71549 41 20 64723 76229 66044 57344 73924 58624 72737 69883 71529 7i5o8 40 11 64746 76210 66o55 75069 57366 73^85 58645 72717 69904 It 32 64768 76192 66088 75o5o 57387 68666 72697 69925 71488 33 tx 76173 66109 75o3o 67409 73865 68688 72677 72657 69945 71468 11 34 76154 66i3i ■ySoii 67430 73846 68709 69966 71447 35 64834 76135 65i53 74992 67452 73826 68730 72637 69987 71437 35 26 64855 76116 66175 74973 74953 67473 73806 68751 72617 70008 71407 34 11 54878 76097 75078 66197 66318 67495 73787 68772 72597 70029 71386 33 64901 74934 67516 73767 68793 68814 72577 70049 71355 33 39 64923 -JtoSq 66340 74^96 67538 73747 72557 70070 71345 3i 3o 64945 76041 66262 67559 73728 68835 72537 70091 71325 3o 3i 64967 76033 66284 74876 7483? 57580 73708 58857 68878 73517 70112 7i3o5 It 33 64989 76003 663o5 67602 73588 73497 70132 71284 33 656iT 75984 65327 67623 73669 58899 73477 70153 71264 27 34 65o33 75965 66349 74818 57645 73649 68920 73457 70174 71243 2i 35 65o55 75946 66371 74799 74700 57665 73529 68941 73437 70195 71233 25' 36 65077 75927 55393 67688 73610 68062 73417 70215 7i3o3 24 u 65o99 75008 70889 66414 74760 57709 73590 68983 72397 70236 71182 23 6S123 65435 74741 67730 73570 69004 ■J'^?'' 70257 71162 22 39 65144 75870 66458 74722 67752 73551 69035 70277 70298 71141 31 40 65i66 75851 66480 74683 67773 73531 69045 72337 71121 30 41 65 1 88 75832 66501 67795 7351 1 6908^ 72317 70319 71100 ;? 43 65210 758i3 66523 74654 578T5 73491 72297 70339 71080 43 65233 75794 66545 74544 67837 73472 69109 72277 7o36o 71059 \i 44 65a54 ]l]t 56565 746a 5 67859 7345a 69130 72257 7o38i 71039 45 55276 56588 74606 67880 73432 69151 72236 70401 71019 i5 46 55398 75738 56610 74586 67901 73412 69172 72316 70433 70998 14 % 65330 75710 66632 74548 67923 73393 69193 73196 70443 70978 i3 65342 75699 65354; 75680 65553 67944 73373 69314 73175 72i55 70463 70957 i> 49 66675 74528 67955 73353 69235 70484 70937 70916 70896 II So 65385J 75661 65697 66718 74309 tn 73333 69255 72135 7o5o5 10 Ji 55408' 75543 74489 73314 69277I 721 16 69298 72095 70535 I 'it 6543o' 75523 66740 74470 74451 68039 73294 70546 7o8?5 S3 554531 75504 66762 68051 73274 69319 73075 1 70567 I 54 55474 75585 60183 658o5 74431 68072 73254 69340 72055 ]l^ 70834 55 55496 75555 74412 68093 73234 69361 73035 70813 S 56 655i8 75547 55540' 75528 66827 66848 74392 0811 5 732i5 69382 73oi5 70638 70793 4 u 74373 68i35 73195 69403 7 "95 70649 70773 3 65563 75S09 66870 74353 68157 73175 69434 71974 70670 70753 a *9 65584 75490 66891 74334 68179 731 55 69445 71934 70690 70731 1 66 656n6 cTsr 75471 66913 74314 68300 73i35 69466 71934 707 1 1 70711 S. s. C. S. S. C.S. S. C.S. S. C.S. 49 I)eg. ! 481 )eg. 47 Deg. 1 4(1 Deg. J 45 Deg. Text-Books in Astronomy Bowen's Astronomy by Observation By Eliza A. Bowen. Boards, quarto, 94 pages. Colored Maps and Illustrations $1 .00 An elementary text-book for schools, and especially adapted for use as an atlas to accompany any other text-book in astronomy. Careful directions are given when, how and where to find the heavenly bodies, and the quarto pages admit star maps and views on a large scale. Gillet and Rolfe's Astronooiies By J. A. Gillet and W. J. Rolfe. First Book in Astronomy. Short Course. 220 pages . $1.00 Astronomy. 415 pages . . .... 1.40 These books have been prepared by practical teachers and contain nothing beyond the comprehension of pupils in secondary schools. Lockyer's Astronomies By J. N. LocKYER, F.R.S. Astronomy. 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