1837 ART~TE ~~~~~OF.M. NOW~~~~~~~~~~~~;-1~ Z7- TUBBOR is-pfINSULP':~: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ IIIIIIII - - - - - - - - - - - - - ~ THE OF~~ PROF ALEANDE MATHEMATICAL TEXTS Edited by PERCEY F. SMITH, PH.D. Professor of Mathematics in the Sheffield Scientific School of Yale University Elements of the Differential and Integral Calculus By W. A. GRANVILLE, PH.D. Elements of Analytic Geometry By P. F. SMITH and A. S. GALE, PH.D. Introduction to Analytic Geometry By P. F. SMITH and A. S. GALE, PH.D. Advanced Algebra By H. E. HAWKES, PH.D. Text-Book on the Strength of Materials By S. E. SLocuM, PH.D., and E. L. HANCOCK, M.Sc. Problems in the Strength of Materials By WILLIAM KENT SHEPARD, PH.D. Plane Trigonometry and Four-Place Tables of Logarithms By W. A. GRANVILLE, PH.D. Plane and Spherical Trigonometry and Four-Place Tables of Logarithms By W. A. GRANVILLE, PH.D. Four-Place Tables of Logarithms By W. A. GRANVILLE, PH.D. PLANE AND SPHERICAL TRIGONOMETRY AND FOUR-PLACE TABLES OF LOGARITHMS BY WILLIAM ANTHONY GRANVILLE, PH.D. SHEFFIELD SCIENTIFIC SCHOOL, YALE UNIVERSITY GINN AND COMPANY BOSTON * NEW YORK. CHICAGO ~ LONDON ENTERED AT STATIONERS' HALL COPYRIGHT, 1909, BY WILLIAM ANTHONY GRANVILLE ALL RIGHTS RESERVED 89.3 Gbte Itbenacum presg GINN AND COMPANY PROPRIETORS ~ BOSTON U.S.A. PREFACE It has been the author's aim to treat the subject according to the latest and most approved methods. The book is designed for the use of colleges, technical schools, normal schools, secondary schools, and for those who take up the subject without the aid of a teacher. Special attention has been paid to the requirements of the College Entrance Board. The book contains more material than is required for some first courses in Trigonometry, but the matter has been so arranged that the teacher can make such omissions as will suit his particular needs. The trigonometric functions are defined as ratios; first for acute angles in right triangles, and then these definitions are extended to angles in general by means of coordinates. The student is first taught to use the natural functions of acute angles in the solution of simple problems involving right triangles. Attention is called to the methods shown in ~~ 23-29 for the reduction of functions of angles outside of the first quadrant. In general, the first examples given under each topic are worked out, making use of the natural functions. A large number of carefully graded exercises are given, and the processes involved are summarized into working rules wherever practicable. Illustrative examples are worked out in detail under each topic. Logarithms are introduced as a separate topic, and attention is called to the fact that they serve to minimize the labor of computation. Granville's Four-Place Tables of Logarithms is used. While no radical changes in the usual arrangement of logarithmic tables have been made, several improvements have been effected which greatly facilitate logarithmic computations. Particularly important is the fact that the degree of accuracy which may be expected in a result found by the aid of these tables is clearly indicated. Under each case in the solution of triangles are given two complete sets of examples, - one in which the angles are expressed in degrees and minutes, and another in which the angles are expressed in degrees and the decimal part of a degree. This arrangement, which is characteristic of this book, should be of great V vi PREFACE advantage to those secondary schools in which college preparation involving both systems is necessary. To facilitate the drawing of figures and the graphical checking of results a combined ruler and protractor of celluloid is furnished with each copy of the book, and will be found on the inside of the back cover. In Spherical Trigonometry some simplifications have been introduced in the application of Napier's rule of circular parts to the solution of right spherical triangles. The treatment of oblique spherical triangles is unique. By making use of the Principle of Duality nearly one half of the work usually required in deriving the standard formulas is done away with, and the usual six cases in the solution of oblique spherical triangles have been reduced to three. An attempt has been made to treat the most important applications of Spherical Trigonometry to Geodesy, Astronomy, and Navigation with more clearness and simplicity than has been.the rule in elementary treatises. The author's acknowledgments are due to Professor John C. Tracy for many valuable suggestions in the treatment of Spherical Trigonometry, to Messrs. L. E. Armstrong and C. C. Perkins for verifying the answers to the problems, and to Mr. S. J. Berard for drawing the figures. the figurs.A W. A. GRANVILLE SHEFFIELD SCIENTIFIC SCHOOL YALE UNIVERSITY CONTENTS PLANE TRIGONOMETRY CHAPTER I TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. RIGHT TRIANGLES rION SOLUTION OF SEC'l 1. 2. 3. 4. 5. 6. 7. 8. Trigonometric functions of an acute angle defined... Functions of 45~, 30~, 60~............... Solution of right triangles.............. General directions for solving right triangles........ Solution of isosceles triangles......... Solution of regular polygons...... Interpolation................. Terms occurring in trigonometric problems... PAGE 1 4 7. 7. 13. 14. 16. 19 CHAPTER II TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 9. Generation of angles........ 10. Positive and negative angles....... 11. Angles of any magnitude.............. 12. The four quadrants.................. 13. Rectangular coordinates of a point in a plane.... 14. Distance of a point from the origin..... 15. Trigonometric functions of any angle defined.... 16. Algebraic signs of the trigonometric functions... 17. Given the value of a function, to construct the angle.. 18. Five of the trigonometric functions expressed in terms of the sixth 19. Line definitions of the trigonometric functions... 20. Changes in the values of the functions as the angle varies. 21. Angular measure....... 22. Circular measure........ 23. Reduction of trigonometric functions to functions of acute angles 24. Functions of complementary angles 25. Reduction of functions of angles in the second quadrant.... 26. Reduction of functions of angles in the third quadrant.. 27. Reduction of functions of angles in the fourth quadrant.... 28. Reduction of functions of negative angles........ 29. General rule for reducing the functions of any angle..... vii.24. 24 25. 25. 26. 27 28. 29. 29. 34 36.38. 43. 43. 47. 47. 47. 51. 53 55 57 viii CONTENTS CHAPTER III RELATIONS BETWEEN THE TRIGONOMETRIC FUNCTIONS SECTION PAGE 30. Fundamental relations between the functions........ 59 31. Any function expressed in terms of each of the other five functions. 60 CHAPTER IV TRIGONOMETRIC ANALYSIS 32. Functions of the sum and of the difference of two angles..... 63 33. Sine and cosine of the sum of two angles....... 63 34. Sine and cosine of the difference of two angles........ 66 35. Tangent and cotangent of the sum and of the difference of two angles 68 36. Functions of twice an angle in terms of the functions of the angle. 69 37. Functions of multiple angles............ 70 38. Functions of an angle in terms of functions of half the angle.. 72 39. Functions of half an angle in terms of the cosine of the angle.. 72 40. Sums and differences of functions........... 73 41. Trigonometric identities...........75 CHAPTER V GENERAL VALUES OF ANGLES. INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC EQUATIONS 42. General value of an angle............. 79 43. General value for all angles having the same sine or the same cosecant 79 44. General value for all angles having the same cosine or the same secant 81 45. General value for all angles having the same tangent or the same cotangent..................... 82 46. Inverse trigonometric functions........... 84 47. Trigonometric equations.............. 89 48. General directions for solving a trigonometric equation.. 90 CHAPTER VI GRAPHICAL REPRESENTATION OF TRIGONOMETRIC FUNCTIONS 49. Variables..................... 93 50. Constants..................... 93 51. Functions.................... 93 52. Graphs of functions............... 93 53. Graphs of the trigonometric functions....... 95 54. Periodicity of the trigonometric functions......... 97 55. Graphs of the trigonometric functions plotted by means of the unit circle 97 CONTENTS ix CHAPTER VII SOLUTION OF OBLIQUE TRIANGLES SECTION PAGE 56. Relations between the sides and angles of a triangle...... 101 57. Law of sines................ 102 58. The ambiguous case......... 104 59. Law of cosines................ 108 60. Law of tangents.................. 111 61. Trigonometric functions of the half angles of a triangle.... 113 62. Formulas for finding the area of an oblique triangle...... 117 CHAPTER VIII THEORY AND USE OF LOGARITHMS 63. Need of logarithms in Trigonometry..........119 64. Properties of logarithms............. 121 65. Common system of logarithms........... 124 66. Rules for determining the characteristic of a logarithm... 125 67. Tables of logarithms................. 128 68. To find the logarithms of numbers from Table I....... 129 69. To find the number corresponding to a given logarithm... 133 70. The use of logarithms in computations....... 135 71. Cologarithms.................. 137 72. Change of base in logarithms............ 138 73. Exponential equations.............. 140 74. Use of the tables of logarithms of the trigonometric functions... 141 75. Use of Table II, angle in degrees and minutes....... 142 76. To find the logarithm of a function of an angle....... 143 77. To find the acute angle corresponding to a given logarithm.... 144 78. Use of Table III, angle in degrees and the decimal part of a degree. 147 79. To find the logarithm of a function of an angle....... 148 80. To find the acute angle corresponding to a given logarithm.... 149 81. Use of logarithms in the solution of right triangles..... 152 82. Use of logarithms in the solution of oblique triangles...... 158 Case I. When two angles and a side are given.... 158 Case II. When two sides and the angle opposite one of them are given (ambiguous case).............. 161 Case III. When two sides and the included angle are given... 164 Case IV. When all three sides are given......... 167 83. Use of logarithms in finding the area of an oblique triangle.... 170 84. Measurement of land areas.............. 172 85. Parallel sailing................... 173 86. Plane sailing................ 174 87. Middle latitude sailing................ 175 x CONTENTS CHAPTER IX ACUTE ANGLES NEAR 00 OR 90~ SECTION PAGE sin x tan x 88. Limits of - and as x approaches the limit zero... 178 x x 89. Functions of positive acute angles near 0~ and 90~.... 179 90. Rule for finding the functions of acute angles near 0...... 180 91. Rule for finding the functions of acute angles near 90~.....181 92. Rules for finding the logarithms of the functions of angles near 0~ and 90.................... 182 93. Consistent measurements and calculations........ 183 CHAPTER X RECAPITULATION OF FORMULAS List of formulas in Plane Trigonometry.......... 189-191 SPHERICAL TRIGONOMETRY CHAPTER I RIGHT SPHERICAL TRIANGLES 1. Correspondence between the parts of a triedral angle and the parts of a spherical triangle.............. 193 2. Properties of spherical triangles........... 194 3. Formulas relating to right spherical triangles........ 195 4. Napier's rules of circular parts........ 199 5. Solution of right spherical triangles........... 200 6. The ambiguous case. Two solutions... 203 7. Solution of isosceles and quadrantal triangles....... 204 CHAPTER II OBLIQUE SPHERICAL TRIANGLES 8. Fundamental formulas............. 206 9. Law of sines............ 206 10. Law of cosines................. 207 11. Principle of Duality............... 208 12. Trigonometric functions of half the supplements of the angles of a spherical triangle in terms of its sides......... 210 13. Trigonometric functions of the half sides of a spherical triangle in terms of the supplements of the angles......... 214 CONTENTS xi SECTION PAGE 14. Napier's analogies.............. 215 15. Solution of oblique spherical triangles........ 216 16. Case I. (a) Given the three sides................ 217 17. Case I. (b) Given three angles........... 218 18. Case II. (a) Given two sides and their included angle..... 219 19. Case II. (b) Given two angles and their included side..... 222 20. Case III. (a) Given two sides and the angle opposite one of them (ambiguous case)......... 224 21. Case III. (b) Given two angles and the side opposite one of them (ambiguous case)............... 226 22. Length of an arc of a circle in linear units........ 228 23. Area of a spherical triangle.............. 229 CHAPTER III APPLICATIONS OF SPHERICAL TRIGONOMETRY TO THE CELESTIAL AND TERRESTRIAL SPHERES 24. Geographical terms............... 231 25. Distances between points on the surface of the earth...... 232 26. Astronomical problems.............. 235 27. The celestial sphere............... 235 28. Spherical coordinates............... 237 29. The horizon and meridian system.......... 238 30. The equator and meridian system.......... 240 31. Practical applications............ 241 32. Relation between the observer's latitude and the altitude of the celestial pole................. 242 33. To determine the latitude of a place on the surface of the earth.. 242 34. To determine the time of day.............. 247 35. To find the time of sunrise or sunset......... 250 36. To determine the longitude of a place on the earth..... 250 37. The ecliptic and the equinoxes........... 253 38. The equator and hour circle of vernal equinox system.. 253 39. The system having for reference circles the ecliptic and the great circle passing through the pole of the ecliptic and the vernal equinox. 255 40. The astronomical triangle........... 258 41. Errors arising in the measurement of physical quantities.... 259 CHAPTER IV RECAPITULATION OF FORMULAS 42. Right spherical triangles............. 262 '43. Relations between the sides and angles of oblique spherical triangles 262 44. General directions for the solution of oblique spherical triangles.. 264 45. Length of an arc of a circle in linear units........ 264 46. Area of a spherical triangle.............. 264 PLANE TRIGONOMETRY CHAPTER I TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES SOLUTION OF RIGHT TRIANGLES 1. Trigonometric functions of an acute angle defined. We shall assume that the student is familiar with the notion of the angle between two lines as presented in elementary Plane Geometry. For the present we will confine ourselves to the consideration of acute angles. D Let EAD be an angle less than 90~, that is, an acute angle. From B, any point in one of the sides of the angle, draw a perpendicular to the other side, thus forming a right tri- / angle, as ABC. Let the capital letters A, B, C denote the angles and the small letters a, b, c the lengths of the corre- A b C E sponding opposite sides in the right triangle.* We know in a general way from Geometry that the sides and angles of this triangle are mutually dependent. Trigonometry begins by showing the exact nature of this dependence, and for this purpose employs the ratios of the sides. These ratios are called trigonometric functions. The six trigonometric functions of any acute angle, as A, are denoted as follows: sin A, read "sine of A"; cos A, read "cosine of A"; tanA, read "tangent of A"; csc A, read "cosecant of A"; sec A, read "secant of A"; cot A, read "cotangent of A." * Unless otherwise stated the hypotenuse of a right triangle will always be denoted by c and the right angle by C. 1 2 PLANE TRIGONOMETRY These trigonometric functions (ratios) are defined as follows (see figure): (1) sin A opposite side / a\ hypotenuse c/ adjacent side / b\ (2) cosA = -; hypotenuse c opposite side / a\ (3) t = adjacent side - b; (4 csc A= hypotenuse / c\ (4) csc A = - = -; (4 sc =opposite side a) hypotenuse c (5) sec A = t — (= b) (5) se A =adjacent side b; adjacent side / a (6) cot A= opposite side a\ The essential fact that the numerical value of any one of these functions depends upon the magnitude only of the angle A, that is, is independent of the point B from which the perpendicular upon the other side is let fall, is easily established.* These functions (ratios) are of fundamental importance in the study of Trigonometry. In fact, no progress in the subject is possible without a thorough knowledge of the above six definitions. They are easy to memorize if the student will notice that the three in the first column are reciprocals respectively of those directly opposite in the second column. For, a 1 1 sin A =- - --; a b 1 1 cos A ==- = _=; c c sec A b a 1 1 tanA = - =; b b cot A a c 1 1 cse A - -=..; a a sin A c c 1 1 sec A = =; b b cos A b 1 1 cot A =-=.-. a a tallA b For, let B' be any other point in AD, and B" any point in AE. Draw the perpendiculars B'C' and B"C" to AE and AD respectively. The three triangles ABC, AB'C', AB"C", are mutually equiangular since they are rightD angled and have a common angle at A. Therefore they are similar, and we have / BC B' B" C" B~/ ABAB B AB' AB" By But each of these ratios defines the sine of A. In the same manner we may prove this property for each of the other functions. This shows that ^^I/ I the size of the right triangle we choose is imma0~C'^ tonterial; it is only the relative and not the actual lengths of the sides of the triangle that are of importance. The student should also note that every one A B", X of these six ratios will change in value when the angle A changes in size. FUNCTIONS OF ACUTE ANGLES 3 If we apply the definitions (1) to (6) inclusive to the acute angle B, there results b c sin B = -; csc B = c b a c cos B =-; secB = -; c a b a tan B= -; cot B = - a b Comparing these with the functions of the angle A, we see that sin A = cos B; csc A = sec B; cos A = sin B; sec A = csc B; tan A = cot B; cot A = tan B. Since A + B' = 90' (i.e. A and B are complementary) the above results may be stated in compact form as follows: Theorem. A function of an acute angle is equal to the co-function " of its comp2lementary acute angle. Ex. 1. Calculate the functions of the angle A in the right triangle where a = 3, b = 4. Solution. ' = Va2 + b2 = /9 + 16 = V25 =5. 5B Applying (1) to (6) inclusive (p. 2), sin A = 3; csc A = -5-; cos A= 4; secA=-; /; tanA= 3; cot A =. Also find all functions of the angle B, and corn- A b C pare results. Ex. 2. Calculate the functions of the angle B in the right triangle where a = 3, c = 4. B __ Solution. b = c2 - a2 = -16 - 9 = V7. sin B= = csc B = 4t 3 4 cosB =; sec B = 4 3' -\/ 7 3 AZ1 V7- 7 tanB= cot B = Also find all functions of the angle A, and compare results. * Sine and cosine are called co-functions of each other. Similarly tangent and cotangent, also secant and cosecant, are co-functions. 4 PLANE TRIGONOMETRY Ex. 3. Calculate the functions of the angle A in the right triangle where a= 2mn, b = m2 - n2 Solution. c = Va2 + b2 =4 m + m4 - 2 m2n2 + n4 / =\/m4 + 2 m2n2 + n4 = m2 + n2 +/, ^./ 2 _nn m2 + n2 G sin A = sm+ —; csSc A = II m2 + n2 2mn >^ 11, 2 - n2 m2 + n2 / cos A = sec A =?2 + n2 m2 _ n2 A~^ --- - (c7 2 mn m2 - n2 b=m2-n2 tanA = 2; cotA =2. m2 - n2 2 mn Ex. 4. In a right triangle we have given sin A = 4 and a = 80; find c. Solution. From (1), p. 2, we have the formula sin A = - c Substituting the values of sin A and a that are given, there results 4 80 5 c and solving, c = 100. Ans. 2. Functions of 45~, 30~, 60~. These angles occur very frequently in problems that are usually solved by trigonometric methods. It is therefore important to find the values of the B trigonometric functions of these angles and to memorize the results. 45 As/ (a) To find the functions of 45~. Draw an isosceles right triangle, as ABC. This makes 45 angle A = angle B = 45~. b=1 Since the relative and not the actual lengths of the sides are of importance, we may assign any lengths we please to the sides satisfying the condition that the right triangle shall be isosceles. Let us choose the lengths of the short sides as unity, i.e. let a = 1 and b = 1. Then c = Va2 + b2 = /2, and we get sin 45~ csc 45~ = V; 1 cos 45~ =; sec 45~ = V/2; tan 45~ = 1; cot 45~ = 1. FUNCTIONS OF ACUTE ANGLES 5 (b) To find the functions of 30~ and 60~. Draw an equilateral triangle, as ABD. Drop the perpendicular BC from B to AD, and consider the triangle ABC, where angle A = 60~ and angle ABC = 30~. Again take the smallest side as unity, i.e. let b = 1. This makes c=AB=AD=2AC= 2b = 2, B and a = Vc2- b2 = V4- 1 = V3. Therefore V33 2 30 sin 60; csc 60~; /30 cos 60~ -; sec 60= 2;/ 2 tan 60~ = V/3; cot 60~ = - 60 Similarly, from the same triangle, A b-1 D sin 30~ =; csc 30~ = 2; cos 30~ =; sec 30~ =; 13 tan 30~ =; cot 30~ = 3. \/3 Writing the more important of these results in tabulated form,* we have ANGLE 30~ 45~ 60~ siln =.50 =.71+ -=.86+ 2 2 1 1 cos -=.86+ =.71+- - 50 2 2 tan. 57 + 1 3=1.73 The cosecant, secant, and cotangent are easily remembered as being the reciprocals of the sine, cosine, and tangent respectively. * To aid the memory we observe that the numbers in the first (or sine) row are respectively /1, /2, V3; each divided by 2. The second (or cosine) row is formed by reversing the order in the first row. The last (or tangent) row is formed by dividing the numbers in the first row by the respective numbers in the second row. 6 PLANE TRIGONOMETRY The student should become very familiar with the 45~ right triangle and the 30~, 60~ right triangle. Instead of memorizing the above table we may then get the values of the functions directly from a mental picture of these right triangles. Ex. 5. Given a right triangle where A = 60~, a = 100; find c. Solution. Since we know A (and therefore also any function of A), and the sine of A involves a, which is known, and c, which is wanted, we can find c by using the formula B sinA = - by (1), p. 2 c Substituting a = 100, and sin A = sin 60~ = - from the above table, we have V/ 100 2 c Clearing of fractions and solving for c, we get ~600~ ~200 200 c c = = 117.6+. Ans. b, VL 1.7+ What is the value of B? Following the method illustrated above, show that b = 58.8+. EXAMPLES Only right triangles are referred to in the following examples. 1. Calculate all the functions of the angle A, having given a = 8, b = 15. Ans. sinA =, cosA = 1, tanA = -, etc. 2. Calculate the functions of the angle B, having given a = 5, c = 7.. /24 5 /24 Ans. sinB = -, cosB = -, tan B =, etc. 7 7 5 3. Calculate the functions of the angle A, having given b = 2, c = 1. 7 2 V7 Ans. T-' -, 2, etc. 4. Calculate the functions of the angle B, having given a = 40, c = 41. Ans. 40 49 etc. 5. Calculate the functions of the angle A, having given a = p, b = q. Ans. _ -_ 7_ -, etc. Vp2~qp2 + q2 2 6. Calculate the functions of the angle A, having given a = /m2 + man, C = m2 + mn +,nmn n+ n2 et Ans.,, -, etc. m, + n m +n n 7. Calculate the functions of the angle B, having given a = m2 + n2, c = m +n. /2mn Vm2+n2 2 mn Ans. -,,, etc. m + n m + n m2 + n2 FUNCTIONS OF ACUTE ANGLES 8. Given sinA = 3, c = 200.5; calculate a. Ans. 120.3. 9. Given cosA =.44, c = 30.5; calculate b. Ans. 13.42. 10. Given tanA -1-, b= 2 7; calculate c. Ans. -1 VT30. 11. Given A = 30~, a = 25; calculate c. Also find B and b. Ans. c= 50, B =60~, b = 25 3. 12. Given B = 30~, c = 48; calculate b. Also find A and a. Ans. b = 24, A = 60~, a=24 V. 13. Given B 45~, b = 20; calculate c. Also find A and a. Ans. c = 20 2, A = 45~, a = 20. 3. Solution of right triangles. A triangle is composed of six parts, three sides and three angles. To solve a triangle is to find the parts not given. A triangle can be solved if three parts, at least one of which is a side, are given.* A right triangle has one angle, the right angle, always given; hence a right triangle can be solved if two sides, or one side and an acute angle, are given. One of the most important applications of Trigonometry t is the solution of triangles, and we shall now take up the solution of right triangles. The student may have noticed that Examples 11, 12, 13, of the last section were really problems on solving right triangles. When beginning the study of Trigonometry it is important that the student should draw the figures connected with the problems as accurately as possible. This not only leads to a better understanding of the problems themselves, but also gives a clearer insight into the meaning of the trigonometric functions and makes it possible to test roughly the accuracy of the results obtained. For this purpose the only instruments necessary are a graduated ruler and a protractor. A protractor is an instrument for measuring angles. On the inside of the back cover of this book will be found a Granville's Transparent Combined Ruler and Protractor, with directions for use. The ruler is graduated to inches and centimeters and the protractor to degrees. The student is advised to make free use of this instrument. 4. General directions for solving right triangles. First step. Draw a figure as accurately as possible representing the triangle in question. Second step. When one acute angle is known, subtract it from 90~ to get the other acute angle. * It is assumed that the given conditions are consistent, that is, that it is possible to construct the triangle from the given parts. t The name Trigonometry is derived from two Greek words which taken together mean "I measure a triangle." 8 PLANE TRIGONOMETRY Third step. To find an unknown part, select from (1) to (6), p. 2, a formula involving the unknown part and two known parts, and then solve for the unknown part. Fourth step. Check the values found by noting whether they satisfy relations different from those already employed in the third step. A convenient numerical check is the relation, a2 = c2-2 ( + b2 ) (c - b). Large errors may be detected by measurement. Since the two perpendicular sides of a right triangle may be taken as base and altitude, we have at once Area of a right triangle = a-b In the last section the functions 30~, 45~, 60~, were found. In more advanced treatises it is shown how to calculate the functions of angles in general. We will anticipate some of these results by making use of the following table where the values* of the trigonometric functions for each degree from 0~ to 90~ inclusive are correctly given to four or five significant figures. In looking up the function of an angle between 0~ and 45~ inclusive, we look for the angle in the extreme left-hand vertical column. The required value of the function will be found on the same horizontal line with the angle, and in the vertical column having that function for a caption at the top. Thus, sin 15~ =.2588, cot 41 = 1.1504, etc. Similarly, when looking up the function of an angle between 45~ and 90~ inclusive we look in the extreme right-hand vertical column. The required value of the function will be found on the same horizontal line with the angle as before, but in the vertical column having that function for a caption at the bottom. Thus, cos 64~ =.4384, sec 85~ = 11.474, etc. When we have given the numerical value of the function of an angle, and wish to find the angle itself, we look for the given number in the columns having the given function as a caption at the top * Also called the natural values of the trigonometric functions in contradistinction to their logarithms (see Tables II and III of Granville's Four-Place Tables of Logarithms). FUNCTIONS OF ACUTE ANGLES 9 TABLE A NATURAL VALUES OF THE.TRIGONOMETRIC FUNCTIONS Angle sin Cos tan cot sec csc 00 1~ 2~ 30 40 50 6~ 70 8~ 90 100 110 12~ 130 140 15~ 160 170 180 19~ 200 210 22~ 230 240 25~ 26~ 270 280 29~ 30~ 31~ 320 330 340 350 36~ 370 380 390 400 410 420 430 440 450.0000.0175.0349.0523.0698.0872.1045.1219.1392.1564.1736.1908.2079.2250.2419.2588.2756.2924.3090.3256.3420.3584.3746.3907.4067.4226.4384.4540-.4695.4848.5000.5150.5299.5446.5592.5736.5878.6018.6157.6293.6428.6561.6691.6820.6947.7071 1.0000.9998.9994.9986.9976.9962.9945.9925.9903.9877.9848.981\6.9781.9744.9703.9659.9613.9563.9511.9455.9397.9336.9272.9205.9135.9063.8988.8910.8829.8746.8660.8572.8480.8387.8290.8192.8090.7986.7880.7771.7660.7547.7431.7314.7193.7071.0000.0175.0349.0524.0699.0875.1051.1228.1405.1584.1763.1944.2126.2309.2493.2679.2867.3057.3249.3443.3640.3839.4040.4245.4452.4663.4877.5095.5317.5543.5774.6009.6249.6494.6745.7002.7265.7536.7813.8098.8391.8693.9004.9325.9657 1.0000 o0 57.290 28.636 19.081 14.301 11.430 9.5144 8.1443 7.1154 6.3138 5.6713 5.1446 4.7046 4.3315 4.0108 3.7321 3.4874 3.2709 3.0777 2.9042 2.7475 2.6051 2.4751 2.3559 2.2460 2.1445 2.0503 1.9626 1.8807 1.8040 1.7321 1.6643 1.6003 1.5399 1.4826 1.4281 1.3764 1.3270 1.2799 1.2349 1.1918 1.1504 1.1106 1.0724 1.0355 1.0000 1.0000 1.0002 1.0006 1.0014 1.0024 1.0038 1.0055 1.0075 1.0098 1.0125 1.0154 1.0187 1.0223 1.0263 1.0306 1.0353 1.0403 1.0457 1.0515 1.0576 1.0642 1.0711 1.0785 1.0864 1.0946 1.1034 1.1126 1.1223 1.1326 1.1434 1.1547 1.1666 1.1792 1.1924 1.2062 1.2208 1.2361 1.2521 1.2690 1.2868 1.3054 1.3250 1.3456 1.3673 1.3902 1.4142 00 57.299 28.654 19.107 14.336 11.474 9.5668 8.2055 7.1853 6.3925 5.7588 5.2408 4.8097 4.4454 4.1336 3.8637 3.6280 3.4203 3.2361 3.0716 2.9238 2.7904 2.6695 2.5593 2.4586 2.3662 2.2812 2.2027 2.1301 2.0627 2.0000 1.9416 1.8871 1.8361 1.7883 1.7434 1.7013 1.6616 1.6243 1.5890 1.5557 1.5243 1.4945 1.4663 1.4396 1.4142 900 890 880 876 86~ 85~ 84~ 830 820 810 800 79o 780 770 76~ 750 740 730 72~ 71~ 70Q 69~ 680 670 660 65~ 64~ 630 620 610 6Q0 590 58~ 570 56~ 550 540 530 520 510 500 49o 48~ 470 460 450 Angle cos sin cot tan csc sec 10 PLANE TRIGONOMETRY or bottom. If we find it in the column having the given function as a top caption, the required angle will be found on the same horizontal line and in the extreme left-hand column. If the given function is a bottom caption, the required angle will be found in the extreme right-hand column. Thus, let us find the angle x, having given tan x =.7536. In the column with tan as top caption we find.7536. On the same horizontal line with it, and in the extreme left-hand column, we find the angle x = 37~. Again, let us find the angle x, having given sin x =.9816. In the column with sin as bottom caption we find.9816. On the same horizontal line with it, and in the extreme right-hand column, we find the angle x = 79~. The following examples will further illustrate the use of the table. Ex. 1. Given A = 35~, c = 267; solve the right triangle. Also find its area. Solution. First step. Draw a figure of the triangle indicating the known and unknown parts. Second step. B 90-A = 90~-35~= 55~. B? Third step. To find a use formula (1), p. 2, namely, a /?I sin A -. 60/^ ein^ Substituting the value of sin A = sin 35~ =.5736 (found from the table) and c = 267, we have a.5736 = 35~ 267,A 'b=? Solving for a, we get a = 153.1.* * Multiplying, sin 35~ =.5736 267 40152 34416 11472 a= 153.1512 Since our table gives not more than the first four significant figures of the sine of an angle, it follows, in general, that all but the first four significant figures of the product are doubtful. The last three figures of the above product should therefore be omitted, for the result will not be more accurate if they are retained. To illustrate this in the above example, suppose we take the sine of 35~ from a five-place table, that is, a table which gives the first tive significant figures of the sine. Then sin 35~ =.57358 267 401506 344148 114716 a= 153.14586 Comparing, we see that the two values of a agree in the first four significant figures only. Hence we take a = 153.1. FUNCTIONS OF ACUTE ANGLES To find b use formula (2), p. 2, namely, b cosA = - c Substituting as before, we have b.8192 = 267 since from the table cosA =cos 35~ =.8192. Hence b = 218.7. Fourth step. By measurements we now check the results to see that there are no large errors. As a numerical check we find that the values of a, b, c satisfy the condition c2 = a2 + b2 To find the area of the triangle we have ab 153. Ix 218.7 Area ab 153.1 x 218.7 = 16,741. 2 2 Ex. 2. A ladder 30 ft. long leans against the side - of a building, its foot being 15 ft. from the building. What angle does the ladder make with the ground? Solution. Our figure shows a right triangle with hypotenuse and side adjacent to the required angle (= x) given. Hence 15 1 cos = = -5= - 1.5 =.5000. This number is found in the column having cos at the bottom and opposite 60~. Hence x = 60~. Ans. We shall now derive three formulas by means of which the work of solving right triangles may be simplified. From (1), (2), (3), p. 2, Sill A -, or, B c a = c sin A; b /c~ ~cos A=- or, b = c cos A; A - b tan A = -, or, a = b tan A. These results may be stated as follows: (7) Side opposite an acute angle = hypotenuse X sine of the angle. (8) Side adjacent an acute angle = hypotenuse X cosine of the angle. (9) Side opposite an acute angle = adjacent side X tangent of the angle. 12 PLANE TRIGONOMETRY EXAMPLES Solve the following right triangles (C = 90~). No. GIVEN PARTS REQUIRED PARTS AREA 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 A = 60~ A = 30~ a=6 a=4 a=2 a =51.303 B = 51~ A = 36~ c=43 b= 9.696 a =137.664 A =75~ A = 25~ B = 55~ B= 15~ a = 36.4 a = 23.315 a= 17.1 A= 10~ A = 20~ B = 86~ B 32~ a = 30.21 a =13.395 b= 93.97 b=4 a=3 c= 12 b=4 c = 2.8284 c= 150 c = 250 c=1 a = 38.313 c =20 c =240 a = 80 a = 30 b= 10 b =20 b= 100 b = 50 = 50 b =30 c = 80 b=.08 c = 1760 c = 33.33 b= 50 c= 100 B = 30~ B = 60~ A = 30~ A = 45~ A = 45~ A = 20~ A =39~ B = 54~ A= 63~ A = 61~ A =35~ B= 15~ B = 65 A =35~ A =75~ A =20~ A 25~ A =20~ B = 80~ B= 70~ A = 4 A = 58~ A = 65~ A= 15~ A = 20~ c=8 c=6 B = 60~ B = 45~ B 45~ B =70~ a= 157.3 a=.5878 B = 27~ B = 29~ B = 55~ b = 21.44 b = 64.336 a = 7.002 a = 74.64 B = 70~ B = 65~ B = 70~ a = 5.289 a = 27.36 a =.0055oQ b = 932.62 B = 25~ B = 75~ B= 70~ a = 6.928 b = 5.196 b =10.39 c = 5.657 b=2 b= 140.95 b =194.3 b.809 b 19.52 a= 17.492 b= 196.6 c = 82.82 c = 70.98 c = 12.206 c = 77.28 c= 106.4 c = 55.17 b = 46.985 c = 30.46 b = 75.176 c =.0802 a = 1492.5 b =14.086 c= 51.77 a =34.2 13.856 7.794 31.18 8 2 3615 15282 2378 373.9 84.8 13532 858 965 35 746 1820 582 402 79 1028.0002 695968 213 335 1607 26. A tree is broken by the wind so that its two parts form with the ground a right-angled triangle. The upper part makes an angle of 35~ with the ground, and the distance on the ground from the trunk to the top of the tree is 50 ft. Find the length of the tree. Ans. 96.05 ft., I illil!;!:MW 27. In order to find the breadth of a river, a disKjiS H 0'J/^l^ tance AB was measured along the bank, the point A being directly opposite a tree C on the other side. If the angle ABC was observed to be 55~ and AB 100 ft., find the breadth of the river. Ans. 142.8 ft. 28. Two forts defending a harbor are 2 mi. apart. From one a hostile battleship is observed due south and from the other 15~ east of south. How far is the battleship from the nearest fort? Ans. 7.464 mi. 29. A vessel whose masts are known to reach 100 ft. above her water line subtends in a vertical plane an angle of 5~ to an observer in a rowboat. How far is the boat from the vessel? Ans. 1143 ft. FUNCTIONS OF ACUTE ANGLES 13 30. The vertical central pole of a circular tent is 20 ft. high, and its top is fastened by ropes 40 ft. long to stakes set in the ground. How far are the stakes from the foot of the pole, and what is the inclination of the ropes to the ground? Ans. 34.6 ft.; 30~. 31. A wedge measures 10 in. along the side and the angle at the vertex is 20~. Find the width of the base. Ans. 3.47 in. 32. At two points A, B, 400 yd. apart on a straight horizontal road, the summit of a hill is observed; at A it is due north with an elevation of 40~, and at B it is due west with an elevation of 27~. Find the height of the hill. Ans. 522.6 ft. 5. Solution of isosceles triangles. An isosceles triangle is divided by the perpendicular from the vertex to the base into two equal right triangles; hence the solution of an isosceles triangle can be made to depend on the solution of one of these right triangles. The following examples will illustrate the method. Ex. 1. The equal sides of an isosceles triangle are each 40 in. long, and the equal angles at the base are each 25~. Solve the triangle and find its area. Solution. B = 180~ - (A + C) = 180~ - 50~ = 130~. Drop the perpendicular BD to AC. B AD =ABcosA =40cos25~ by (8), p. 11 = 40 x.9063 25 252 = 36.25. A C D Therefore AC = 2 AD = 72.50 in. To find the area we need in addition the altitude BD. BD = AB sinA = 40 sin 25~ by (7), p. 11 = 40 x.4226 = 16.9. Check: BD = AD tan25~ = 36.25 x.4663 = 16.9. by (9), p. 11 Also, Area = -AC x BD = 603.6 sq. in. Ex. 2. A barn 60 ft. wide has a gable roof whose rafters are 30 V2 ft. long. What is the pitch of the roof, and how far above the eaves is the ridgepole? B Solution. Drop a perpendicular from B to AD. Then AC 30 _1 cosx= ~ - AB 30 V2 \2 Hence x = 45~ = pitch of the roof. 30X 30| ^Also, BC =ABsinx by (8), p. 11 I 30 ------------— _ -= 3 0 - V 1 = 30 ft. = height of the ridgepole above the eaves. Check: AB = /AC2 + BC2 = (30)2 + (30)2 = 1800 30 V2. 14 PLANE TRIGONOMETRY EXAMPLES 1. The equal sides of an isosceles triangle are each 12 in. long, and the angle at the vertex is 120~. Find the remaining parts and the area. Ans. Base = 20.78 in.; base angles = 30~; area = 62.35 sq. in. 2. The equal angles of an isosceles triangle are each 35~, and the base is 393.18 in. Find the remaining parts. Ans. Vertex angle = 110~; equal sides = 240 in. 3. Given the base 300 ft. and altitude 150 ft. of an isosceles triangle; solve the triangle. Ans. Vertex angle = 90~; equal angles = 45~; equal sides = 212.13 ft. 4. The base of an isosceles triangle is 24 in. long and the vertical angle is 48~; find the remaining parts and the area. Ans. Equal angles = 660; equal sides = 53.5 in.; area = 586.5 sq. in. 5. Each of the equal sides of an isosceles triangle is 50 ft. and each of its equal angles is 40~. Find the base, the altitude, and the area of the triangle. Ans. Alt. = 32.14 ft.; base = 76.6 ft.; area = 1231 sq. ft. 6. The base of an isosceles triangle is 68.4 ft. and each of its equal sides is 100 ft. Find the angles, the height, and the area. Ans. 40~, 70~; 93.97 ft.; 3213.8 sq. ft. 7. The base of an isosceles triangle is 100 ft. and its height is 35.01 ft. Find its equal sides and the angles. Ans. 61.04 ft.; 35~, 110~. 8. The base of an isosceles triangle is 100 ft. and the equal angles are each 65~. Find the equal sides, the height, and the area. Ans. 118.3 ft.; 107.23 ft.; 5361.5 sq. ft. 9. The ground plan of a barn measures 40 x 80 ft. and the pitch of the roof is 45~; find the length of the rafters and the area of the whole roof, the horizontal projection of the cornice being 1 ft. Ans. 29.7 ft.; 4870 sq. ft. <^ c i r ~ 6. Solution of regular polygons. Lines drawn from the center of a regular polygon of n sides to the vertices are the radii of the circumV/By \: scribed circle and divide the poly/V \ A gon into n equal isosceles triangles. The perpendiculars from the center \\ / r arm // to the sides of the polygon are the radii of the inscribed circle and ^^^^'^ '^^ ~ divide these n equal isosceles triangles into 2 n equal right triangles. Hence the solution of a regular polygon depends on the solution of one of these right triangles. FUNCTIONS OF ACUTE ANGLES 15 360~ From Geometry we know that the central angle ABC =; hence in the right triangle ABD the 180~ angle x = n Also, AD = - = half the length of one side, 2 AB = R = radius of circumscribed circle, BD = r = radius of inscribed circle, p = no = perimeter of polygon, pr 2 = area of polygon. EXAMPLES 1. One side of a regular decagon is 10 in.; find radii of inscribed and circumscribed circles and area of polygon. Solution. Since n = 10, in this example we have 1 80~ 1800~ n 10 Then R= 5= 16.18 in., sin 18~.3090 5 5 and r = = = 15.39 in. tan 18~.3249 Check: r = R cos 18 = 16.18 x.9511 = 15.39. Also, p = 10 x 10 = 100 in. / \ = perimeter of polygon; hen pr 100 x 15.39 7. hence - - = 769.5 sq. in. 2 2 = area. 2. The side of a regular pentagon is 24 ft.; find R, r, and area. Ans. 20.42 ft.; 16.52 ft.; 991.2 sq. ft. 3. Find the remaining parts of a regular polygon, having given (a) n = 9, c = 12. Ans. R = 17.54; r = 16.48; area = 889.9. (b) n = 18, R = 10. r = 9.848; c = 3.472; area = 307.8. (c) n = 20, R = 20. r = 19.75; c = 6.256; area = 1236. (d) =- 12, r=8. R=8.28; c=4.29; area =206. 4. The side of a regular hexagon is 24 ft. Find the radii of the inscribed and circumscribed circles; also find the difference between the areas of the hexagon and the inscribed circle, and the difference between the areas of the hexagon and the circumscribed circle. Ans. R = 24 ft.; r = 20.8 ft.; 139.1 sq. ft.; 311 sq. ft. 5. If c be the side of a regular polygon of n sides, show that 1 180~ 1 180~ 1 = - ccsc and r = -c cot 2 n 2 n 16 PLANE TRIGONOMETRY 6. If r be the radius of a circle, show that the side of the regular inscribed 180~ polygon of n sides is 2 r sin —, and that the side of the regular circumscribed 180~ n polygon is 2 r tan -. n 7. Interpolation. In the examples given so far we have needed the functions of such angles only as were explicitly given in our table; that is, the number of degrees in the angle involved was given by a whole number. It is evident that such will not always be the case. In general, our problems will involve angles expressed in degrees and parts of a degree, as 28.4~, 5.63~, 10~ 13', 72~ 27.4', 42~ 51' 16", etc. In order to find from the table the numerical value of the function of such an angle not given in the table, or to find the angle corresponding to a given numerical value of some function not found in the table, we use a process called interpolation. This is based on the assumption that a change in the angle causes a proportional change in the value of each function, and conversely, provided these changes are small.*' To illustrate; from the table we have sin 38~ =.6157 sin 37~ =.6018 Subtracting,.0139 = difference for one degree; that is, at 37~ a change of one degree in the angle causes a change in the value of the sine of.0139. If, then, x is any other small change in the angle from 37~, and d the corresponding change in the value of the sine, we must have, near 37~, 1~: x:.0139: d,.. =.0139 x, if x is expressed in the decimal parts of a degree. For example, let us tabulate the values of the sines of all angles from 37~ to 38~ at intervals of 0.1 of a degree. x d 0.1~.0014.-. sin37.1~ =.6018 +.0014 =.6032 0.2~.0028.-. sin 37.2~ =.6018 +.0028 =.6046 0.3~.0042... sin 37.3~=.6018 +.0042 =.6060 0.4~.0056.. sin 37.4~ =.6018 +.0056 =.6074 0.5~.0070.-. sill 37.5~ =.6018 +.0070 =.6088 0.6~.0083.-. sin37.6~ =.6018 +.0083 =.6101 0.7~.0097.. sin 37.7~ =.6018 +.0097 =.6115 0.8~.0111.. sin 37.8~ =.6018 +.0111 =.6129 0.9~.0125.-. sin 37.9~ =.6018 +.0125 =.6143 * This condition is most important. The change in value of the cotangent for one degree is very large when the angle is very small. In this case the table would therefore lead to very inaccurate results if interpolation was used for cotangents of small angles (see Chapter IX, p. 178). FUNCTIONS OF ACUTE ANGLES 17 The following examples will further illustrate the process of interpolating. (a) iTo find the function of a given angle when the angle is not found in the table. Ex. 1. Find sin32.8~. Solution. The sine of 32.8~ must lie between sin 32~ and sin 33~. From the table on p. 9, sin 33 =. 5446 sin 32~ =.5299.0147 = difference in the sine (called the tabular difference) corresponding to a difference of 1~ in the angle. Now in order to find sin 32.8~, we must find the difference in the sine corresponding to.8~ and add it to sin 32~, for the sine will be increased by just so much when the angle is increased from 32~ to 32.8~. Denoting by d the difference corresponding to.8~, we have 1~:.8::.0147: d, or, d =.0118. Hence sin 32 =.5299 d =.0118 = difference for.8~.. sin 32.8~ =. 5417. Ans. Ex. 2. Find tan 470 25'. Solution. The tangent of 47~ 25' must lie between tan 47~ and tan 48~. From the table, tan 48~ = 1.1106 tan 47~ = 1.0724.0382 = tabular difference corresponding to a difference of 60'(=1~) in the angle. Denoting by d the difference corresponding to 25', we have 60':25'::.0382: d, or, d =.0159. Hence tan 47 = 1. 0724 d =.0159 = difference for 25'.... tan47~ 25' = 1.0883. Ans. Ex. 3. Find cos 68.57~. Solution. The cosine of 68.57~ must lie between cos 68 and cos 69~. From the table, cos 68~ =.3746 cos 69 =.3584.0162 = tabular difference corresponding to a difference of 1~ in the angle. Denoting by d the difference corresponding to.57~, we have 10:.57~::.0162: d, or, d =.0092. 18 PLANE TRIGONOMETRY Since the cosine decreases as the angle increases, this difference must be subtracted * from cos 68~ in order to get cos 68.57~. Hence cos 68 =.3746 d =.0092 = difference for. 57~... cos 68.57 =.3654. Ans. (b) To find an angle when the given numerical value of a function of the angle is not found in the table. Ex. 4. Find the angle whose tangent is.4320. Solution. This problem may also be stated: Having given tan x =.4320, to find the angle x. We first look up and down the columns with tan at top or bottom, until we find two numbers between which.4320 lies. These are found to be.4245 and.4452, the former being tan 23~ and the latter tan 24~. We then know that the required angle x must lie between 23~ and 24~. To find how far (= y) beyond 23~ the angle x lies, we first find the difference between tan 23~ and tan x; thus, tan x =.4320 tan 23~ =.4245.0075 = difference in the tangent corresponding to the excess of the angle x over 23~; denote this excess by y. Also, tan 24~ =.4452 tan 23~ =.4245.0207 = tabular difference corresponding to a difference of 1~ in the angle. Then, as before, 1~: y::.0207:. 0075, or, y =.36~. Hence x = 23~ + y = 23.36~. Ans. In case we want the angle expressed in degrees and minutes, we can either multiply.36~ by 60, giving 21.6' so that the required angle is 23~ 21.6', or else we can find y in minutes at once by using instead the proportion 60':y::.0207:.0075, or, y = 21.6'. Hence x = 23~ + y = 23~ 21.6'. Ans. EXAMPLES 1. Verify the following: (a) sin 51.6~ =.7836. (f) csc 80.3~ = 1.0145. (k) sec 25~ 2.5'= 1.1038. (b) tan 27.42~ =.5188. (g) sin 430 18' =.6858. (1) csc 72~ 54' = 1.0463. (c) cos 79.9~ =.1753. (h) cos84~ 42' =.0924. (m) sin 580 36.2' =.8536. (d) cot 65.62~ =.4532. (i) tan 31~ 7.8'.6041. (e) sec 12.37~ = 1.0238. (j) cot 11~ 43.4' = 4.8278. * In the case of the sine, tangent, and secant this difference is always added, because these functions increase when the angle increases (the angle being acute). In the case of the cosine, cotangent, and cosecant, however, this difference is always subtracted, because these functions decrease when the angle increases. It is always the function of the smaller of the two angles that this difference is added to or subtracted from. TERMS IN TRIGONOMETRIC PROBLEMS 19 2. Find the angle x, having given (a) sin x =.5280. (b) tan x =.6344. (c) sec x = 1.2122. (d) cos x =.9850. (e) cot x = 3.5249. (f) csc x = 1.7500. (g) sin x.9425. (h) cos z =.2118. (i) tanx = 1.1652. (j) cot x =.0803. (k) sec x = 4.6325. (1) csc x = 1.2420. (m) sin x =.7100. (n) cos x =.9999. (o) tan x=.9845. (p) cot x =8.6892. Ans. x = 31.87~. x = 32.39~. x = 34.41~. x = 9.93~. x = 15.85~. x = 34.85~. x = 70~ 28.8'. x = 77.77~. x = 49~ 21.6'. x = 85.41~. x = 77.51~. x = 53.63~. x = 45~ 14.4'. x = 0~ 30'. x = 44~ 33'. x = 6~ 36'. 8. Terms occurring in trigonometric problems. The vertical line at a point is the line which coincides with the plumb line through that point. A horizontal line at a point is a line which is perpendicular to the vertical line through that point. A vertical plane at a point is a plane which contains the vertical line through that point. The horizontal plane at a point is the plane which is perpendicular to the vertical line through that point. A vertical angle is one lying in a vertical plane. A horizontal angle is one lying in a horizontal plane. The angle of elevation of an object above the horizontal plane of the observer is the ver- D - tical angle between the line drawn from the observer's e n [ eye to the object, and a hori- of e Horizontal Line zontal line through the eye. The angle of depression of an object below the horizontal plane of the observer is the vertical IHorizontal Line n/////ta- L angle between the line drawn espression from the observer's eye to the object, and a horizontal line X||~ nM ~through the eye. ~H /////W///g///The// horizontal distance between two points is the distance from one of the two points to the vertical line drawn through the other. 20 PLANE TRIGONOMETRY The vertical distance between two points is the distance from one of the two points to the horizontal plane through the other. ~~~~D A,~B Thus, let BC be the vertical line at B, and let the horizontal plane at A cut this vertical line in C; then AC is called the horizontal distance between A and B and BC the vertical distance. The Mariner's Compass is divided into 32 equal parts; hence each part = 360 -- 32 = 11k~. The following figure shows how the different divisions are designated. North, south, east, and west are called the cardinal points, and on paper these directions are usually taken as upward, downward, to the right, and to the left respectively. The direction of an object from an observer at C may be given in several ways. Thus, A in the figure is said to bear N.E. by E. from C, or from C the bearing of A is N.E. by E. In the same way the bearing of C from A is S.W. by W. The point A is 3 points north of east and 5 points east of north. Also, E. 333~ N. means the same as N.E. by E. In order to illustrate the application of the trigonometric functions (ratios) to the solution of practical examples, we shall now give a variety of problems on finding heights, dis- tances, angles,areas,etc. ' X 1 o ' In solving these prob- ' ' 4 4, lems it is best to follow. d.\\7// some definite plan. In z.t-. ^ general we may proceed as follows: IVEST r- ^^ ) B^ EAIST (a) Constructadraw- EST EA ing to some convenient 11 7. b~ s scale which will show f.s ^'.V the relations between / / \ the given and the re- ' - quired lines and angles. 4 ' (b) If necessary draw o any auxiliary lines that will aid in the solution, and decide on the simplest steps that will solve the problem. (c) Write down the formulas needed, make the calculations, and check the results. EXAMPLES 21 EXAMPLES Solve the following right triangles (C = 90~). NO. GIVEN PARTS REQUIRED PARTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 a =-60 a = 16.98 a 147 A = 34~ 15' A =31~ 14.2' B = 47.26~ A =23.5~ A = 28~ 5' B = 43.8~ B = 6~ 12.3' a =.624 a 5 a = 101 A = 43.5~ B = 68~ 50' A = 58.65~ B 10.85~ a = 24.67 B = 21~ 33' 51" A = 74~ 0' 18" A = 64~1.3' b =.02497 b = 1.4367 c= 100 c = 18.7 c= 184 a = 843.2 c = 2.934 c = 4.614 c = 627 c = 2280 b = 50.94 c = 3721 c =.91 b=2 b=116 c 11.2 a= 729.3 c= 35.73 c =.7264 b = 33.02 a =.821 c 275.62 b = 200.05 c =.04792 c= 3.4653 A = 36~ 52' A = 65 14' A = 53~ 2' B = 55~ 45' B = 58~ 45.8' A =42.74~ B = 66.5~ B = 61~ 55' A = 46.2~ A = 83~ 47.7' A = 430 18' A = 68~ 12' A = 41~ 3' B = 46.5~ A = 21~ 10' B = 31.35~ A = 79.15~ A = 36~ 46' A = 68 26' 9" B = 15~ 59' 42" B = 25~ 58.7' A = 58~ 36' A = 65~ 30' B = 53~ 8' B = 24~ 46' B = 36~ 58' c = 1498.5 a = 1.522 a = 3.131 a = 250 a = 1073 a= 53.18 a = 3699 B = 46~ 42' B = 21~ 48' B = 48~ 57' a = 7.71 b = 1884 a = 30.51 a =.7133 B = 53~ 14' b =.3244 a = 264.9 a = 410.5 B = 31~ 24' B = 24~ 30' b= 80 b = 7.834 b =110.67 b = 1238 b = 2.509 b = 3.387 b= 575 b = 2011 c = 73.6 b = 402.2 b =.6622 c = 5.385 c= 153.8 b = 8.124 c = 2020 b = 18.59 b =.1367 c = 41.22 c =.8827 b = 75.93 c = 456.7 a =.0409 a = 3.153 24. The length of a kite string is 250 yd., and the angle of elevation of the kite is 40~. Find the height of the kite, supposing the line of the kite string to be straight. Ans. 160.7 yd. 25. At a point 200 ft. in a horizontal line from the foot of a tower the angle of elevation of the top of the tower is observed to be 60~. Find the height of the tower. Ans. 346 ft. 26. A stick 10 ft. in length stands vertically on a horizontal plane, and the length of its shadow is 8.391 ft. Find the angle of elevation of the sun. Ans. 50~. 27. From the top of a rock 2 that rises vertically 80 ft. out of the water the angle of depression S of a boat is found to be 30~; find.9 J1 the distance of the boat from the foot of the rock. Ans. 138.57 ft. 28. Two ships leave the same o d dock at the same time in direc- f.. tions S.W. by S. and S.E. by E. at rates of 9 and 9.5 mi. per hour S respectively. Find their distance apart after 1 hr. Ans. 13.1 mi. 22 PLANE TRIGONOMETRY 29. From the top of a tower 120 ft. high the angle of depression of an object on a level with the base of the tower is 27~ 43'. What is the distance of the object from the top and bottom of the tower? Ans. 258 ft., 228 ft. 30. A ship is sailing due east at the rate of 7.8 mi. an hour. A headland is observed to bear due north at 10.37 A.M. and 33~ west of north at 12.43 P.M. Find the distance of the headland from each point of observation. Ans. 25.22 mi., 30.07 mi. 31. A ship is sailing due east at a uniform rate of speed. At 7 A.M. a lighthouse is observed bearing due north, 10.32 mi. distant, and at 7.30 A.M. it bears 18~ 13' west of north. Find the rate of sailing of the ship and the bearing of the lighthouse at 10 A.M. Ans. 6.79 mi. per hour, 63~ 14' W. of N. 32. From the top of a tower the angle of depression of the extremity of a horizontal base line 1000 ft. in length, measured from the foot of the tower, is observed to be 21~ 16' 37". Find the height of the tower. Ans. 389.5 ft. 33. The length of the side of a regular octagon is 12 in. Find the radii of the inscribed and circumscribed circles. Ans. 14.49 in., 15.68 in. 34. What is the angle of elevation of an inclined plane if it rises 1 ft. in a horizontal distance of 40 ft.? Ans. 1 26'. 35. A ship is sailing due N.E. at the rate of 10 mi. an hour. Find the rate at which she is moving due north. Ans. 7.07 mi. per hour. 36. A ladder 40 ft. long may be so placed that it will reach a window 33 ft. high on one side of the street, and by turning it over without moving its foot it will reach a window 21 ft. high on the other side. Find the breadth of the street. Ans. 56.68 ft. 37. At a point midway between two towers on a horizontal plane the angles of elevation of their tops are 30~ and 60~ respectively. Show that one tower is three times as high as the other. 38. A man in a balloon observes that the bases of two towers, which are a mile apart on a horizontal plane, subtend an angle of 70~. If he is exactly above the middle point between the towers, find the height of the balloon. Ans. 3770 ft. 39. In an isosceles triangle each of the equal angles is 27~ 8' and each of the equal sides 3.088. Solve the triangle. Ans. Base = 5.496. 40. What is the angle of elevation of a mountain slope which rises 238 ft. in a horizontal distance of one eighth of a mile? Ans. 19~ 50'. 41. If a chord of 41.36 ft. subtends an arc of 1450 37', what is the radius of the circle? Ans. 21.65 ft. 42. If the diameter of a circle is 3268 ft., find the angle at the center subtended by an arc whose chord is 1027 ft. Ans. 36~ 49'. 43. From each of two stations east and west of each other the angle of elevation of a balloon is observed to be 45~, and its bearings N.W. and N.E. respectively. If the stations are 1 mi. apart, find the height of the balloon. Ans. 3733 ft. EXAMPLES 23 44. In approaching a fort situated on a plain, a reconnoitering party finds at one place that the fort subtends an angle of 100, and at a place 200 ft. nearer the fort that it subtends an angle of 15~; How high is the fort and what is the distance to it from the second place of observation? Hint. Denoting the height by y and the distance by x, we have y = x tan 150, by (9), p. 11 also, y = (x + 200) tan 10~. by (9), p. 11 Solve these two simultaneous equations for x and y, substituting the values of tan 150 and tan 100 from the table on p. 9. Ans. x - 385 ft., y = 103 ft. oal | --- —200 --- —-- 45. A cord is stretched around two wheels with radii of 7 ft. and 1 ft. respectively, and with their centers 12 ft. apart. Prove that the length of the cord is 12 -3+ 107rft. 46. A flagstaff 25 ft. high stands on the top | of a house. From a point on the plain on which the house stands, the angles of elevation of the 50 top and the bottom of the flagstaff are observed to be 600 and 450 respectively. Find the height of the house. Ans. 33.94 ft. 47. A man walking on a straight road observes at one milestone a house in a direction g making an angle of 300 with the road, and at the next milestone the angle is 60~. How far is the house from the road? Ans. 1524 yd. 48. Find the number of square feet of pavement required for the shaded portion of the streets shown in the figure, all the streets being x 50 ft. wide. 28750 Ans. + 7500 = 24094. V13 CILAPTER II TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 9. Generation of angles. The notion of an angle, as usually presented in Elementary Geometry, is not general enough for the purposes of Trigonometry. We shall have to deal with positive and negative angles of any magnitude. Such a conception of angles may be formed as follows: An angle may be considered as generated by a line which first coincides with one side of the angle, then revolves about the vertex, and finally coincides with the other side. B B a0 Ainis, X e; initial side 0 initial side i nitial side B This line is called the generating line of the angle. In its first position it is said to coincide with the initial side of the angle, and in its final position with the terminal side of the angle. Thus, the angle A OB is generated by the line OP revolving about 0 in the direction indicated from the initial side OA to the terminal side OB. 10. Positive and negative angles. In the above figures the angles were generated by revolving the generating line counter-clockwise; mathematicians have agreed to call such angles positive. Below are le B B^ -' i nitianitial side initial side P p P angles having the same initial and terminal sides as those above, but the angles are different since they have been generated by revolving the generating line clockwise; such angles are said to be negative.* * The arcs with arrowheads will be drawn full when indicating a positive angle, and dotted when indicating a negative angle. 24 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 25 11. Angles of any magnitude. Even if angles have the same initial and terminal sides, and have been generated by rotation in the same direction, they may be different. Thus, to generate one right angle, the generating line rotates into the position OB as shown in Fig. a. If, however, the generating line stops in the position OB after making one complete revolution, as shown in Fig. b, then we have generated an angle of magnitude five right angles; or, if two complete revolutions were first made, as shown in Fig. c, then we have B B B o A A A FIG. a FIG. b FIG. C generated an angle of magnitude nine right angles; and so on indefinitely. This also shows that positive angles may have any magnitude whatever. Similarly, by making complete revolutions clockwise, it is seen that negative angles may have any magnitude.* 12. The four quadrants. It is customary to divide the plane about the vertex of an angle into four parts called quadrants, by passing two mutually perpendicular lines through the vertex. Thus, if 0 is the vertex, the different quadrants are named as indicated in the figure below, the initial side being horizontal and drawn to the right. An angle is.said to be (or lie) in a certain quadrant when its ter-1~ ~ Second First minal side lies in that quadrant. uadrant Quadrant In the figures shown on the previous page, only the least positive 0 An l se o initial side and negative angles having the Third Fourth given initial and terminal sides Quadrant Quadrant are indicated by the arcs. As a matter of fact there are an infinite number of positive and negative angles in each case which have the same initial and terminal sides, all differing in magnitude by multiples of 360~. The following examples will illustrate the preceding discussion. * Thus, the minute hand of a clock generates - 4 rt. X every hour, i.e. - 96 rt. A every day. 26 PLANE TRIGONOMETRY EXAMPLES 1. Show that 1000~ lies in the fourth quadrant. Solution. 1000~ = 7200 + 280~ = 2 x 360~ + 2800. Hence we make two complete revolutions in the positive direction and 280~ beyond, and the terminal side of 280~ lies in the fourth quadrant. 2. Show that - 568~ lies in the second quadrant. Solution. - 568~ = - 360~ - 208~. Hence we make one complete revolution in the negative direction and 208~ beyond in the negative direction, and the terminal side of 208~ lies in the second quadrant. 3. In what quadrants are the following angles? (a) 225~. (e) 6510. (i) 5400. (m) 15000. (b) 1200. (f) - 150~. (j) 420~. (n) 810~. (c) - 315~. (g) - 75~. (k) - 910~. (o) - 540~. (d) -2400. (h) - 12000. (1) - 3000. (p) 537O. 13. Rectangular coordinates of a point in a plane. In order to define the functions of angles not acute, it is convenient to introduce the y notion of coordinates. Let X'X be a horizontal line and Y'Y a line per1N a P(ab) pendicular to it at the point O. Any b point in the plane of these lines (as P) is determined by its distance 0I' _____ _ _ and direction from each of the perpendiculars X'X and Y'Y. Its distance from Y'Y (as NP = a) is called the Ir abscissa of the point, and its distance from X'X (as MPP = b) is called the ordinate of the point. Abscissas measured to the right of Y'Y are positive. Abscissas measured to the left of Y'Y are negative. Ordinates measured above X'X are positive. Ordinates measured below X'X are negative. The abscissa and ordinate taken to- gether are called the coordinates of the II I point and are denoted by the symbol (-,+)' (+,+) (a, b). The lines X'X and Y'Y are called the x' 0 axes of coordinates, X'X being the axis of _, abscissas or the axis of X, and Y'Y the axis I IV of ordinates or the axis of Y; and the point 0 is called the origin of coordinates. Y The axes of coordinates divides the plane into four parts called quadrants (just as in the previous section), the figure indicating the proper signs of the coordinates in the different quadrants, TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 27 To plot a point is to locate it from its coordinates. The most convenient way to do this is to first count off from 0 along X'X a number of divisions equal to the abscissa, to the right or left according as the abscissa is positive or negative. Then from the point so determined count off a number of divisions equal to the ordinate, upward or downward according as the ordinate is positive A x' x or negative. The work of plotting points is much simplified by the use of coordinate or plotting paper, con- (4,-3) structed by ruling off the plane into equal squares, the sides being parallel to the axes. Thus, to plot the point (4, - 3), count off four divisions from 0 on the axis of X to the right, and then three divisions downward from the point so determined on a line parallel to the axis of Y. Similarly, the following figures show the plotted points (- 2, 3), (- 3, - 4), (0, 3). Y, Y,,(0,3) (-2,3) X' OX ' FIX' O, % 0 ~ 2 X 0 0 (-3,-4) Y ~ Y'0 Y 14. Distance of a point from the origin. Represent the abscissa of a point P by a and the ordinate by b, and its distance from the origin by h. Then. P(a,b) h = a2 + b2, h since h is the hypotenuse of a right b triangle whose sides are a and b. _____ x Although h may be either positive M or negative, it will be sufficient for our purposes to treat it as being Y always positive. In order to become familiar with the notion of coordinates, the student should plot a large number of points. 28 PLANE TRIGONOMETRY EXAMPLES 1. (a) Plot accurately the points (5, 4), (- 3, 4), (- 2, - 4), (5, - 1), (6, 0), (-5, 0), (0, 4), (0, - 3). (b) What is the distance of each point from the origin? Ans. 41, 5, 2 5, etc. 2. Plot accurately the points (1, 1), (- 1, - 1), (- 1, 1), (V/3, 1), (-3, - 1), ( —3, - 1), and find the distance of each one from the origin. 3. Plot accurately the points (V/2, 0), (- 5, - 10), (3, - 2 /2), (10, 3), (0, 0), (0, — v/5), (3, - 5), (- 4, 5). 15. Trigonometric functions of any angle defined. So far the six trigonometric functions have been defined- only for acute angles (~ 1, p. 2). Now, however, we shall give a new set of definitions which will apply to any angle whatever, and which agree with the definitions already given for acute angles. B p y B P *x 6 Y Q Q G Y Angle in first quadrant Angle in second quadrant Q Q B YF U Angle in fourth quadrant Angle in fourth quadrant Angle in third quadrant Take the origin of coordinates at the vertex of the angle and the initial side as the axis of X. Draw an angle XOB in each quadrant. From any point P on the terminal side OB of the angle draw PQ perpendicular to the initial side, or the initial side produced. In every case OQ is the abscissa and QP the ordinate of the point P. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 29 Denoting by XOB any one of these angles, their functions are defined as the following ratios: Qp ordinate OP hypotenuse (10) sin OB = OP =; (13) csc XOB ordinate op hypotenuse QP ordinate OQ abscissa OP hypotenuse (11) cosXOB= hypotenuse; (14) secXOB abscissa OP hypotenuse OQ abscissa QP ordinate OQ abscissa (12) tanXOB=- =-.; (15) cotXOB=OQ abscissa (15) otXOB ordinate To the above six functions may be added the versed sine (written versin) and coversed sine (written coversin), which are defined as follows: versin XOB = 1- cos XOB; coversin XOB = 1- sin XOB. 16. Algebraic signs of the trigonometric functions. Bearing in mind the rule for the algebraic signs of the abscissas and ordinates of points given in ~ 13, p. 26, and remembering that the hypotenuse OP is always taken as positive (~ 14, p. 27), we have at once, from the definitions of the trigonometric functions given in the last section, that: In I Quadrant, all the functions are positive. In II Quadrant, sin and csc are positive; all the rest are negative. In III Quadrant, tan and cot are positive; all the rest are negative. In IV Quadrant, sec and cos are positive; all the rest are negative. These results are also exhibited in the following RULE FOR SIGNS sin + all + csc + \ tan + cos + cot-+ sec + ~III<_ /V All functions not indicated in each quadrant are negative. This rule for signs is easily memorized if the student remembers that reciprocal functions of the same angle must necessarily have the same sign, i.e. sin and csc have the same sign, cos and sec have the same sign, tan and cot have the same sign. 17. Having given the value of a trigonometric function, to construct geometrically all the angles which satisfy the given value, and to find the values of the other five functions. Here we will make use of the * As in acute angles it is seen that the functions in one column are the reciprocals of the functions in the other. 30 PLANE TRIGONOMETRY notion of coordinates, assuming as before that each angle has its vertex at the origin, and its initial side coinciding with the axis of X. It remains, then, only to fix the terminal side of each angle, or, what amounts to the same thing, to determine one point (not the origin) in the terminal side. When one function only is given, it will appear that two terminal sides satisfying the given condition may be constructed. Thus, B iif we have given tan x = 2, we may write 2 - 2 ordinate 2 tan x --, (12), p. 29 -X —2 3 tan 3 -3 abscissa ( 0i 3 Q and hence, taking tan x= 2, one terminal side is determined by the origin ~y~ ' ~ and (3, 2), giving the angle XOB (in FIG. a the first quadrant). — 2 The other terminal side, taking tan x = -3, is determined by the origin and (- 3, - 2), giving the angle XOB' (in the third quadrant). Hence all the angles x ` which satisfy the condition YJ.a~ i tan x= 2 have the initial side OX Fi and the terminal side OB, ' Q' K3 o>; 'or, have the initial side OX S-2 and the terminal side OB'.. B' Yl Let us now determine the values of all the FIG. b functions. From Fig. a, OP = -oQ2 + QP2 = V9 4 = V13 (always positive). Hence by ~ 15, p. 29, from Fig. a, sin XOB = —; csc XOB = 2; 3 +13 cosXOB- 1 secXOB = 3 2 3 tan XOB =; cot'XOB =- 3 2 * It is evident that, corresponding to each figure, there are an infinite number of both positive and negative angles differing by multiples of 360~ which satisfy the given condition. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 31 Similarly, from Fig. b, 2 \/13 sin XOB =; csc XOB'- 2 cos XOB' = - sec XOB' -- V13 3 2 3 tan XOB' = cot XOB = 3 2 Or, denoting by x any angle which satisfies the given condition, we may write down these results in more compact form as follows: 2 3 sin x = -~; cs x = t2; 3 cos x =; sec x= <l 313 2 3 tan x =; cot x= - 3 2 The method is further illustrated in the following examples: Ex. 1. Having given sin x = -, construct the angle x. Also find the values of the other five functions. Solution. Here we may write, by (10), p. 29, 1 - 1 ordinate sin = - - = (hypotenuse always positive). 3 3 hypotenuse Since abscissa =+ / (hypot.)2 - (ord.)2 - — 1 = 2 /2, one terminal side is determined by the origin and (- 2 V2, - 1), y giving the angle XOB in the third quadrant. Here sin XOB =- cscXOB =- 3; 3 3' Y,, Q-2V 2 x —\ -2 -- cosXOB=- 22 secXOB=- 3 -tanXOB = cotXOB =2 2. 2 /, The other terminal side is determined by the origin and (2 '/2, - 1) giving y the angle XOB' in fourth quadrant. Here 1 sin XOB' =- csc XOB' = - 3; 3' 3 ^ cos XOB' -=;see XOB' = — '- tBX 3 2V2 tan XOB' =- -; cot XOB' = - 2 V2. Y? 2 V/2 32 PLANE TRIGONOMETRY Or, denoting by x any angle which satisfies the given condition, we have sin x =-; csc X =- 3; 3 22 3 cos x = -; sec x =; 3 2-\/2V tanx =; cot x = ~2 2. 2V2 m Ex. 2. Having given cot x = -, find all the other functions of x. n Solution. Here we may write, by (15), p. 29, cot m - m abscissa cot X = _ = _ n -n ordinate and hypotenuse = n/m2 + n2. Hence one terminal side is determined by the origin and (m, n), and the other terminal side by the origin and (- m, - n). Therefore n /nm2 + n2 sin x =; cscx =; Nm2 + n2 n m -\/m2 + n2 cosx =; secx =; Vmi2 + n2 m n m tan x =- cot x = - m' n EXAMPLES In each of the following examples construct geometrically the angle x, and compute the values of all the functions of x. Given. 3 a 1. sinx=-. 6. tan x = - 11. tan x = —/;7. 5 b 1 2 2. cos x -. 7. sin x = c. 12. sin x = - 3 3 a2 _ b2 3. cotx=-3. 8. cosx = a-b 13. tanz =2.5. a2 + b2 5 4. secx = —. 9. csc x=-V-. 14. sec x = p. 3 13 mn 5. cscx =-. 10. cosx=-. 5 c * When m and n have the same sign, x represents angles in the first and third quadrants. When m and n have opposite signs, x represents angles in the second and fourth quadrants. t cot x =3 =- 3 1 -1 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 33 ANSWERS 1. 2. 3. 4. 5. 6. 4 3 4 3 1 2V2 1 2V2 -3 -3 3 4 3 4 12 5 12 5 b a i1 - C2 +-c2 C c a2 - b2 2 ab ~ i2d 7. 8. 9. 10. 11. m + _VC2 - t2 1 +5 2 12. 13. 2 5 14. 1 p2/i- - 1 34 PLANE TRIGONOMETRY 18. Five of the trigonometric functions expressed in terms of the sixth. For this purpose it is again convenient to use the definitions of the functions which depend on the notion of coordinates (~ 13, p. 26). The following examples will illustrate the method. Ex. 1. Express, in terms of sin x, the other five functions of x. Solution. Since sin = sin x ordinate by (10), p. 29 1 hypotenuse abscissa = /(hypotenuse)2 - (ordinate)2 a =- 1-sin2x. 2X Hence, by definitions,* ~+ /s1 t 2 X sin x = sin x; sc x = - siln x cos x = V V1- sin2 x; sec x 1 V/1-sin2x sin x V1 - sin2 x tan x =; cot x = ~ V/1- sin 2 sin x Ex. 2. Express, in terms of tan x, the other five functions of x. tan x ordinate /v^ Solution. Since tan = tan = or t^^y^t7/S~~ H 1 abscissa hypotenuse = i V(abscissa)2 + (ordinate)2 x^^K_____z ~Vi += V1 + tan2 x. ~1 ~ Hence tan x V/1 + tan2 x sin x =; csc x =; V/1 + tan2 x tan x cos x = - 1; sec x = V/1 + tan2 x; V/1 + tan2 x tan x = tan x; cot x - tan x Ex. 3. Having given sec x = 4, find the values of the other five functions. 5 hypotenuse Solution. Since secx = - = ypo is by (14), p. 29 4 abscissa ordinate = V /(hypotenuse)2 - (abscissa)2 +3 = V25 - 16 = 3. Hence 4 sin x = ~ -; csc x = i ~; os X = 4; sec x = 4; -5 4, tanx = ~ 4; cotx = ~-. * It is convenient to draw a right triangle (as above) to serve as a check on the numerical part (not the algebraic signs) of our work. We then refer to the definitions of the functions of an acute angle (p. 2) where adjacent side corresponds to the abscissa, and opposite side corresponds to the ordinate. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 35 EXAMPLES 1. Express, in terms of cos x, the other five functions of x. Ans. sin x = 1- cos2x; cscx = OOnzOXC OS2 X cos a = cos x; sec x = - cos x 1/ - cos2 x cos x tan x = C --- cot x = ~ - - cosX V '-cos2x 2. Express, in terms of cot x, the other five functions of x. Ans. sin x = - -; csc x = ~ V1 + cot2x; 1 + cot2 x cotx V/1 +cot2x cosx= -; secx=; 1 + cot2 x cotx tan x = cot x = cot;. cot X 3. Express, in terms of sec x, the other five functions of x. /sec2 x -1 sec x Ans. sin x = +; cscx =; sec x V/sec2 x-1 1 cos = --; sec a = sec x; sec x tan x = ~ /sec2x - 1; cotx = ~ /sec2 x -1 4. Express, in terms of csc x, the other five functions of x. Ans. sinx= - csc =cscx; csc x Vcsc2 x-1 csc x cosx=+; secx=4 csca; X csc2x-1 tan x = 1; cot x =~ /sc2 x-1. csc2 x - 1 5. Having given sec x =- 17-, find the values of the other five functions of x. Ans. sin x = 1; cs x = -+; cos x = -; sec x = - 7 tan x = F -15-; cot x =F T8. 6. Having given sin x = a, find the values of the other functions of x. Ans. sin x =a; cscx = -; a cosx = ~/1-a2; secx=~ n/1- a2 a /- a2 tan x = -+; cot x = -- [1-a2 a 36 PLANE TRIGONOMETRY 7. Having given cot x = V2, find the values of the other functions of x. Ans. sin x = -; cscx= V3; cosx = t;sec x=A; tan x =; cot x = 2. 19. Line definitions of the trigonometric functions. The definitions of the trigonometric functions given in ~ 15, p. 29, are called the ratio definitions. From these we shall now show how the functions of any Angle in first quadrant Angle in second quadrant Angle in third quadrant Angle in fourth quadrant angle may be represented by the numerical measures of the lengths of lines drawn as shown above in connection with a unit circle (i.e. a circle with radius unity). TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 37 Applying these ratio definitions, we get QP sin AOP = QP QP; OP (= ) OQ cos AOP = OP ( - OQ; OP (= 09 QP AT T tanAOP Q = = AT; oQ OA (=-l) OP OT X secAOP=o = OT; OQ OA (=-1) oQ BC t cot AOP- Qp — B()= BC; OP OC t csc AOP = - - - OC. QP OB (=1) From these results the so-called line definitions of the trigonometric functions may be stated as follows: The sin equals the length of the perpendicular drawn from the extremity of the terminal radius to the horizontal diameter. The cos equals the length of the line drawn from the center to the foot of this perpendicular. The tan equals the length of a line drawn tangent to the circle fronm the right-hand extremity of the horizontal diameter and meeting the terminal radius produced. The sec equals the distance from the center to the point of intersection of this tangent with the terminal radius produced. The cot equals the length of a line drawn tangent to the circle from the upper extremity of the vertical diameter and meeting the terminal radius produced. The csc equals the distance from the center to the point of intersection of this cotangent with the terminal radius produced. Algebraic signs must, however, be attached to these lengths so as to agree with the rule for the signs of the trigonometric functions on p. 29. We observe that sin and tan are positive if measured upward from the horizontal diameter, and negative if measured downward; cos and cot are positive if measured to the right of the vertical diameter, and negative if measured to the left; sec and csc are positive if measured in the same direction as the terminal side of the angle, and negative if measured in the opposite direction. * Since triangles OQP and OAT are similar. t Since triangles OQP and OBC are similar. 38 PLANE TRIGONOMETRY 20. Changes in the values of the functions as the angle varies. (a) The sine. Let x denote the variable angle A OP. As x decreases, the sine decreases through the values Q1P1, Q2P2, etc., and as x approaches zero as a limit, the sine approaches zero as a limit. This is written sin = 0 As x increases from 0~ and approaches 90~ as a limit, the sine is positive, and increases from zero through the values Q3P3, Q4P4, etc., and approaches OB (= 1) as a limit. This is written sin 90~ = 1. As x increases from 90~ and approaches 180~ as a limit, the sine is positive and decreases from OB (= 1) through Q5P5, etc., and approaches zero as a limit. This is written sin 180~ = 0. As x increases from 180~ and approaches 270~ as a limit, the sine is negative and increases in nuB p merical value from zero through 3 /^ps Q6P6, etc., and approaches the p / limit OB'(=-1). This is written PI /P ~ / y^.^\sin 270~=-1. As x increases from 270~ X _AQ — 0 / 7 - \ 4 (. Q Q a and approaches 360~ as a limit, the sine is negative and decreases in numerical value from \ / =1 \ / OB' (=-1) through Q7P7, etc., and approaches the limit zero. tPs ^^ 7>~~~ ~ This is written B' sin 360~ = 0. (b) The cosine. Using the last figure, we see that as x decreases, the cosine increases through the values OQ1, OQ2, etc., and as x approaches zero as a limit, the cosine approaches the limit OA (= 1). This is written cos 0 = 1. As x increases from 0~ and approaches 90~ as a limit, the cosine is positive and decreases from OA (= 1) through the values OQ3, OQ4, etc., and approaches the limit zero. This is written cos 90 = 0. TRIGONOETRIRI FUNCTIONS OF ANY ANGLE 39 As x increases from 90~ and approaches 180~ as a limit, the cosine is negative and increases in numerical value from zero through OQ5, etc., and approaches the limit OA'(= - 1). This is written cos 180~ = - 1. As x increases from 180~ and approaches 270~ as a limit, the cosine is negative and decreases in numerical value from OA'(= - 1) through OQ6, etc., and approaches the limit zero. This is written cos 270~ = 0. As x increases from 270~ and approaches 360~ as a limit, the cosine is positive and increases from zero through OQ7, etc., and approaches the limit OA (= 1). This is written cos 360~ = 1. (c) The tangent. Let x denote the variable angle AOT. As x decreases, the tangent decreases through the values A T1, A T1, etc., and as x approaches zero as a limit, the tangent approaches the limit zero. This is written tan 0~ = 0. [4 As x increases from 0~ and approaches 90~ as a limit, the tangent is positive and // B increases from zero through the values T AT3, AT4, etc., without limit, i.e. beyond any numerical value. This is written \T tan 90~ = + co.* Now suppose the angle x to be equal \ A to the angle A OP and let it approach 90~ as a limit; then the corresponding T tangent A T. is negative and increases in numerical value without limit. This B' is written To tan 90~ =- o We see, then, that the limit of the tangent will be + oo or - o according as x is increasing or decreasing as it approaches the limit 90~. As one statement these last two results are written tan 90~ = o, when, as in this book, no distinction is made for the manner in which the angle approaches the limit 90~. * + oo is read plus infinity. - oo is read minus infinity. co is read simply infinity. 40 PLANE TRIGONOMETRY As x increases from 90~ and approaches 180~ as a limit, the tangent is negative and decreases in numerical value from -oo through AT6, ATs5, etc., and approaches the limit zero. This is written tan 180~ = 0. As x increases from 180~ and approaches 270~ as a limit, the, tangent is positive and increases from zero through A T3, A T4, etc., without limit. This is written tan 270~ = o. As x increases from 270~ and approaches 360~ as a limit, the tangent is negative and decreases in numerical value from -oo through AT6, AT5, etc., and approaches the limit zero. This is written tan 360~ = 0. (d) The secant. Using the last figure, we see that as x decreases, the secant decreases through the values OT1, OT2, etc., and approaches OA (= 1) as a limit. This is written sec 0~ = 1. As x increases from 0~ and approaches 90~ as a limit, the secant is positive and increases from OA (= 1) through OT3, 074, etc., without limit. This is written sec 90~ = o. As x increases from 90~ and approaches 180~ as a limit, the secant is negative and decreases in numerical value from - oo through OT6, OT6, etc., and approaches minus OA (= - 1) as a limit. This is written sec 180~ =-1. As x increases from 180~ and approaches 270~ as a limit, the secant is negative and increases in numerical value from minus OA (=- 1) through OT3, OT4, etc., without limit. This is written sec 270~ = o. As x increases from 270~ and approaches 360~ as a limit, the secant is positive and decreases from + oo through OT6, OT,5 etc., and approaches the limit OA (= 1). This is written sec 360~ =1. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 41 (e) The cotangent. Let x denote the variable angle AOC. As x decreases, the cotangent increases through the values BC1, BC2, etc., and as x apCs 0.45 B Ca Co. C 0' Ca proaches 0~ as a limit, the c C4 C C 2 cotangent increases without limit. This is written cot 0~ =-00. As x increases from 0~ and approaches 90~ as a limit, the cotangent is posi- \ tive and decreases from TB' + oo through the values BC3, BC4, etc., and approaches the limit zero. This is written cot 90~ = 0. As x increases from 90~ and approaches 180~ as a limit, the cotangent is negative and increases in numerical value from zero through BCs, BC6, etc., without limit. This is written cot 180~ = o. As x increases from 180~ and approaches 270~ as a limit, the cotangent is positive and decreases from + oo through BC3, BC4, etc., and approaches the limit zero. This is written cot 270~ = 0. As x increases from 270~ and approaches 360~ as a limit, the cotangent is negative and increases in numerical value from zero through BC5, BC6, etc., without limit. This is written cot 360~ = o. (f) The cosecant. Using the last figure, we see that as x decreases, the cosecant increases through the values OC1, OC2, etc., and as x approaches 0~ as a limit, the cosecant increases without limit. This is written csc 00 = XO. As x increases from 0~ and approaches 90~ as a limit, the cosecant is positive and decreases from + oo through OC8, 0C4, etc., and approaches the limit OB (= 1). This is written csc 90~ = 1. 42 PLANE TRIGONOMETRY As x increases from 90~ and approaches 180~ as a limit, the cosecant is positive and increases from OB (= 1) through OC5, OC6, etc., without limit. This is written csc 180~ = o. As x increases from 180~ and approaches 270~ as a limit, the cosecant is negative and decreases in numerical value from - oo through OC3, OC4, etc., and approaches the limit minus OB (=- 1). This is written csc 270~ = - 1. As x increases from 270~ and approaches 360~ as a limit, the cosecant is negative and increases in numerical value from minus OB (=- 1) through OC5, OC6, etc., without limit. This is written csc 360~ = oo. These results may be written in tabulated form as follows: 0~ 90~ 180~ 270~ 360~ sin o0 1 0 0 cos 0 1 0 1 tan O oo O o6 o cot 0o 0 co 0 co sec 1 c< I c 1 csC c o 1 oo It is of importance to note that as an angle varies its sine and cosine can only take on values between - 1 and + 1 inclusive; tangent and cotangent can take on any values wzhatever; secant and cosecant can take on any values whatever, except those lying between - 1 and + 1. EXAMPLES 1. Prove the following: (a) sin 0~ + cos 90~ = 0. (f) cos 00 + sin 90~ = 2. (b) sin 180~ + cos 270~ = 0. (g) cos 180~ + sin 270~ = - 2. (c) cos 0~ + tan 0~ = 1. (h) sec 0~ + csc 90~ = 2. (d) tan 180~ + cot 90~ = 0. (i) sec 180~ - sec 0 = - 2. (e) sin 270~ - sin 90 = - 2. (j) cos 90~ - cos 270~ = 0. (k) sin 90~ + cos 90~ + csc 90~ + cot 90~ = 2. (1) cos 180~ + sec 180~ + sin 180~ + tan 1800 =- 2. (m) tan 360~ - sin 270~ - csc 270~ + cos 360~ = 3. * The above table is easily memorized if the student will notice that the first four columns are composed of squares of four blocks each, in which the numbers on the diagonals are the same; also the first two columns are identical with the next two if 1 be replaced by -1; also the first and last columns are identical. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 43 2. Compute the values of the following expressions: (a) a sin 0~ + b cos 90~ - c tan 180~. Ans. O. (b) a cos 90~ - b tan 180~ + c cot 90~. 0. (c) a sin 90~ - b cos 360~ + (a - b) cos 180~. 0. (d) (a2 - b2) cos 360~ - 4 ab sin 270~. a2 + 4 ab - b2. 21. Angular measure. There are two systems in general use for the measurement of angles. For elementary work in mathematics and for engineering purposes the system most employed is Degree measure, or the sexagesimal system. * The unit angle is one degree, being the angle subtended at the center of a circle by an arc whose length equals 3-6 of the circumference of the circle. The degree is subdivided into 60 minutes, and the minute into 60 seconds. Degrees, minutes, and seconds are denoted by symbols. Thus 63 degrees 15 minutes 36 seconds is written 63~ 15' 36". Reducing the seconds to the decimal part of a minute, the angle may be written 63~ 15.6'. Reducing the minutes to the decimal part of a degree, the angle may also be written 63.26~.t It has been assumed that the student is already familiar with this system of measuring angles, and the only reason for referring to it here is to compare it with the following newer system. 22. Circular measure. The unit angle is one radian, being the angle subtended at the center of a circle by an arc whose length equals the length of the radius of the circle. Thus, in the figure, if the length of the B arc AB equals the radius of the circle, then angle AOB = 1 radian. X The circular measure of an angle is its ~0 magnitude expressed in terms of radians. This system was introduced early in the last century. It is now used to a certain extent in practical work, and is universally used in the higher branches of mathematics. Both of the above systems will be used in what follows in this book. $ * Invented by the early Babylonians, whose tables of weights and measures were based on a scale of 60. This was probably due to the fact that they reckoned the year at 360 days. This led to the division of the circumference of a circle into 360 degrees. A radius laid off as a chord would then cut off 60 degrees. f To reduce seconds to the decimal part of a minute we divide the number of seconds by 60. Similarly, we reduce minutes to the decimal part of a degree. See Conversion Tables on p. 17 of Granville's Four-Place Tables of Logarithms. * A third system is the Centesimal or French System. The unit is one grade, being 1I5 of a right angle. Each grade is divided into 100 minutes and each minute into 100 seconds. This system has not come into general use. 44 PLANE TRIGONOMETRY Now let us find the relation between the old and new units. From Geometry we know that the circumference of a circle equals 2 7r; and this means that the radius may be measured off on the circumference 2 wr times.* But by the above definition each radius measured off on the circumference subtends an angle of one radian at the center, and we also know that the angles about 0 equal 360~. Therefore 2 7r radians = 360~, 7r radians = 180~, 180~ 180~ 1 radian = 3 or, 7r 3.1416 (16) 1 radian = 57.2957~ +. It therefore follows at once that: To reduce radians to degrees, multiply the number of radians by 57.2957 (= 180). 7T To reduce degrees to radians, divide the number of degrees by 57.2957 (= 180). Since 360 degrees = 2 7r radians, 7r 3.1416 1 degree =. radian ad, or, 180 r -0i180 (17) 1 degree =.01745 radian. Hence the above rules may also be stated as follows: To reduce radians to degrees, divide the number of radians by 01745 (= 180) To reduce degrees to radians, multiply the number of degrees by.01745 (= r) The student should now become accustomed-to expressing angles in circular measure, thus: 360~ = 2 7r radians, 60~ = - radians, 3 180~ = wr radians, 30~ = - radians, 90~= - radians, 45~= radians, 270~ = - radians, 15 = - radians, etc. The student should carefully observe that 12e do not lay o these radii as chords. " The student should carefully observe that we do not lay off these radii as chords. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 45 When writing the trigonometric functions of angles expressed in circular measure it is customary to omit the word " radians," thus: sin (7r radians) is written simply sin wr and = sin 180~, tan(- radians) is written simply tan - and = tan 90~, cot radians)is written simply cot- and = cot 135~, /5t 377.\ _ and_ = cot 135' cos( radians) is written simply cos and = cos 150~, csc (1 radian) is written simply csc 1 and = csc 57.29~, sec(- radian) is written simply sec I and = sec 28.69~, etc. Since the number of times that the radius of a circle can be measured off on an arc of the same circle determines the number of radians in the angle subtended at the center by that arc, we have length of subtending arc (18) Number of radians in angle = le h of length of radius Hence, knowing any two of the three quantities involved, the third may easily be found. Ex. 1. What is the circular measure of the angle subtended by an arc of length 3.7 in. if the radius of the circle is 2 in.? Also express the angle in degrees. Solution. Substituting in (18), we have 3.7 Number of radians = -= 1.85. Ans. 2 To reduce this angle to degrees, we have, from (16), 1.85 x 57.29570 = 105.997~. Ans. Ex. 2. What is the radius of a circle in which an arc of length 64 in. subtends an angle of 2.5 radians? Solution. Substituting in (18), 2.5 = -, R R = 25.6 in. Ans. EXAMPLES 1. In what quadrant does an angle lie * if its sine and cosine are both negative? if sine is positive and cosine negative? if sine is negative and cosine positive? if cosine and tangent are both negative? if cosine is positive and tangent negative? if sine and cotangent are both negative? if sine is negative and secant positive? 2. What signs must the functions of the acute angles of a right triangle have? Why? * That is, in what quadrant will its terminal side lie? 46 PLANE TRIGONOMETRY 3. What functions of an angle of an oblique triangle may be negative? Why? 4. In what quadrant do each of the following angles lie? 57r 7r 77r 147r 117r. 157r. 7 + 2 37r+w 2 1 5 12 6; 3 ' 4; 16' 6 ' 5 2;4;-1;' 2 5. Determine the signs of the six trigonometrical functions for each one of the angles in the last example. 6. Express the following angles in degrees: 1 27r 37r7+1r 1 3 + 2 1.32; -2.; - -; -3; - -2.8; Ans. 74.4844~; 28.6478~; 120~; - 143.239~; - 67.5~; 39.549~; - 171.887~; - 160.4279~; 130.92~. 7. Express the following angles in circular measure: 22-~; 60~; 135~; -720~; 990~; -120~; -100.28~; 45.6~; 142~ 43.2'; -243.87~; 125~ 23' 19". Ans. 0.3926; 1.0472; 2.3562; -12.56; 17.2788; -2.0944; -1.75;.7958; 2.4909; -4.2563; 2.1884. 8. Express in degrees and in radians: (a) Seven tenths of four right angles. (b) Five fourths of two right angles. (c) Two thirds of one right angle. 7w7 5w w Ans. (a). 252~,; (b) 2250, (c) 600, r. 5 4 3 9. Find the number of radians in an angle at the center of a circle of radius 25 ft., which intercepts an arc of 37- ft. Ans. 1.5. 10. Find the length of the arc subtending an angle of 4- radians at the center of a circle whose radius is 25 ft. Ans. 112- ft. 11. Find the length of the radius of a circle at whose center an angle of 1.2 radians is subtended by an arc whose length is 9.6 ft. Ans. 8 ft. 12. Find the length of an arc of 800 on a circle of 4 ft. radius. Ans. 53 ft. 13. Find the number of degrees in an angle at the center of a circle of radius 10 ft. which intercepts an arc of 5 7r ft. Ans. 90~. 14. Find the number of radians in an angle at the center of a circle of radius 32T inches, which intercepts an arc of 2 ft. Ans. 4|. 15. How long does it take the minute hand of a clock to turn through - 13 radians? 50 Ans. - min. 7/ 16. What angle in circular measure does the hour hand of a clock describe in 39 min. 221 sec.? 7 r ~~~2 Anrs. - -rad. 64 17. A wheel makes 10 revolutions per second. How long does it take to turn through 2 radians, taking 7r = _22? Ans. 2- - sec. 18. A railway train is traveling on a curve of half a mile radius at the rate of 20 mi. per hour. Through what angle has it turned in 10 sec.? Ans. 6-4 degrees. 19. The angle subtended by the sunr at the eye of an observer is about half a degree. Find approximately the diameter of the sun if its distance from the observer be 90,000,000 mi. Ans. 785,400 mi TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 47 23. Reduction of trigonometric functions to functions of acute angles. The values of the functions of different angles are given in trigonometric tables, such, for instance, as the one on p. 9. These tables, however, give the trigonometric functions of angles between 0~ and 90~ only, while in practice we sometimes have to deal with positive angles greater than 90~ and with negative angles. We shall now show that the trigonometric functions of an angle of any magnitude whatever, positive or negative, can be expressed in terms of the trigonometric functions of a positive angle less than 90~, that is, of an acute angle. In fact, we shall show, although this is of less importance, that the functions of any angle can be found in terms of the functions of a positive angle less than 45~. In the next eighteen sections x and y denote acute angles. 24. Functions of complementary angles. To make our discussion complete we repeat the following from p. 3. Theorem. A function of an acute angle is equal to the co-function of its complementary acute angle. Ex. Express sin 72~ as the function of a positive angle less than 45~. Solution. Since 90~ - 72 = 18~, 72~ and 18~ are complementary, and we get sin 72~ = cos 18~. Ans. EXAMPLES 1. Express the following as functions of the complementary angle: (a) cos 68~. (e) cot 9.167~. (i) csc 52~ 18'. (b) tan 48.6~. (f) sin 72~ 51' 43". ) c2 -*r ' (j) cot2-. (c) sec810~ 16'. g) os 5 (d) sin.. c6 (k) sin 1.2. 3 (h) sec 19~ 29.8'. (1) tan 66~ 22.3'. 2. Show that in a right triangle any function of one of the acute angles equals the co-function of the other acute angle. 3. If A, B, C are the angles of any triangle, prove that sin A = cos I (B + C). 25. Reduction of functions of angles in the second quadrant. First method. In the unit circle whose center is 0 (see figure on next page), let A OP' be any angle in the second quadrant. The functions of any such angle are the same as the corresponding functions of the positive angle A OP' = 180 ~- P'OQ'. Let x be the measure of the acute angle P'OQ', and construct A OP = P'OQ' = x. 48 PLANE TRIGONOMETRY Now draw the lines representing all the functions of the supplemental angles x and 180 ~- x. From the figure angle QOP = angle P'OQ', by construction OP = OP'. equal radii Therefore the right triangles OPQ and OP'Q' are equal, giving OQ' OQ. T But OQ'= cos (180~- x) and )^.^\ M X OQ = cos x; hence cos (180~- x) 8c/ -</~ | \I \ equals cos x in numerical value. Since they have opposite Qr O\ Q signs, however, we get cos (180 - x)= - cos x. Rl, a r Also, from the same triangles, Q'p' = QP. But Q'P' = sin (180 ~- x) and QP = sin x, and since they have the same sign, we get same sign, we get sin (180~ - x) = sin x. Similarly, the two right triangles OTA and OT'A may be proven equal, giving AT' = AT and OT' = OT, or, tan(180 - x)= - tan x and sec (180 - x)=- sec x. In the same manner, by proving the right triangles OBC and OBC' equal, we get BC' =BC and OC' = OC, or, cot (180~ - x) = - cot x and csc (180 - x) = csc x. Collecting these results, we have sin (180~ - x) = sin x; csc (180 - x) = csc x; cos (180 - x) = - cos x; see (180~ - x) = - sec x; tan (180~- x) = - tan x; cot (180~ - x) =- cot x. Hence we have the Theorem. The functions of an angle in the second quadrant equal numerically the same-named functions of the acute angle between its terminal side and the terminal side of 180~. The algebraic signs, however, are those for an angle in the second quadrant. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 49 Ex. 1. Express sin 123~ as the function of an acute angle, and find its value. Solution. Since 180~ - 123~ = 57~, sin 123~ = sin (180~ - 57~) = sin 57~ =.8387 (p. 9). Ans. Ex. 2. Find the value of se — Solution. sec - = sec 150~ = sec (180~ - 300) = -.se A0ns. Ex. 3. Find tan 516~. Solution. 516~ is an angle in the second quadrant, for 516~ - 360~ = 156~. Hence tan 516~ = tan 156~* = tan (180~ - 24~) = - tan 24~ -.4452. Ans. Second method. The angle A OP' may also be written 90~ + y, where y measures the acute angle BOP'. Since the angles BOP' and P'OQ' are complementary, we have, from theorem on p. 47, sin x = cosy; csc x = sec y; cos x = sin y; sec x= cscy; tan x =coty; cot x = tan y. Since 180~ - x = 90~ + y, we get, combining the above results with the results on the previous page, sin (90~ + y) = cosy; csc (90~ + y) = sec y; cos (90~ + y) = - sin y; sec (90~ + y) = — csc y; tan (90~ + y) =- cot y; cot (90 + y) =- tan y. Hence we have the Theorem. The functions of an angle in the second quadrant equal numerically the co-named functions of the acute angle between its terminal side and the terminal side of 90~. The algebraic signs, however, are those for an angle in the second quadrant. Ex. 4. Find the value of cos 109~. Solution. Since 109~ = 90~ + 19~, cos 109~ = cos (90~ + 19~) - sin 19~ =-.3256. Ans. 197r Ex. 5. Find the value of cos -- 4 19wr Solution. = 855~ = 720~ + 135~. 4 Therefore 19wr 1 cos - = cos 855~ = cos 135~ = cos (90~ + 45~) = - sin 45~ - -. Ans. 4 V2 The above two methods teach us how to do the same thing, namely, how to find the functions of an angle in the second quadrant in terms of the functions of an acute angle. The first method is generally to be preferred, however, as the name of the function does not change, and hence we are less likely to make a mistake. * The above theorem was proven for an angle of any magnitude whatever whose terminal side lies in the second quadrant. The generating line of the angle may have made one or more complete revolutions before assuming the position of the terminal side. In that case we should first (if the revolutions have been counter-clockwise, i.e. in the positive direction) subtract such a multiple of 360~ from the angle that the remainder will be a positive angle less than 360~. 50 PLANE TRIGONOMETRY EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 0~ to lQ^O no:n.n -n_ OnO IOUV ai intlervals 01 oU-. Ans. 00 300 600 900 120~ 1500 180~ 1 1 _ sin 0 - -- 1 -- - 0 2 20 cos! 1 0... 1 2 2 2 2 tan 0 - 0 - 1 A/3- V_ 2. Construct a table of sines, cosines, and tangents of all angles from 90~ to 180~ at intervals of 15~, using table on p. 9. 900 105~ 120~ 1350 1500 165~ 180~ Ans. sin 1.0000.9659.8660.7071.5000.2588 0.0000 cos 0.0000 -.2588 -.5000 -.7071 -.8660 -.9659 -1.0000 tan 00 - 3.7321 -1.7321 -1.0000 -.5774 -.2679 0.0000 3. Construct a table of sines, cosines, and tangents of all angles from 90~ to 135~ at intervals of 5~. 4. Express the following as functions of an acute angle: (a) sin 138~. (b) tan 883~. (c) cos 165~ 20'. (d) sec 102~ 18'. 47w (e) cot -. (f) cot 170.48~. (g) csc 317~. 13 7 (h) sin - (i) cos 2.58. (j) tan 1.5. 5. Find values of the following: (a) sin 128~ =.788. (b) cos 160~ =-.9397. (c) tan 135~= - 1. 2w7 (d) sec — =- 2. 3 (e) cot = — 1. 4 (f) csc 835~= 1.1034.. 87r (g) sin -. 7 (h) tan 108~ 15'. (i) cos 173~ 9.4'. 5w7 (j) tan. 6 (k) cos 496.7~. (1) sec 168.42~. (m) cot 95~ 14'. (n) csc126~ 42.8'. 7 7 (o) sin9' (p) cos 500~. (q) tan 870~. (r) sec 1.9~. (s) tan 1. 6. Express the following as functions of an acute angle less than 45~: (a) sin 1060 = cos 16~. 117r (b) cos 148.3~ = - cos 31.7. c 12 (c) tan 862~. 23 7r (d) sec794~ 52'. (f) cos9 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 51 26. Reduction of functions of angles in the third quadrant. First method. In the unit circle whose center is 0, let A OP' be any angle in the third quadrant. The functions of any such angle are the same as the corresponding functions of the positive angle AOP'=180~+ Q'OP'. Let x be the measure of the acute angle Q'OP', and construct A OP = Q'OP' = x. Now drawing the lines representing all the functions of the angles x and 180~ + x, we get, just as in the previous case, sin (180 + x) = - sin x; csc (180~ + x) = - csc x; cos (180~ + x) = - os x; sec (180~ + x) = - se x; tan (180~ + x) = tan x; cot (180~ + x) = cot x. Hence we have the Theorem. The functions of an C angle in the third quadrant equal / \ numerically the same-named functions of the acute angle between its 0 terminal side and the terminal side Q' x of 180~. The algebraic signs, how- \ ever, are those for an angle in the third quadrant. p =1 Ex. 1. Express cos217~ as the function of an acute angle, and find its value. B' Solution. Since 217~ - 180~ = 37~, cos 217 = cos(180~ + 37~) =- cos 37 = -.7986. Ans. Ex. 2. Find value of csc 225~. Solution. csc 225~ = csc (180~ + 45~) =- csc 45~ =- /2. Ans. Ex. 3. Find value of sin 600~. Solution. 600~ is an angle in the third quadrant, for 600~ - 360~ = 240~. Hence sin 600~ = sin 240~ = sin (180~ + 60~) = -sin 60~ = -- Ans. 2 Second method. The angle A OP' may also be written 270~ - y, where y measures the acute angle P'OB'. Since the angles P'OB' and Q'OP' (= A OP) are complementary, we have, from theorem on p. 47, combined with the above results, remembering that 180~ + x = 270~ - y, sin (270~ - y) =- cos y; csc (270~ - y) =- sec y; cos (270 - y) = - sin y; sec (270~ - y) = - csc y; tan (270~ - y) = cot y; cot (270~ - y) = tan y. Hence we have the Theorem. The functions of an angle in the third quadrant equal numerically the co-named functions of the acute angle between its terminal side and the terminal side of 270~. The algebraic signs, however, are those of an angle in the third quadrant. 52 PLANE TRIGONOMETRY Ex. 4. Find sin 259~. Solution. Since 2700 - 11~ = 259~, sin 259~ = sin (270~ - 11~) = - cos 11 =-.9816. Ans. As in the last case, the first method is generally to be preferred. EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 0~ to 270~ at intervals of 45~. 0~ 45~ 90~ 135~ 180~ 225~ 270~ sin 0 1 0 - -- Ans. V2 \2 V2 1 1 1 cos 1 0 - - -1 -- 0 ta n2 2 12o 1 tan 0 1 oo — 1 0 1 2. Construct a table of sines, cosines, and tangents of all angles from 180~ to 270~ at intervals of 15~, using table on p. 9. 180~ 1950 210~ 225~ 240~ 255~ 270~ Ans. sin 0 -.2588 -.5000 -.7071 -.8660 -.9659 - 1.0000 cos- 1.0000 -.9659 -.8660 -.7071 -.5000 -.2588 0 tan 0.2679.5774 1.0000 1.7321 3.7321 o 3. Construct a table of sines, cosines, and tangents of all angles from 135~ to 270~ at intervals of 5~. 4. Express the following as functions of an acute angle: (a) tan 200~. 7 7r (e) cot - (b) sin 583~. 5 (c) cos224~ 26'. (f) csc4.3. (d) sec 260~ 40'. (g) sin 128~. 5. Find values of the following: (a) tan 235~ = 1.4281. (b) cot 1300~ = 1.1918. (c) sin 212~ 16'. 47w 1 (d) cos — 3 2 77r 2 (e) sec- 7- 2_. 6 <3 13rr (g) cos —. 12 (h) tan 4. 29 rr (i) cot9 -. 9 21 wr (j) CSG-. 4 (h) cos 998.7~.. 167r (i) sin-. 5 87r (j) cos-. 3 (m) cot 185~ 52'. (n) cos 587~. 8rr (o) csc- -. 7 (p) sin 262~ 10'. (q) cos 204.86~. 9w (r) tan. 8 (f) sin 609~. (k) sin 228.4~. (1) tan 255~ 27.8'. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 53 6. Express the following as functions of an acute angle less than 45~: 17 rr (c) sin 594~. 23 wr (a) cos 1 —. 5 (e) sin 12 5 7 (d) sec —. (b) tan 236.5~. 4 (f) cos 260~ 53.4' 27. Reduction of functions of angles in the fourth quadrant. First method. As before, let quadrant. The functions of any such angle are the same as the corresponding functions of the positive angle A OP' =360~ - P'OQ. Let x be the measure of the acute angle P'OQ, and construct A OP = P'OQ = x. Now, drawing the lines representing all the functions of the angles x and 360~- x, we get, just as in the previous cases, A OP' be any angle in the fourth ~B> II R=1 BI sin (360~ - x)=- sin x; cos (360 - x) = cos x; tan (360~ - x) = - tan x; cse (360 - x) = - cse x; sec (360 - x)= sec x; cot (360~ - x)= - cot x; Hence we have the Theorem. The functions of an angle in the fourth quadrant equal numerically the same-named functions of the acute angle between its terminal side and the terminal side of 360~. The algebraic signs, however, are those for an angle in the fourth quadrant. Ex. 1. Express sin 327~ as the function of an acute angle, and find its value. Solution. Since 360~ - 327~ = 33~, sin 327~ = sin (360 - 33~) = - sin 33 = -.5446. Ans. 5wr Ex. 2. Find value of cot -. 3 5wr Solution. cot - = cot 300~ = cot (360~ - 60~) = - cot 60~ =- 1 Ans. 3 V Ex. 3. Find value of cos 1000~. Solution. This is an angle in the fourth quadrant, for 1000~ - 720~ = 280~. Hence cos 1000~ = cos 280~ = cos (360~ - 80~) = cos 800 =.1736. Ans. 54 PLANE TRIGONOMETRY Second method. The angle AOP' may also be written 270 + y, where y measures the acute angle B'OP'. Since the angles B'OP' and P'OQ are complementary, we have, from theorem on p. 47, combined with the above results, remembering that 360~ - x = 270~ + y, sin (270~ + y) = - cos y; csc (270~ + y) = - sec y; cos (270~ + y) = sin y; sec (270~ + y) = csc y; tan (270~ + y) =- cot y; cot (270~ + y) =- tan y. Hence we have the Theorem. The functions of an angle in the fourth quadrant equal numerically the co-named functions of the acute angle between its terminal side and the terminal side of 270~. The algebraic signs, however, are those of an angle in the fourth quadrant. 11r Ex. 4. Find value of cos - 6 Solution. cos - = cos 330~ = cos (270~ + 60) = sin 60~ Ans. 6 2 As before, the first method is generally to be preferred. EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 180~ to 360~ at intervals of 30~. Ans. 1800 210~ 2400 2700 300~ 330~ 3600 * 0 1 1 - 3 1 0 sin 0 - -/3 -1 - - - 0 2 2 2 2 1/ 3 1 1 V3 cos -1 - - - 0 - 1 2 2 2 2 — ~ 1 tan 0 1 00 -V - 0 V a1 3 2. Construct a table of sines, cosines, and tangents of all angles from 270~ to 360~ at intervals of 15~, using table on p. 9. 270~ 285~ 300~ 315~ 330~ 345~ 360~ Ans. sin -1.0000 -.9659 -.8660 -.7071 -.5000 -.2588 0 cos 0.2588.5000.7071.8660.9659 1.0000 tan 0o -3.7321 -1.7321 -1.0000 -.5774 -.2679 0 3. Construct a table of sines, cosines, and tangents of all angles from 270~ to 360~ at intervals of 5~. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 55 4. Express the following as functions of an acute angle: (a) sin 289~. (e) sin 655~. (b) cos 322.4~. 9 r (c) tan 295~ 43'. ) c5 (d) cot 356~ 11'. (g) sin 275.5~. 5. Find values of the following: (a) sin 275 = -.9962. (b) cos 336~ =.9135. (c) tan 687~ = -.6494. (d) cot 1055~. (e) sec 295~ 52.6'. 17rr (f) sin —. 9 (g) csc 5.2. 15 7r (h) cos --- 11 7r (i) csc —. 6 5wr (j) tan-. 3 18 w (h) cos —. 5 (i) sec 246~. 3r (j) tan. (k) sin 275~ 22'. (1) cot 348~. (m) tan 660~. 13 r (n) sec —. 8 57r (o) sin-. 6 28. Reduction of functions of negative angles. Simple relations exist between the functions of the angle x and - x where x is any angle whatever. It is evident that x and - x will lie, one in the first quadrant and the other in the fourth quadrant, as angles A OP and A OP' in the right-hand figure; or, one will lie in the second quadrant and the other in the third quadrant, as the angles A OP and A OP' in the left-hand figure. In either figure, remembering the rule for signs (~ 16, p. 29), we get QP =- QP',.'. sin x = - sin (- x); OQ = OQ,.'. cos x = cos (- x); AT = -AT',.'. tan x = - tan (- x); OT = OT',.'. see x = see(- x); BC =-BC',.'. cot x = - cot(- x); OC =-OC',.'. csc x =- csc(-x). We may write these results in the form sin (- x) =- sin x; cos(-x)= cosx; tan (- x) =- tan x; csc (- x) = - csc x; sec(- x)= sec x; cot (- x) = - cot x. 56 PLANE TRIGONOMETRY Hence we have the Theorem. The functions of - x equal numerically the same-named functions of x. The algebraic sign, however, will change for all functions except the cosine and secant.* Ex. 1. Express tan (- 29~) as the function of an acute angle, and find its value. Solution. tan (- 29~) = - tan 29~ =-.5543. Ans. Ex. 2. Find value of sec (- 135~). Solution. sec (-135~) = sec 135~ = sec (180~ - 45) =- sec 45~ = -. Ans. Ex. 3. Find value of sin (- 540~). Solution. sin (-540)= - sin 540 = - sin (360~+ 180~) = -sin 180= 0. Ans. EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 0~ to - 360~ at intervals of 30~. Ans. Angle sin cos tan 2. Find values of the following: 00 0 1 0 (a) tan (- 33~)= -.6494. -1 V 1 (b) sin (-600) = —. 1 - 600 - 1 - 3 (c) cos(-145) = —. 2 2 Va/24 -90~ -1 0 oo (d) cot (-259). 3 (e /e) c - 120~ - -- /3 (e) se 6-). 2 2_ 1 \/ 1 (f) sin (-12310). - 1500 - 2 (g) cos (-. -1800 -1 0 1 3 1 (h) sin (-1000). -210~ 2 2 V _3 (i) cos (-2.3). (-3 57rt/n Adin 240~ -2 g /3 (j) cot(- )2 2. 2 2 \ 8 / - 2700 1 0 oo (k) sin(-176.9~0). 1 (1) cos (- 880 12.7') - 3002 - 3 (m) tan 4 - ~ 3300 -1 _ _1 2 2 V3 (n) cot (- 8420). - 300~ 0 1 0 * Another method for reducing the functions of a negative angle consists in adding such a multiple of + 3600 to the negative angle that the sum becomes a positive angle less than 3600. The functions of this positive angle will be the same as the functions of the given negative angle, since their terminal sides will coincide. To illustrate: Ex. Find value of cos (- 2400). Solution. Adding + 3600 to - 2400 gives + 1200. Hence cos (- 2400) = cos 120~ = cos (1800 - 600) = - cos 600 = -. Anrs. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 57 29. General rule for reducing the functions of any angle to the functions of an acute angle. The results of the last seven sections may be stated in compact form as follows, x being an acute angle.* General Rule. I. Whenever the angle is 180~ ~ x or 360~ ~ x, the functions of the angle are numerically equal to the same-named functions of x. II. Whenever the angle is 90~ ~ x or 270~ ~ x, the functions of the angle are numerically equal to the co-namedfunctions of x. III. In any case the sign of the result is the same as the sign of the given function taken in the quadrant where the given angle lies. The student is advised to use I wherever possible, since the liability of making a mistake is less when the name of the function remains unchanged throughout the operation. Work out examples from pp. 50-56, applying the above general rule. EXAMPLES 1. Construct a table for every five degrees from 90~ to 180~. Angle sin cos tan cot- sec csc -.082 -.0875. 1.0038 Ans. 90~ 950 100~ 1.0000.9962.9848 e 0.0000 -.0872 -.1736 so -11.430 -5.6713 0.0000 -.0875 -.1763,. *o o00 - 11.474 -5.7588... 1.0000 1.0038 1.0154.. 2. Construct a table as in Ex. 1 for every 150 from 180~ to 270~. 3. Construct a table as in Ex. 1 for every 10~ from 0~ to - 90~. 4. Reduce the following to functions of x: (a) sin (x- 90) = - cosx. t (b) cos (x - r) =- cos x. (c) tan( —x -- = cot x. (d) cot (x - 2 r) = cot x. (e) sec (x - 180~) = - sec x. (f) csc(- -2 ) =- se x. \ 2/ (g) sin (x - 270~). (h) cos (- x - 7r). (i) tan -- ). (j) cot(- x-8 7r). (k) see (x - 6300). (1) csc (x - 720~). 4 In case the given angle is greater than 360~ we assume that it has first been reduced to a positive angle less than 3600 by the subtraction of some multiple of 360~. Or, if the given angle is negative, we assume that it has been reduced to a positive angle by the theorem on p. 56. t Since x is acute, x - 90~ is a negative angle. Hence sin (x - 900) = - sin (90~ - x ) = - cos x. 58 PLANE TRIGONOMETRY 5. Find values of the following: (a) cos 420~. (i) cot 549~ 39'. (b) sin 768. (j) csc387~ 58'. (c) se(- (k) s(- 1 ). (d) cot(-240). (1) tan(- ) 13 7rr ( \ 4 / (e) csc —. 3 (m) sin(- 830~). (f) tan 7.5. 9 r (g) sin(- 2.8). (n)cos (h) cos 952.8~. (o) cot 1020~. 13 7T (p) sec- -. (q) sin(- 5.3). 23 7r (r) cos (- 12/' (s) tan (- 7)r. (t) sec(- 123.8~). (u) sin (- 256~ 19.6'). (v) cos(- 98~ 31'). 6. Prove the following: (a) sin 420~0 cos 390~ + cos (- 300~). sin (- 330) = 1. (b) cos 570~ sill 510~ - sin 330~ cos390~ = 0. (c) a cos (900 - x) + b cos (90~ + x) = (a - b) sin x. (d) m cos (-x). sin(-x) = m sin x cos x. (e) (a - b) tall (90~0- x) + (a + b) cot (90~ + x) = (a - b) cot x - (a + b) tan x. (f) sill (- + x)sin (r + x) + cos ( + x) cos ( - x) = 0. (g) cos(7r+x) cos(3 -) - sin( x)sin -y) =cosxsiny-sinx cosy. (h) tan x + tan (- y) - tan (7r - y) = tan x. (i) cos (90~ + a) cos (2700 - a) - sin (1800 - a) sin (360~ - a) = 2 sin2 a. sin (1800 - y) t___1___ 9 + y (j) tan (900 + y) +3 - " 1 - seC2y. sin (270~ - y) sin2 (270~ - y) (k) 3 tan 210~ + 2 tan 120~ = - -. (n) tan 7 (2 +x) = tan x. (1) sec2 136~ - 6 cot2300 = -8. (o) csc (x - 2 7r) - sec I x. (m)cos(x -270) = sin ~ x, (p) cos ( (y - 810) = -sin I y. CHAPTER III RELATIONS BETWEEN THE TRIGONOMETRIC FUNCTIONS 30. Fundamental relations between the functions. From the definitions (and footnote) on p. 29 we have at once the reciprocal relations (19) sin x, =, csc x =; csc x sin x 1 1 (20) cos x = -, sec x = -; sec x Cos X 1 1 (21) tanx = -, cotx = - cot x tan x Making use of the unit circle, we shall now derive five more very important relations between the functions. In the right triangle QPO B Cot QP OQ tan x-, and cotx=. T = 6 Substituting the functions equivalent to QP and OQ, we get A sinx cos \ (22) tanx= —, cotx=. cosx sinx R= Again, in the same triangle, - 2 2 2 Qp + OQ2= P2, or, (23) sin2 x + cos2 x = 1. 5 2 2 2 In triangle OA T, OA +AT = OT2 or, (24) 1 +tan2x= sec2x. j 2 2 2 In triangle OCB, OB + BC = OC2, or (25) 1 + cot2 x CaC2x While in the above figure the angle x has been taken in the first quadrant, the results hold true for any angle whatever, for the above 59 60 PLANE TRIGONOMETRY proofs apply to any one of the figures on p. 36 without the change of a single letter. While it is of the utmost importance to memorize formulas (19) to (25), p. 59, as they stand, the student should also learn the following formulas where each one of the functions is expressed explicitly in terms of other functions. (26) sin x = -1 Csc X (27) sin x = d /1 - cos2 x.* (28) cos x =. sec x (29) cos x = 4 V1 - sin2 x. (30) tanx = - cot x (19), p. 59 Solving (23), p. 59, for sin x (20), p. 59 Solving (23), p. 59, for cos x (21), p. 59 (31) tan x = + V/sec2 x-1. Solving (24), p. 59, for tan x sin x sin x H V - cos2 (32) tanx= = = cos x VI-/ - sin2 x cos x [From (22), P. 59; also (29) and (27).] (33) csc x = -1-. sin x (34) csc x = = /1 + cot2 x. (35) sec x = —. cos x (36) sec x = d= /1 + tan2 x. (37) cot x = 1 tan x (38) cot x = d4= csc2 x-1. cos x cos x (19), p. 59 Solving (25), p. 59, for csc x (20), p. 59 Solving (24), p. 59, for sec x (21), p. 59 Solving (25), p. 59, for cot x - -/1-sin2x (39) cot x = --- = -- =cs sinx x l/-cos2x sin x [From (22), p. 59; also (27) and (29).] 31. Any function expressed in terms of each of the other five functions. By means of the above formulas we may easily find any function in terms of each one of the other five functions as follows: * The double sign means that we get two values for some of the functions unless a condition is given which determines whether to choose the plus or minus sign. The reason for this is that there are two angles less than 360~ for which a function has a given value. TRIGONOMETRIC RELATIONS 61 Ex. 1. Find sin x in terms of each of the other five functions of x. (a) sinx =, from (26) CsC X (b) sin x = V/1 -cos2, from (27) (c) sin x =, substitute (34) in (a):1V+ cot2 x 1 + a/see2 x - 1 (d) sin x = - = 1 sx-, substitute (28) in (b) sec2x sec x (e) sin x = 1 tan Substitute (37) in (c) l +tan21 ~ /tan2x + 1 \ tanI2 x Ex. 2. Find cos x in terms of each of the other five functions. (a) cosx =, from (28) sec x (b) cosx = qV 1-sin2x, from (29) (c) cos =, substitute (36) in (a) -/1 + tan2 x 1 _ 4-csc2x -1 (d) cos x = i~ 1- /csc2xc substitute (26) in (b) csc2 x csc X 1 cot x (e) cosx = 1 = cox Substitute (30) in (c) (e )X Vcot2 + 1 1 A: cot2x+1 cot2 a Ex. 3. Find tan x in terms of each of the other five functions. (a) tan x = -, from (30) cot x (b) tanx =-\/sec2 - 1, from (31) sin x (c) tan =, from (32) V1- sin2 x - x/ - cos2 a (d) tanx = -- c, from (32) COs X (e) tanx =. Substituting (38) in (a) ~- csc2 - 1 Ex. 4. Prove that sec x - tan x ~ sin x = cos x. Solution. Let us take the first member and reduce it by means of the formulas (26) to (39), p. 60, until it becomes identical with the second member. Thus 1 sin x see x - tan x * sin x = -.sin x cos x cos x |c.1 sin x*] Since sec x = and tan x = cos x cos x 1 -sin2x cos2 (.cos.x - Cos (23), p. 59 cos x cos xa = cos X. Ans. * Usually it is best to change the given expression into one containing sines and cosines only, and then change this into the required form. Any operation is admissible thatdoes not change the value of the expression. Use radicals only when unavoidable. 62 PLANE TRIGONOMETRY Ex. 5. Prove that sin x (see x + csc x) - cos x (sec x - csc x) = sec x csc x. Solution. sin x (sec x + csc x) - cos x (sec x - csc x) 1 = sin x + -C — cos x.os ) \cos sin x cos smin [Since sec = and csc x =. cos x sln x sin x cos X -- q- 1-1 lcos x sin x sin x cos x cos x sin x sin2 x + cos2 X 1p. = ------- = -— (2 ), r. 59 cos x sin x cos x sin x 1 1 -= - =secx cscx. Ans. cos x sin x 'EXAMPLES 1. Find sec x in terms of each of the other five functions of x. 1 Ans., ~,/ 1 tan2cx ot2 x + 1 csc X Ans. 1, + V t a cos x - /1 - sin2 x cot x csc2x - 1 2. Find cot x in terms of each of the other five functions of x. Ans. -, - VC Xcsc2 X-, - - tan x sin x I +1-cos2 x V/sec2 x-1 3. Find csc x in terms of each of the other five functions of x. A 1 s x + cot2 I 1 + tan2 x +1 sec x Ans. I,: 1 + cot2, 7 smn x q- /1 -cos2 x tan x ~ sec2x - 4. Prove the following: (a) cosx tan = sinx. (h) cosA cscA = cotA. (b) sin x sec x = tan x. (i) cos2A - sin2A = 1- 2 sin2A. (c) sin y cot y = cosy. (j) CC2A - sin2A = 2 cos2A - 1. (d) (1 + tan2 y) cos2 y = 1. (k) (1 + cot2 B) sin2 B = 1. (e) sin2A + sin A tan2A = tan2A. (1) (csc2A - 1) csc2A = cos2A. (f) cotA - cos2A = cot2A cos2A. (m) sec2A + csc2A = sec2A csc2A. (g) tan A + cotA = secA cscA. (n) cos4 C sin4 C + 1 = 2 cos2 C. (o) (sin x + cos x)2 + (sin x - cos x)2 = 2. (p) sin3 x cos x + cos3 x sin x = sin x cos x. (q) tanA + cotA = secA cscA. sin y (r) coty + s = cscy. '1 cos y (s) cos B tan B + sin B cot B = sin B + cos B. (t) sec x csc x (cos2 x sin2 x) = cot x - tan x. cos C sin C (u) CsC s+ i = sin C + cos C. 1 - tan C 1 - cot C sin z 1 - cos z (v) n z + s = 2 csc z. 1 - cos z sin CHAPTER IV TRIGONOMETRIC ANALYSIS 32. Functions of the sum and of the difference of two angles. We now proceed to express the trigonometric functions of the sum and difference of two angles in terms of the trigonometric functions of the angles themselves.* The fundamental formulas to be derived are: (40) sin (x + y) = sin x cos y + cos x sin y. (41) sin (x - y) = sin x cos y - cos x sin y. (42) cos (x + y) = cos x cos y - sin x sin y. (43) cos (x - y) = cos x cos y + sin x sin y. 33. Sine and cosine of the sum of two angles. Proofs of formulas (40) and (42). Let the angles x and y be each a positive angle less than 90~. In the unit circle whose center is 0, lay off the angle A OP = x and the angle POQ = y. Then the angle AOQ = x + y. Q P / F D A. A 0 E C ' E In the first figure the angle x + y is less than 90~, in the second greater than 90~. Draw QC perpendicular to OA. Then (a) sin (x + y)= C Q, and (b) cos (x + y) = OC. Draw QD perpendicular to OP. Then (c) sin y = DQ, and (d) cosy= OD.t * Since x and y are angles, their sum x + y and their difference x - y are also angles. Thus if x = 61~ and y = 23~, then x + y =84~ and x - y = 38~. The student should observe that sin (x + y) is not the same as sin x + sin y, or cos(x - y) the same as cos x - cos y, etc. t The student will see this at once if the book is turned until OP appears horizontal. 63 64 PLANE TRIGONOMETRY Draw DE perpendicular and DF parallel to OA. Then angle DQF = angle A OP (= x), having their sides perpendicular each to each. From (a), (e) sin(x + y) = CQ = CF + FQ = ED + FQ. ED being one side of the right triangle OED, we have But from (d), (f) ED = OD sin x. OD = cos y. Therefore ED = sin x cos y. from (7), p. 11 FQ being one side of the right triangle QFD, we have But from (c), (g) FQ= DQ. cos x. DQ = siny. Therefore FQ = cos x sin y. from (8), p. 11 Substituting (f) and (g) in (e), we get (40) sin(x + y) = sinx cos y + cosx siny. To derive (42) we use the same figures. From (b), (h) cos (x + y) = OC = OE - CE = OE - FD.* OE being one side of the right triangle OED, we have But from (d), (i) OE = OD cos x. OD = cos y. Therefore OE = cos x cos y. from (8), p. 11 FD being a side of the right triangle QFD, we have But from (c), (j) FD = DQ sin x. DQ = sin y. Therefore FD = sin x sin y. from (7), p. 11 Substituting (i) and (j) in (h), we get (42) cos (x + y) = cos x cosy - sin x siny. In deriving formulas (40) and (42) we assumed that each of the angles x and y were positive and less than 90~. It is a fact, however, that these formulas hold true for values of x and y of any magnitude whatever, positive or negative. The work which follows will illustrate how this may be shown for any particular case. * When x + y is greater than 900, OC is negative. TRIGONOMETRIC ANALYSIS 65 Show that (42) is true when x is a positive angle in the second quadrant and y a positive angle in the fourth quadrant. Proof. Let x = 900 + x' and y = 2700 + y'*; thenx + y = 360~ + (x' + y') and (k) x'= x - 900, y' = y - 270~, x'+ y' = x + y - 360~. cos (x + y) = cos [3600 + (x' + y')] = cos (x' + y') by ~ 29, p. 57 = cos x' cos y' - sin x' sin y' by (42) = cos (x - 900) cos (y - 270) - sin (x - 900) sin (y - 270~) from (k) = sin x (- sin y) - (- cos x cos y) by ~ 29, p. 57 = cos x cos y - sin x sill y. QE.D. Show that (40) is true when x is a positive angle in the first quadrant and y a negative angle in the second quadrant. Proof. Let x = 90 - x' and y =-180 - y'; then x + y =-900 -(x'+ y') and (1) x' = 90 - x, y' =- 180 - y, x' + y' =-900 - (x + y). sin (x + y) = si[- 90 - (x' + y')] =- cos (x' + y') by ~ 28, p. 56 = - [cos x' cos y' - sin x' sin y'] by (40) =- [cos(900-x) cos (-1800-y)-sin (900-x) sin(-1800-y)] from (1) = - [sin x (- cos y) - cos x sill y] by ~ 29, p. 57 = sin xs c osy + sin y. Q.E.D. EXAMPLES 1. Find sin 75~, using (40) and the functions of 450 and 30~. Solution. Since 750 = 450 + 30~, we get from (40) sin 750 = sin (450 + 300) = sin 450 cos 30~ + cos 45~ sin 30~ 1 v3 1 1 --..+.1 from p. 5 +/ 2 + 2 2 /3+1 -= — * Ans. 2 V2 2. Find cos(x + y), having given sinx = 3 and sin y = -5, x and y being positive acute angles. Solution. By the method shown on p. 30 we get first sinx=, CS X = 4, s = 1, cos y 12 Substituting these values in (42), we get cos (X + y)= * 3 = 35 Ans. 6 T Y 5 S - T -j - -g. * The student should note that x' and y' are acute angles. 66 PLANE TRIGONOMETRY 3. Show that cos 75~ =, using the functions of 45~ and 30~. 2V2 4. Prove that sin 90~ = 1 and cos 90~ = 0, using the functions of 60~ and 30~. 5. If tan x = 3 and tan y = -, find sin (x + y) and cos (x + y) when x and y are acute angles. Ans. sin (x + y) =, cos (x + y) = 3 6. By means of (40) and (42) express the sine and cosine of 90~ + x, 180~ + x, 270~ + x, in terms of sin x and cos x. 7. Verify the following: (a) sin (450 + x) = + si (c) sin (y 30) cosA - \3 sin A cosB - sin B (b) cos(60 +A)= - (d) cos (B + 45) = cos - 8. Find sin (A + B) and cos (A + B), having given sin A = - and sin B = 2. Ans. sin(A + B)= 2 /3 cos(A + B) =2 6 6 34. Sine and cosine of the difference of two angles. Proofs of formulas (41) and (43). It was shown in the last section that (40) and (42) hold true for values of x and y of any magnitude whatever, positive or negative. Hence (41) and (43) are merely special cases of (40) and (42) respectively. Thus, from (40), sin (x + y) = sin x cos y + cos x sin y. Now replace y by - y. This gives (a) sin (x - y) = sin x cos (- y) + cos x sin (- y). But cos (- y) = cos y, and sin (- y)= - sin y. from p. 55 Substituting back in (a), we get (41) sin (x- y) = sin x cos y - cos x sin y. Similarly, from (42), cos (x + y) = cos x cos y - sin x sin y. Now replace y by - y. This gives (b) cos (x - y) = cos x os (- y) - sin x sin(-y). But cos (- y) = cos y, and sin (- y) =- sin y. from p. 55 Substituting back in (b), we get (43) cos (x - y) = cos x cos y + sin x sin y. TRIGONOMETRIC ANALYSIS 67 EXAMPLES 1. Find cos 15~, using (43) and the functions of 45~ and 30~. Solution. Since 15~ = 45~ - 30~, we get from (43) cos 15~ = cos (45~ - 30~) = cos 45~ cos 30~ + sin 45~ sin 30~ 1 V/ 1 1 +/ 2 '/ i 2 _/3+1 =- ' Ar8s. 22 Work out the above example, taking 15~ = 60~ - 450. 2. Prove sin (60~ + x) - sin (60~ - x) = sin x. Solution. sin (60~ + x) = sin 60~ cos x + cos 60~ sin x. by (40) sin (600 - x) = sin 60 cos x - cos 60~ sin x. by (41).. sin (60~ + x) - sin (60~ - x) = 2 cos 60~ sin x by subtraction = 2 in sin= sinx. Ans. v3 -1 3. Show that sin 15~ =, using the functions of 45~ and 30~. 2V2 4. Find sin(x - y) and cos(x - y), having given sin = and sin = 3, x and y being acute angles. 2V -Vi5 2V30 +1 Ans. sin(x - y) =, cos(x -y) = 1 12 12 5. Find sin (x - y) and cos (x - y), having given tan x = 4 and tan y = 3, x and y being acute angles. Ans. sin (x - y) = 7, cos (x - y).= 2 4 6. By means of (41) and (43) express the sine and cosine of 90~ - x, 180~ - x, 270~ - x, 360~ - x, in terms of sin x and cos x. 7. Verify the following: cosx - sinx VFx ~ y - cos y (a) sin(450 - x) = s (c) sin (y - 30) = -V J~~~2 2 cos A + -\3sin A cosB - sinB (b) cos(60 -A)= cA + (d) cos(B + 45) 2 '\/2 (e) sin (60~ + x) - sin x = sin (60~ - x). (f) cos (30~ + y) - cos (300 - y) = - sin y. (g) cos (45~ + x) + cos (45~ - x) = 2 cos x. (h) sin (45~ + P) - sin (45~ - P) = /2 sinP. (i) cos (Q + 45~) + sin (Q - 45) = 0. (j) sin (x + y) sin (x - y) = sin2x - sin2y. (k) sin(x + y + z) = sin x cosycosz + cos xsin ycosz + cos x cos y sin z - sin x sin y sin z. Hint. sin(x + y + z)= sin (x + y) cos z + cos (x + y) sin z. 68 PLANE TRIGONOMETRY 35. Tangent and cotangent of the sum and of the difference of two angles. From (22), p. 59, and (40) and (42), p. 63, we get t a nsin (x + y) sin x cos y + cos x sin y ) cos (x + y) cos x cos y - sin x sin y Now divide both numerator and denominator by cos x cos y. This gives sin x cos y cos x sin y tan (x cos x co) CO cos x cos y tan (x + y) = cos x cos y sin x sin y cos x cos y cos x cos y sin x siny cos x cos y sinx siny cos x cos y sin x sin y Since s= tan x and -s ='tan y, we get cos x cos y tan x + tan y (44) tan (x + y) = 1 t — tan y 1 - tanxtany In the same way, from (41) and (43) we get A\~(45) t/ ( tan x - tan y (45) tan (x- y)= 1 tan x tan y From (22), p. 59, and (40) and (42), p. 63, we get cos (x + y) cos x cos y - sin x sin y sin (x + y) sin x cos y + cos x sin y Now divide both numerator and denominator by sin x sin y. This gives cos x cos y sin x sin y ot ( si sn x sin y sin x sin y sin x cos y cos x sin y sinx siny sin x sin y cos x cos y sin x sin y cos y + cos x y siny sinx cos X cos y Since cos = cot x, and co = cot y, we get sin x sin y cot x coty - 1 (46) cot (x + y) = coty + cot — cot y + cot x TRIGONOMETRIC ANALYSIS 69 In the same way, from (41) and (43) we get (47) cot x cot y + 1 cot (x - y) = cot y -cot cot y - cot x Formulas (40) to (47) may be written in a more compact form as follows' follows: sin (x y) = sin x cos y i cos x sin y, cos (x ~ y) = cos x cos y: sin x sin y, tan x ~ tan y tan(x y)= 1:F tan x tan y cot x cot y 5F 1 cot (x i M)=- ) cot y ~ cot x The formulas derived in this chapter demonstrate the Addition Theorem for trigonometric functions, namely, that any function of the algebraic sum of two angles is expressible in terms of the functions of those angles. EXAMPLES 1. Find tan 15~, using (45) and the functions of 60~ and 45~. Solution. Since 15~ = 60~ - 45, we get from (45) tan 600 - tan 450 1a/3. -ns tan 15~=tan (60~-45~)= tan60~-t 45~ =2-V. Ans. 1 + tan 60~ tan 45~ 1 + -V3 Work out above example, taking 15~ = 45~ - 30~. 2. Find tan (x + y) and tan (x - y), having given tan x = - and tan y = 4. Ans. tan (x + y) = 6, tan (x - y) =. 3. Find tan 75~ from the functions of 45~ and 30~. Ans. 2 + /3. 4. Verify the following: 1 ~ tan x (a) tan (45? + x) = 1 tan 1 + cot y (b) cot (y- 450) = 1 - coty tan A - \/3 (c) tan (A - 60~)= 1V-3ta 1 + V tan A (d) cot (B + 300) = /cotB-1 cot B + V (e) tan (x + 45~) + cot (x TF 45~) = 0. sin (x + y) tan x + tan y sin (x - y) tan x - tan y sin (x + y) (g) tan x + tan y = ( + cos x cos y (h) cotA - cot B A) sin A sin B cos (x + y) (i) 1-tan x tan y = — Cs cos x cos y (j) cotPcotQ - os(P+Q) sin P sin Q 36. Functions of twice an angle in terms of the functions of the angle. Formulas (40) to (47) were shown to hold true for all possible values of x and y; hence they must hold true when x equals y. To find sin 2 x we take (40), sin (x + y) = sin x cos y + cos x sin y. 70 PLANE TRIGONOMETRY Replace y by x. This gives sin (x + x) = sin x cos x + cos x sin x, or (48) sin 2 x= 2 sin x cos x. To find cos 2 x we take (42), cos (x + y) = cos x cos y - sin x sin y. Replace y by x. This gives cos (x + x) = cos x os s - sin x sin x, or (49) cos 2 x = cos2x - sin2x. Since cos2x = 1 - sin2x, (49) may be written (49 a) cos 2 x = 1- 2 sin2x. Or, since sin2x = 1 - cos2x, (49) may also be written (49 b)- cos 2 x = 2 cos2x -1. To find tan 2 x we take (44), tan x + tan y tan (x + y) = — tall x tan y Replace y by x. This gives tan x + tan x tan(x + x) = -tanx tanx or 2- tan x tan 1 - tan2 x (50) tan2x= 1 tan2x 37. Functions of multiple angles. The method of the last section may readily be extended to finding the functions of nx in terms of the functions of x. To find sin 3 x in terms of sin x we take (40), sin (x + y) = sin x cos y + cos x sin y. Replace y by 2 x. This gives sin (x + 2 x) = sin x cos 2 x + cos x sin 2 x, or sin 3 x = sin x (cos2x - sin2x) + cos x (2 sin x cos x) by (49), (48) = 3 sin x cos2x -sin3 x = 3 sin x (1 - sin2 ) - sin3 x by (23), p. 59 = 3 sin x - 4 sin3x. Ans. To find tan 4 x in terms of tan x, we take (44), (50), 2 tan 2 x 4 tan x ( —tan2 ) A tan 4 x = tan(2 x + 2 ) = ta22 x 6 tan2x tan4x A TRIGONOMETRIC ANALYSIS 71 EXAMPLES 1. Given sinx =, x lying in the second quadrant; find sin 2x, cos 2x, tan 2 x. 5 29 Solution. Since sin x = - and x lies in the second quadrant, we get, using method on p. 30, 2 1 sin x = -, cosx = —, tanx = -2. V5 5 Substituting in (48), we get si2 sin 2 = 2 sin xcos x = 2 - / 5 = —. Similarly, we get cos 2 x =- - by substituting in (49), and tan 2 x = by substituting in (50). 2. Given tanx = 2, x lying in the third quadrant; find sin 2x, cos 2x, tan 2x. Ans. sin 2 =, cos 2 x = - tan 2 x =-. a 3. Given tan x = -; find sin 2 x, cos 2 x tan 2 x. b 2ab b2 - a2 2ab A a2 + b2' b2 + a2 b2 - a2 4. Show that cos 3 x = 4 cos3 x - 3 cos x. 3 tan A - tan3 A 5. Show that tan 3 A =3 tan 1- 3 tan2 A 6. Show that sin 4 x = 4 sin x cos x - 8 sin3x cos x. 7. Show that cos 4 B = 1 - 8 sin2 B + 8 sin4 B. 8. Show that sin 5 x = 5 sin x - 20 sin3x + 16 sin5x. 9. Show that tan (45~ + x) - tan (45~ - x) = 2 tan 2 x. 10. Show that tan (45~ + C) + tan (45~ - C) = 2 sec 2 C. 11. Verify the following: 2 tan x csc2 X (a) sin 2 x tan '(f) sec 2 x= s 1 + tan2z X csc2x -2 1( - tan2 x (g) 2 csc 2 s = sec s csc s. (b) cos 2 x = 1 + tan2 x (h) cot t-tant = 2 cot 2 t. (c) tanP + cotP = 2 csc 2P. () cos2 2- sec2x (d) cos 2 x = cos4 - sin4 x. sec2 X (e) (sin x + cos x)2 = 1 + sin 2 x. (j) (sin - cos x)2 = 1- sin 2 x. 12. In a right triangle, C being the right angle, prove the following: sin2A - sin2 B 2 ab (a) cos2B= iA~si2B (h) tan 2A=-a2 v /shA + sinj2 B b2 - a2 (b) sin (A - B) + cos 2 A= 0. (i) sin 2 A = sin 2 B. (c) sin(A - B) + sin(2A + C) = 0. sin (d) (sinA - sinB)2 + (cosA + cosB)2 = 2. c2 b / a -b a_ 2 sinA ( A b2 -a2 (e^ II~ / -- 4 / -- -. -*(k) cos 4 2 A - - (e) a-b + a -b Ncos2B c2 (f) tanB= cotA+cosC. ()sin3A 3ab2 -a3 (g) cos 2 A + cos 2 B= 0. C3 72 PLANE TRIGONOMETRY 38. Functions of an angle in terms of functions of half the angle. From (48), p. 70, sin 2 = 2 sin x cos x. Replace 2 x by x, or, what amounts to the same thing, replace x by ~. This gives x x (51) sin x = 2 sin cos2 From (49), p. 70, cos 2 x = cos2x - sin2x. Replace 2 x by x, or, what amounts to the same thing, replace x x by - This gives x x2 (52) cos x = cos2 - sin - ( 2 2 tan x From (50), p. 70, tan 2x =1 tan2 Replace 2 x by x, or, what amounts to the same thing, replace x by 2. This gives 2 tan x (53) tan x = 1 - tan2 X 2 39. Functions of half an angle in terms of the cosine of the angle. From (49a) and (49b), p. 70, we get 2 sin2x 1 - cos 2 x, and 2 cos2x = 1 + cos 2 x. Solving for sin x and cos x, sin x 1- cos 2x 2 cos- 1+ cos 2 x and cos X = ~ 2 2 Replace 2 x by x, or, what amounts to the same thing, replace x x by 2- This gives (54) sin x =:- cosix X 11 + cosx COS - = = \ 2 2 (55) TRIGONOMETRIC ANALYSIS 73 To get tan2 we divide (54) by (55). Thils gives x - cos x x 2 ~ 2.. j c, ) 1 - cos -_ tan 2 = = or, 2 + c os x (56) tan- = =1: 2 1 + cos x Multiplying both numerator and denominator of the right-hand member by V + cos x, we get * x sin x (57) tan-= — 2 1 + cosx or, multiplying both numerator and denominator by V/- cosx, we get x 1 -cos x (58) tan -= --. 2 sin x Since tangent and cotangent are reciprocal functions, we have at once from (56), (57), and (58), x 1+ cos x (59) cot - = /2 N1 +cs- cos x (60) cot x 1 + cos x 2 sin x x sin x (61) cot-=2 1 -cosx 40. Sums and differences of functions. From p. 63, (40) sin(x + y)= sin x cos y + cos x sin y. (41) sin(x - y)= sin x cos y -cosx sin y. (42) cos (x + y) = cos x cos y - sin x sin y. (43) cos (x - y) = cos x cos y + sin x sin y. l1l-cosx V/1 co+ l-cos2x sinx 1+ cos' Vl + cosx 1+ cosx 1+ cos x The positive sign only of the radical is taken since 1 + cos x can never be negative and tan and sin x always have like signs. 2 74 PLANE TRIGONOMETRY First add and then subtract (40) and (41). Similarly, (42) and (43). This gives (a) sin(x+y)+sin(x —y)= 2sinxcosy. Adding(40)and(41) (b) sin (x + y) - sin (x - y) = 2 cos x sin y. Subtracting (41) from (40) (c) cos(x +y) +cos(x —y)= 2 cos x cosy. Adding(42)and(43) (d) cos (x + y) - cos (x - y) =- 2 sinx siny. Subtracting(43)from (42) Let x + y =A x + y =A and x-y=B x - y=B Adding, 2 x = A + B Subtracting, 2y =A -B x = -(A+B). y=- (A-B). Now replacing the values of x + y, x - y, x, y in terms of A and B in (a) to (d) inclusive, we get (62) sinA + sin B = 2 sin (A + B) cos 1 (A -B). (63) sin A - sin B = 2 cos 1 (A + B) sin (A - B). (64) cos A + cosB = 2 cos 1 (A + B) cos (A - B). (65) cos A - cos B = - 2 sin (A + B) sin (A - B). Dividing (62) by (63), member for member, we obtain sin A +sin B 2 sin (A + B) cos (A -B) sinA -sinB 2 cos - (A + B) sin 2 (A -B) _sin (A- B) cos (A —B) cos (A +B) sin (A -B) = tan - (A + B)cot - (A -B). But cot (A -B) = 1 Hence B2( tan-(-B) sin A + sinB tan (A - B) (66) 2 = T 2 sinA-sin B tan (A-B) EXAMPLES 1. Find sin 22-~, having given cos 45~= 2 1/2 x /1 -cosx Solution. From (54), sin = + - 2 = 2 Let x = 45~, then x = 221~, and we get 2i1- 2 A sin 22~ =2-V2. Ans. 2 2 2 TRIGONOMETRIC ANALYSIS 2. Reduce the sum sin 7 x + sin 3 x to the form of a product. Solution. From (62), sin A + sin B = 2 sin I (A + B) cos I (A - B). Let A = 7 x,B = 3x. Then A + B =10x, and A-B = 4x. Substituting back, we get sin7 xsin 3 x 2sin 5x cos 2x. Ans. 3. Find cosine, tangent, and cosecant of 22-~. 75 Ans. 2 /2 + 2, 2 -1, 2 2 ~2 V2-V2 V3 4. Find sine, cosine, and tangent of 150, having given cos 30~ = — I 1 ~~~~~~~2 5. Verify the following: (a) sin 82~ + sin 28~ = cos 2~. (b) sin 50~ + sin 10~ = 2 sin 30~ cos 20~. (c) cos 80~- cos 20~= - 2 sin 50~ sin 300. (d) cos 5 x + cos 9 x = 2 cos 7 x cos 2 x. sin 7 x - sin 5 x (e) = tan x. cos 7 x + cos 5 x sin 330 + sin 30 cos 33~ + cos 3~ 6. Find sine, cosine, and tangent of -, quadrant. As 3-1 /3+1 3-1 Ans. I, _, 2 /2 '2 /2 3 + 1 (g) sin - - cos ) = 1- sin x. 2 2) x sin - x (h) tan-= 2 -4 1 + cos I x x sin I x (i) cot- = 4 1-cos xa; sinA + sinB 1 (j) A = tan - (A + B). cosA + cos B 2 if cos x = and x lies in the first A3 /V V2 Ans.,,.2 2 2 7. Find sine, cosine, and tangent of -, if cos x = a. II - l + a Ans.:-a I /1-a 2 2 Xi a 8. Express sine, cosine, and tangent of 3 x in terms of cos 6 x. - cos 6 x /1 +- cos 6 x - cos 6 x Ans. ~V1~7- ~ 1-cosx 2 2 1 - cos 6 x 9. In a right triangle, C being the right angle and c the hypotenuse, prove the following: B c-a a-b A-B (a) sin2- (d) a tan - 2 2c a+b 2 AA 2A\2 a~c A a (b) (cos- + sin_ = ll (e) tans\ 2 2/ c 2 b+c A b+c (c) cos2 2 2c 41. Trigonometric identities. A trigonometric identity is an expression which states in the form of an equation a relation which holds true for all values of the angles involved. Thus, formulas (26) to (39), p. 60, are all trigonometric identities, since they hold true for all 76 PLANE TRIGONOMETRY values of x; also formulas (62) to (66), p. 74, are identities, since they hold true for all values of A and B. In fact, a large part of the work of this chapter has consisted in learning how to prove identities by reducing one member to the form of the other, using any known identities (as in Ex. 4, p. 61). Another method for proving an identity is to reduce both members simultaneously, step by step, using known identities, until both members are identical in form. No general method can be given that will be the best to follow in all cases, but the following general directions will be found useful. General directions for proving a trigonometric identity.* First step. If multiple angles, fractional angles, or the sums or differences of angles are involved, reduce all to functions of single angles only t and simplify. Second step. If the resulting members are not readily reducible to the same form, change all the functions into sines and cosines. Third step. Clear of fractions and radicals. Fourth step. Change all the functions to a single function. In case the second step has been taken, this means that we change to sines only or to cosines only. The two members may now easily be reduced to the same form. Ex. 1. Prove the identity 1 + tan 2 x tan x = sec 2 x. 2 tan x 1 1 Solution. Since tan 2 x = ta and sec 2 x = =, the equation becomes 1- tan2x cos2x 2x - sin2x equation becomes: 2 tan2 x 1 First step. 1 + 2 t 1 - tan2 x cos2 x - sin2 x 1 + tan2 x 1 or, simplifying, = 1 - tan2 x cos2 x - sin2 x sin2 x cos2 x 1 Second step. -- C =, (22), p. 59 sin2 cos2 - sin2 cos2x cos2 x + sin2 x 1 or, simplifying, = cos2 x - sin2 x cos2 - sin2 x Third step. cos2 x + sin2 x 1, or 1 = 1. from (23), p. 59 * In working out examples under this head it will appear that it is not necessary to take all of the steps in every case, nor will it always be found the best plan to take the steps in the order indicated. 2 tan x t For instance, replace sin 2 x by 2 sin x cos x, tan 2 x by - tan cos (x + y) by cos x cos y - sin x sin y, etc. TRIGONOMETRIC ANALYSIS 77 Ex. 2. Prove sin (x + y) tanx + tan y sin (x - y) tan x - tan y Solution. Since sin (x + y) = sin x cos y + cos x sin y, and sin (x - y) = sin x cos y - cos x sin y, the equation becomes: sin x cos y + cos x sin tan x + tan y First step. sin x cos y - cosx sin y tan x - tan y sin x sin y sin x cos y + cos x sin y cos x cos y Second step. ---- sin x cos y - cos x sin y sin x sin y cos x cos y sin x cos y + cos x sin y sin x cos y + cos x sin y Simplifying, = sin x cos y - cos x sin y sin x cos y - cos x sin y or, 1=1. EXAMPLES Prove the following identities: 1. tan x sinx + cosx = secx. 2. cot x - sec x csc x (1 -2 sin2 x) = tan x. 3. (tan x + cot x) sin x cosx = 1. sin y 1 -cos y 1 + cos y sin y 1+ — sin A 5. =1 i - secA- tanA. 6. tan x sin x cos x + sin x cos x cot x = 1. 7. cot2x = cos2x + (cotx cosx)2. 8. (sec y + csc y) (1 - cot y) = (sec y - csc y) (1 + cot y). 9. sin2 z tan z + cos2z cot + 2 sin z cos z = tan z + cot z. 10. sin6 x + cos6x = sin4 + cos4 x - sin2 x cos2 x. 11. sin B tan2 B + csc B sec2 B = 2 tanB sec B + csc B - sin B. 12. Work out (a) to (v) under Ex. 4, p. 62, following the above general directions. 13. cos(x + y) cos(x - y) = cos2x - sin2y. 14. sin (A + B) sin (A - B) = cos2 B - cos2 A. 15. sin (x + y) cos y - cos (x + y) sin y = sin x. 16. cos (x - y) 1 + tan x tan y cos (x + y) 1 - tan x tan y 17. tanA -tanB sin (A-B) cos A cos B sin (x + y) 18. cotx + coty= sin sin x sin y 19. sin x cos (y + z) - sin y cos (x + z) = sin (x - y) cos z. 78 PLANE TRIGONOMETRY 20 tan ( - 4) + tan 0 = tan 0 20. =(tan+ 6. 1 - tan ( - 0) tan 0 21. cos(x - y + ) = cos x cos y cosz + cos x sin y sin z - sin x cos y sin z + sin x sin y cos z. 22. sin (x + y - z) + sin (x + z - y) + sin (y + z - x) = sin (x + y + z) + 4 sin x sin y sin z. 23. cos x sin (y - z) + cos y sin (z - x) + cos z sin (x - y) = 0. 24. Work out (a) to (k) under Ex. 7, p. 67, and (a) to (j), under Ex. 4, p. 69, following the above general directions. 25. 1 +sin2 =tanx +12 29. cot (A + 45) 1-sin 2A 1 - sin 2 x \tanx - 1 cos 2 A 2. c sin 2 x cos3 x + sin3 x 2 - sin 2 x 26. tan x = -- 30. = — 1 + cos 2 x cos x +- sin x 2 sin 2 x sin 3x - sill x 27. cota-x si= 31. -=tan x. 1 - cos2x cos 3 x + cos 28. cot A - 1 I - sin 3 x-sin x ct 2 28. --- \ - — 32. -- = cot 2 a. cotA + 1 + sin 2A cos x -cos 3 x 33. sin 3 x = 4 sin x sin (60~ + x) sin (60~ - x)..sin 4 x 2co2x sin 2 x 35. sin 2 x + sin 2 y + sin 2 z = 4 sin x sin y sin z. 36. directi 37. 38. 39. 40. 41. 42. 43. 44. 45. Work out (a) to (j) under Ex. 11, p. 71, following the above general ons. sin 9x - sin 7 x = 2 cos 8x sin x. 1 -tan cos 7 x + cos 5x = 2 cos 6x cos x. 46. 2 46. -cos x. sin 5x - sin 2x cot7 x 1 + tan2X 2 cos 2x - cos 5x 2 x sin- cos - 1 +- sinx. 2 + secy 2 c1 + tan2 -1 2 C-t2 2 sec y 2 x 48. 1 + tan x tan - = sec x. sin A + sinB_ 1 2 - =- cot-(A - B). X X cosA-cosB 2 49. tan - 2 sin2- cotx =sinx. 60 2 2. 1 + tan- x 2 cos 0 2 250. 1 + cot2 - 2 X X tan2 x_ 1- cot2 x cos3cr-cos7a=2 sin5csin2c. tan2 - cot2 + 2 51. 2 2 x X X X 2 cos x cot- + tan = 2 csc x. tan2 -x cot2 x cos 2 2 2 2 52. Work out (a) to (f) under Ex. directions. 5, p. 75, following the above general CHAPTER V GENERAL VALUES OF ANGLES. INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC EQUATIONS 42. General value of an angle. Since all angles having the same initial and terminal sides have the same functions, it follows that we can add 2 wr to the angle or subtract 2 wr from the angle as many times as we please without changing the value of any function. Hence all functions of the angle x equal the corresponding functions of the angle 2 nr +, where n is zero or any positive or negative integer. The general value of an angle having a given trigonometric function is the expression or formula that includes all angles having this trigonometric function. Such general values will now be derived for all the trigonometric functions. 43. General value for all angles having the same sine or the same cosecant. Let x be the least positive angle whose sine has the given value a, and consider first the case when a is positive. Construct the angle x (= XOP), a,s on p. 31, and also the angle v - x (= XOP'), having the same value a for its sine. Then P, every angle whose terminal s side is either OP or OP' has a a its sine equal to a, and it is ' a evident that all such angles x ~ Ia2 x are. found by adding even multiples of vr to, or subtracting them from, x and w - x. Let n denote zero, or any Y positive or negative integer. When n is even, nwr + x includes all the angles, and only those, which have the same initial and terminal sides as x (= XOP). Therefore, when n is even, (A) n7r + x = n'r + (- l)nx.* * The factor (- 1)n is positive for all even values of n and negative for all odd values of n. 79 80 PLANE TRIGONOMETRY Again, when n is odd, n -1 is even, and (n -1) 7r + (r - x) includes all the angles, and only those, which have the same initial and terminal sides as 7r - x (= XOP'). But when n is odd, (B) (n - 1) 7r + (7r - X) = r - X = n7r + (- 1)x. From (A) and (B) it follows that the expression nrr +(- 1)nx for all values of n includes all the angles, and only those, which have the same initial and terminal sides as x and r - x. In case a is negative, 7r - x will be negative, as shown in x the figure, but the former line > of reasoning will still hold X^ o ^^ ~~ true in every particular. Since sine and cosecant P.' are reciprocal functions, it yI~, ~follows that the expression rYI~ ~ for all angles having the same cosecant is also nr + (-1)n x. Hence (67) nr + (- 1)"x is the general value of all the angles, and only those, which have the same sine or cosecant as x.* This result may also be expressed as follows: sin x = sin [nT7 + (- 1)nx], csc x = csc[ n7r + (- 1)nx]. 37r Ex. 1. Find the general value of all angles having the same sine as -- 3,r Solution. Let x = - in (67). This gives ~~~~4 nr + (- l)n -. Ans. Ex. 2. Find the four least positive angles whose cosecant equals 2. Solution. The least positive angle whose cosecant = 2 is - Let x = in (67). This gives 6 6 n7r + (- l)-. * In deriving this rule we have assumed x to be the least positive angle having the given sine. It follows immediately from the discussion, however, that the rule holds true if we replace x by an angle of any magnitude whatever, positive or negative, which has the given sine. The same observation applies to the rules derived in the next two sections. GENERAL VALUES OF ANGLES 81 When n = 0, we get -= 30~ 6 When n = 1, we get 7r - - = 150~. 6 When n = 2, we get 2 7r + 7 = 3900. When n = 3, we get 37 - - = 510~. Ans. 6 44. General value for all angles having the same cosine or the same secant. Let x be the least positive angle whose cosine has the given value a, and consider first the case when a is positive. Construct the angle x (= XOP), and also the angle - x (= XOP'), having the same value a for its p cosine. Then every angle whose terminal side is either VT-1-a2 OP or OP' has its cosine equal - to a, and it is evident that all 0 Xa -Vl-a F such angles are found by adding 1a even multiples of wr to, or sub- ' tracting them from, x and - x. Let n denote zero, or any Y positive or negative integer. For any value of n, (A) 2 nr + x includes all the angles, and only those, which have the same initial and terminal sides as x (= XOP). Similarly, Y) (B) 2 n -x includes all the angles, and ' only those, which have the Yx -same initial and terminal sides xi ____ as -x (= XOP') 2X ^ J-x X In case a is negative, the same line of reasoning still PY/~',I~ ~holds true. Since cosine and secant are Y reciprocal functions, it follows that the same discussion holds for the secant. Hence, from (A)and(B), (68) 2 n7r =4 x is the general value of all the angles, and only those, which have the same cosine or secant as x. 2n is even and 2 n - 1 is odd for all values of n. 82 PLANE TRIGONOMETRY This result may also be expressed as follows: cos x = cos (2 nwr - x), sec x = sec (2 nmr 4 x). Ex. 1. Given cos A -; find the general value of A. Also find the five least positive values of A. /2 1 3wr Solution. The least positive angle whose cosine = - is - If we let 3/~ 4 37 /V2 4 x = - in (68), we get 4 37r A = 2 n7r ~ When n = 0, A = = 135~. 4 37w When n = 1, A = 2 ~ 37r = 225~ or 495~. 4 37r When n = 2, A = 47 = 585~ or 855~. Ans. 45. General value for all angles having the same tangent or the same cotangent. Let x be the least positive angle whose tangent has the given value a, and consider first the case when a is positive. Construct the angle x (= XOP), and also the angle 7r + x (= XOP') having the same value a for its tangent. Then every angle whose terminal side is either OP or Or^'Y )OP' has its tangent equal to a, and it is evident that all such angles are found by adding 7r+X a even multiples of v to, or -1 /K x+ subtracting them from, x and IXI > X1' _ 1 X 7r + X. -a^S' Hi~Let n denote zero, or any p, Apositive or negative integer. When n is even, (A) n7r + x includes all the angles, and only those, which have the same initial and terminal sides as x (= XOP). Again,,when n is odd, n -1 is even, and (B) (n - 1) r + (7r + x) = nr + x includes all the angles, and only those, which have the same initial and terminal sides as 7r + x (= XOP'). In case a is negative, the same line of reasoning still holds true. GENERAL VALUES OF ANGLES 83 Since tangent and cotangent are reciprocal functions, it follows that the same discussion holds for the cotangent. Hence, from (A) and (B), for all values of n, (69) rnr + x is the general value of all the P angles, and only those, which, have the same tangent or cotangent as x. X' 7r X This result may also be stated as follows: tan x = tan (n7r + x), cot x = cot (nrr + x). Y EXAMPLES 1. Given sinA = -; find the general value of A. Also find the four least positive values of A. 7' positive values of A. ns. n7r + (- 1)n 30~, 150~, 390~, 510~. 2. Given cosA =; find the general value of A. Also find all values of A 7r r 11 7r numerically less than 2 rr. Ans. 2 nr -; - — 6 6 6 6 3. Given tanA = 1; find the general value of A. Also find the values of A numerically less than 4 r. 7r rr 37r 57r 77r 9rr 11r 13 r 15 r Ans. r + -; - - - - - - - - -- - 4'4- 4 4 4 4 4 4 4 I nrr,r 4. Given sin 2 =; show that x = + (- )n —. 2 2 12 1 2wnr 7r 5. Given cos 3 x —; show that x = + -- - 2 3 9 6. In each of the following examples find the general values of the angles, having given (a) sinA = 1. Ans. A = nr(- l)n (- - = nr - 7r 1 rr (b) cot x a=.x = n7r. (c) cosy= ~. y=n7r ~. 2 3 (d) tanB = 1. B = n7r - 4 (e) cseC= ~/2. Cn = n r 4 (f) secA=~- A= nrwr-. V3 6 84 PLANE TRIGONOMETRY 1 1 7. Given sin x = -- and tan x = find the general value of x. 2 Solution. Since sin is - and tan x is +, x must lie in the third quadrant. The smallest positive angle in the third quadrant which satisfies the condition 1. 7r 1 sinx = — is -, and this angle also satisfies tan x =. 26si~7n Hence x = 2n7r + - Ans. 6 8. In each of the following examples find all the positive values of x less than 2 7r which satisfy the given equations. 1 7 77r (a) cosx= - Ans. -, VJ~~~~2 ~4 4 1 7r 3 r 57r 77r (b) sin x = - -, -. -- 4 4 4 4 7 27r 4 7r 57r (c) tan x=+ 3., -_, 3, 3 3 3 3 7r 37r 57r 77r (d) cotx =. 1 - -. 5 7 7r (e) cosx —. - 2 6 6 wr 5 7r (f) secx =2. 73 3 3 - 1 7r 57r 7 7r 11 7r (g) sin x = I -, -7 11 2 6 6 6= 6 46. Inverse trigonometric functions. The value of a trigonometric function of an angle depends on the value of the angle; and conversely, the value of the angle depends on the value of the function. If the angle is given, the sine of the angle can be found; if the sine is given, the angle can be expressed. It is often convenient to represent an angle by the value of one of its functions. Thus, instead of saying that an angle is 80~, we may say (what amounts to the same thing) that it is the least positive angle whose sine is h. We then consider the angle as a function of its sine, and the angle is said to be an inverse trigonometric function, or anti-trigonometric, or inverse circular function, and is denoted by the symbol in or arsin, or, ar sin read "inverse sine of I," or, "arc (or angle) whose sine is,." Similarly, cos- x is read "' inverse cosine of x," tan-ly is read " inverse tangent of y," etc. If a is the value of the tangent of the angle x, we are now in a position to express the relation between a * Symbol generally used in Continental books. INVERSE TRIGONOMETRIC FUNCTIONS 85 and x in two different ways. Thus, tan x = a, meaning the tangent of the angle x is a; or, x tan- a, meaning x is an angle whose tangent is a. The student should note carefully that in this connection -1 is not an algebraic exponent, but is merely a part of the mathematical symbol denoting an inverse trigonometric function. tan-la does not denote (tan a)- =tan a but does denote each and every angle whose tangent is a. The trigonometric functions (ratios) are pure numbers, while the inverse trigonometric functions are measures of angles, expressed in degrees or radians. Consider the expressions tan x = a, x =tan-la. In the first we know that for a given value of the angle x, tan x (or a) has a single definite value. In the second we know from (69), p. 83, that for a given value of the tangent a, tan-l a (or x) has an infinite number of values. Similarly, for each of the other inverse trigonometric functions. Hence: The trigonometric functions are single valued, and the inverse trigonometric functions are many valued. The smallest value numerically of an inverse trigonometric function is called its principal value.* For example, if tan x =1, then the general value of x is, by (69), p. 83, 7r x = tan-l1 = n7r + where n denotes zero or any positive or negative integer, and - 450 4 is the principal value of x. * Hence, if sin x, csc x, tan x, or cot x is positive, the principal value of x lies between 0 and; if sin x, csc x, tan x, or cot x is negative, the principal value of x lies between 0 2 7r and - 7r 7r If cos x or sec x is positive, the principal value of x lies between 2 and - 2, preference being given to the positive angle. 86 PLANE TRIGONOMETRY Similarly, if cos x =, then by (68), p. 81, 1 7r x = cos-1 = 2 nr ~ -, 2 3 7/ where the principal value of x is - = 60~. 3 Since the sine and cosine of an angle cannot be less than - 1 nor greater than + 1, it follows that the expressions sin-l a and cos-la have no meaning unless a lies between- 1 and + 1 inclusive. Similarly, it is evident that the expressions se-l a and csc-1a have no meaning for values of a lying between — 1 and +1. Any relation that has been established between trigonometric functions may be expressed by means of the inverse notation. Thus, we know that cos x= V1- sin2x. (29), p. 60 This may be written (A) x = cos- VI - sin2 x. Placing sin x = a, then x = sin-la, and (A) becomes sin-' a = cos-1 VI - a2. Similarly, since cos 2 x = 2 cos2x -1, (49 b), p. 70 we may write (B) 2x = cos-1(2 cos2x -). Placing cos x = c, then x = cos-l, and (B) becomes 2 cos-lc = cos- (2 C2 — 1). Since we know that the co-functions of complementary angles are equal, we get for the principal values of the angles that 7T' sin-l a + cos-l a = r, 0 < a -1 tan- b + cot-l b =, 0 < b see c+ csc-l - < sec-~1 c csc~1 c = - I < C 2 INVERSE TRIGONOMETRIC FUNCTIONS 87 We shall now show how to prove identities involving inverse trigonometric functions for the principal values of the angles. Ex. 1. Prove the identity (a) tan-lm + tan- n = tan-1 -n. 1- mn Proof. Let (b) A= tan- m and B = tan-in. (c) Then tanA = m and tan B =n. Substituting from (b) in first member of (a), we get nm + n A + B= tan- --—, 1- mn or, what amounts to the same thing, (d + f (d) tan (A + B)= 1 - mnn But from (44), p. 68, tan A + tan B (e) tan (A + B) = tA ta 1 - tanA tan B Substituting from (c) in second member of (e), we get m+n (f) tan (A + B) - 1 - mnz Since (d) and (f) are identical, we have proven (a) to be true. Ex. 2. Prove that (g) sin-13 + co- s 15 = sin-1 7 7 Proof. Let (h) A = sin-13 and B = cos-l 1 (i) Then sinA = 3 and cos-B = 1. (j) Also cosA = 4 and sinB = -$ 5 17 Substituting from (h) in first member of (g), we get A + B =sin-1 7 or, what amounts to the same thing, (k) sin (A + B)= - But from (40), p. 63, (1) sin (A + B) = sinA cos B + cos A sin B. Substituting from (i) and (j) in second member of (1), we get (m) sin (A + B) = 5 ~._ 75 + * = - 5 -Since (k) and (m) are identical, we have proven (g) to be true. The following example illustrates how some equations involving inverse trigonometric functions may be solved. * Found by method explained on p. 30. 8 01 01 8 PLANE TRIGONOMETRY Ex. 3. Solve the following equation for x: tan-12x + tan-13x=. 4 Solution. Take the tangent of both sides of the equation. Thus 77 tan (tan-1 2 x + tan-1 3 x)= tan - 4 tan (tan-12 x) + tan (tan-1 3 x) c~ ~ ~ - --- —, --- = 1, from (44), p. 68 1 - tan (tan-1 2 x) tan (tan-1 3 x) 2x + 3x ram',~~ 1-2x.3x Clearing of fractions and solving for x, we get x = - or — 1. x -= satisfies the equation for the principal values of tan-1 2 x and tan-1 3 x. x - 1 satisfies the equation for the values tan-(- 2) = 116.57~, tan- (-3) =- 71.57~. EXAMPLES 1. Express in radians the general values of the following functions: I 71' (a) sin-1. Ans. nwr + (- 1)n - V2 4 (b) sin-l- - ~ - (-1)n.2(c) cos-1 2 3 2 6 /(C) COS —1 —' 2% 27w7 -(d) cos( —. 2mw r2 2. Prove the following: a-b (a) tan- a - tan- b= tan-1 a-b I+ab 2a (b) 2 tan-' a = sin- 12 a 1 + a2 (c) 2sin-1a = cos-l(- 2 a2). (d) tan-la = sin-1 a V/ + a2 m rn - n wr (e) tan- 1 - tan- - =- -- n m+n 4 4 3 27 (f) cos-1 - + tan-1 - = tan-' - 5 5 11 2 12 (g) 2 tan- 12 = tan3 5 (e) tan-1 Ans. nwr + 7 V/3 6 77 (f) tan-' n 3). nw i (g) cot-'l( 1). n7 -r 4 (h) cot-(. nr + 2a (h) 2 tan-1 a = tan- 1. 1 - a2 (i) sin-1a = cos-1 1- a2. (j) sin- a = tan- a V1 - a' 1 (k) tan-1a = cos-1- V + a2. 3. 8 77 (1) sin-1 - sin- sin- i -. 5 17 85 4 12 33 (m) cos-1 + cos- -= cos-' 5 13 65 1 1 = tanl2 (n) tan-1- + tan-1 -= tan-1 7 13 9 * The student should remember that tan-' 2 x and tan-1 3 x are measures of angles. TRIGONOMETRIC EQUATIONS 89 3. Solve the following equations: (a) tan-lx + tan-l(1 - x) = tan-1 '). Ans. x =1 27r2 (b) tan-lx + 2 cot-lx =. - =/3. 3 (c) tan-1 + tan-l + 1 =T. = x a+2 x + 2 42 x2 - I 2x 127 (d) cos- 1 + tan-1 2 x 27r x2 + 1 x2 -1 3 x+1 x-1 (e) tan-1 + tan-1 = tan- (- 7). x = 2. x-1 x 8 1 (f) tan-l(x + 1) + tanl —l(x -1)= tan-1-. x = - 8,-. (g) sin -lx + sin-12 x =-. x = + _ 3 14 5 1-2 -- (h) sin — - + sin-' -. x = 13. x x 2 4. Find the values of the following: (a) sin (tan- - ). Ans. -t (d) cos(2cos-la). Ans. 2a2-1. \ 12 13 b) cotj sin-l7 2a (bh) cot2sinl I. + 2 (e) tan (2 tan-la). \ t 4 1i- a2 1 q- a2 (c) sinl(tan-I + tan-l-). ~ 1 (f) cos(2 tan-la). 1- a2 47. Trigonometric equations. By these we mean equations involving one or more trigonometric functions of one or more angles. For instance, 2 cos2x + V3 sin x +1 = 0 is a trigonometric equation involving the unknown angle x. We have already worked out many problems in trigonometric equations. Thus, Examples 1-8, pp. 83, 84, are in fact examples requiring the solution of trigonometric equations. To solve a trigonometric equation involving one unknown angle is to find an expression (the student should look up the general value of an angle, p. 85) for all values of the angle which satisfy the given equation. No general method can be given for solving trigonometric equations that would be the best to follow in all cases, but the following general directions (which are similar to those given on p. 76 for proving identities) will be found useful. 90 PLANE TRIGONOMETRY 48. General directions for solving a trigonometric equation.* First step. If multiple angles, fractional angles, or the sums or differences of angles are involved, reduce all to functions of a single angle,t and simplify. Second step. If the resulting expressions are not readily reducible to the same function, change all the functions into sines and cosines. Third step. Clear of fractions and radicals. Fourth step. Change all the functions to a single function. Fifth step. Solve algebraically (by factoring or otherwise) for the one function now occurring in the equation, and express the general value of the angle thus found by (67), (68), or (69). Only sutch values of the angle which satisfy the given equation are solutions. Ex. 1. Solve the equation cos 2 x sec x + sec x + 1 = 0. Solution. Since cos 2 x = cos2x - sin2, we get First step. (cosgx - sin2x) secx + sec x + 1 = 0. Second step. Since secx = --, this becomes cos X cosx - sin2 x 1 + — +1=0. Cos X COs X Third step. cos2x - sin2x + 1 + cos = 0. Fourth step. Since sin = 1 - cos2 x, we have cos2x - 1 + cos2 + 1 + cos = 0, or, 2 cos2x + cos = 0. Fifth step. cos x (2 cosx + 1) = 0. Placing each factor equal to zero, we get cos X = 0, or, from (68), p. 81, x = cos-10 = 2 nr ~. Also, 2 cosx + 1 = 0, COS X =- 1 27 or, x = cos-' -) = 2 r 2n - - \2/ -3 Hence the general values of the angles which satisfy the equation are 7r 2w 2 n7r +- and 2nT7r 2 3 The positive angles less than 2 7r which satisfy the equation are then 7r 27r 4w 37r 2' 3, 3 2 * In working out examples under this head it will appear that it is not necessary to take all of the steps in every case, nor will it always be found the best plan to take the steps in the Order indicated. tFor instance, replace cos 2 x by cos2 x - sin2 x, sin + by, etc. \ 4/ 4 TRIGONOMETRIC EQUATIONS 91 Ex. 2. Solve the equation 2 sin2x + /3 cos + 1 = 0. Solution. Since sin2X = 1 - cos2x, we get Fourth step. 2 - 2 cos2x + 3 cos x + 1 = 0, or, 2 cos2x - v3 cos x - 3 = 0. Fifth step. This is a quadratic in cos x. Solving, we get cos X-/3 or -. 2 Since no cosine can be greater than 1, the first result, cos x = /-3, cannot be used. From the second result, x=cos-1 = 2 n7r 5. Ans. EXAMPLES Solve each of the following equations: 1. sin2x = 1. Ans. x=n7+( —1)n (~-*=n7w7+. 2. csc2x = 2. =n7+(-l)n( '4 =n77 ^ 3. tan2x = 1. x = n7Z ~q - 77 4. cot2x = 3. x = nr -t - 4 3 5. cos2x xn7r ' 4 7T 6. sec2x = -- X = n7 —. 7. 2 sin2x + 3 cosx =. x = 2 ni - 3 8. cos2r- sin2cr =-. c = n77 + (- 1)n. _. 2 6 9. 2 cos2a = sin. C = n7 + (- 4 11. sinA+cosA=V/2. W=27r+4* 12. 4 sec2y - 7 tan2y = 3. y = nr - -rr~~,-7/' 7r 13. tan + cot=2. =n7r + 4r + 14. tan2x - (1 /3) tan x + V3 =0. x = n + r- ~, n72r + -. * Since the principal value of x = sin-1 1= - and of x = sin- (- 1) = - 2 2 92 PLANE TRIGONOMETRY 1 5w7 2wr 15. cot2X + 3 + ) cotx +1=0. Ans. x = nr + 6, nwT + 37 V_3/ 6 3 16. tan2x + cot2x =2. x = n7r +. 17. tan (x + - = + sin 2. x = nr, n7r- -. 4/ 4 18. cscx cotx = 2. x = 2n7r - a;X~~~~~ 6 19. sin- = cs x - cot x. x = 2n7r. 2 7r 20. csc y+cot y = 3. y =2 n7r+. 3 21. 3 (sec2 a + cot2 a) = 13. = n, n7r. 6 3 Find all the positive angles less than 360~ which satisfy the following equations: 22. cos 2 x + cos x = - 1. Ans. x = 90~, 120~, 240~, 270~. 23. sin 2 x - cos 2 x - sin x + cos x = 0. x = 0, 90~, 180~. 24. sin (60~ - x) - sin (60~ + x) = * x = 240~, 300~. 25. sin (30~ + x)-cos(60~+ x) =- 2 x = 210~, 330~. 2 26. tan (450 - x) + cot (450 - x) = 4. x = 30~, 150~, 210~, 330~. 27. cos 2 x = cos2. x = 0~, 180~. 28. 2 sin y = sin 2 y. y = 0~, 180~. 29. sin x + sin 2 x + sin 3 x = 0. x = 0~, 90~, 120~, 180~, 240~, 270~. 30. tan x + tan 2 x = tan 3 x. x = 0~, 600~,120~,180~, 240~, 300~. '31. sec x - cotx = cscx - tanx. x = 45~, 135~, 225~, 315~. 32. sin 4x - cos 3x = sin 2 x. x =30~,90~,150~,210~,270~,330~. 33. V1 + sin x - w/1-sinx = 2 cos x. 34. sin4x + cos4x = - 8 35. sec (x+120~) +sec(x -120~) = 2 cosx. 36. sin (x + 120~) + sin (x + 60~) - 3. 2 37. sin y + sin 3 y = cos y - cos 3 y. Find the general value of x that satisfies the following equations: 1 w 38. cosx = -- and tanx =1. Ans. x = (2 n + 1)7r + V/2 4 39. cotx =- V3 and csc x =- 2. x = 2 n7r - 6 40. Find positive values of A and B which satisfy the equations cos(A - B) = and sin(A + B)=. Ans. 7 and 7r 2 2 12 4 41. Find positive values of A and B which satisfy the equations 2 25 7 19w tan (A - B) = 1 and sec (A + B)= Ans. and _V/3_ 24 24 CHAPTER VI GRAPHICAL REPRESENTATION OF TRIGONOMETRIC FUNCTIONS 49. Variables. A variable is a quantity to which an unlimited number of values can be assigned. Variables are usually denoted by the later letters of the alphabet, as x, y, z. 50. Constants. A quantity whose value remains unchanged is called a constant. Numerical or absolute constants retain the same values in all problems, as 2, 5, /;7, wr, etc. Arbitrary constants are constants whose values are fixed in any particular problem. These are usually denoted by the earlier letters of the alphabet, as a, b, c, etc. 51. Functions. A function of a variable is a magnitude whose value depends on the value of the variable. Nearly all scientific problems deal with quantities and relations of this sort, and in the experiences of everyday life we are continually meeting conditions illustrating the dependence of one quantity on another. Thus, the weight a man is able to lift depends on his strength, other things being equal. Hence we may consider the weight lifted as a function of the strength of the man. Similarly, the distance a boy can run may be considered as a function of the time. The area of a square is a function of the length of a side, and the volume of a sphere is a function of its diameter. Similarly, the trinomial x2 -7x - 6 is a function of x because its value will depend on the value we assume for x, and sinA, cos2A, tan2 are functions of A. 52. Graphs of functions. The relation between the assumed values of a variable, and the corresponding values of a function depending on that variable, are very clearly shown by a geometrical representation where the assumed values of the variable are taken as the abscissas, and the corresponding values of the function as the 93 94 PLANE TRIGONOMETRY ordinates of points in a plane (see ~ 13, p. 26). A smooth curve drawn through these points in order is called the graph of the function. Following are General directions for plotting the graph of a function. First step. Place y equal to the function. Second step. Assume different values for the variable (= x) and calculate the corresponding values of the function (= y), writing down the results in tabulated form. Third step. Plot the points having the values of x as abscissas and the corresponding values of y as ordinates. Fourth step. A smooth curve drawn through these points in order is called the graph of the function. Ex. 1. Plot the graph of 2x- 6. Solution. First step. Let y = 2 x - 6. Second step. Assume different values for x and compute the corresponding values of y. Thus, if x=0, y=-6; x=-l, y=-8; x=1, y= —4; x=-2, y=-10; x=2, y=-2; etc. Arranging these results in tabulated form, the first two columns give the corresponding values of x and y when we assume positive values of x, and the x y x y -6 -6 1 -4 - - 8 2 -2 -2 -10 3 0 -3 -12 4 2 -4 -14 5 4 -5 -16 6 6 -6 -18 etc. etc. etc. etc. last two columns when we assume nega41f/~~ tive values of x. For the sake of symY\ metry x 0 is placed in both pairs of columns. Third step. Plot the points found. Fourth step. Drawing a smooth curve through these points gives the graph of the function, which in this case is a straight line. GRAPHICAL REPRESENTATION 95 Ex. 2. Plot the graph of x2-2 x- 3. Solution. First step. Let y =x2-2x- 3. Second step. Computing y by assuming values of x, we find the following table of values. x y x y 0 -3 0 -3 1 -4 -1 0 2 -3 -2 5 3 0 -3 12 4 5 -4 21 5 12 etc. etc. 6 21 ---- X etc. etc. Third step. Plot the points found. Fourth step. Drawing a smooth curve through these points gives the graph of Y the function. 53. Graphs of the trigonometric functions. To find the graph of a trigonometric function x y x y we assume values for the angle; the circular 00 0 0 0 0 0 measures of these angles. -30 7 -.50 30~ ~.50 - 300.50 are taken as the ab- 6 6 7r 7r scissas, and the corre- 600.86 - 60 -.86 3 2 spending values of the function found from 900 - 1.00 - 900 - 1.00 2 2 the table on p. 9 are 2r.86 120 27r -.8 -,, *L, 120~ -.86_120o - -.86 taken as the ordinates 3 3 of points on the graph. 51 50r - 150o.150 - 1500.5O 6 6 Ex. 1. Plot the graph 180~ 7r 0 - 180~ - r 0 of sinx. 7w 7w of sin x. 2100 7r -.50 -210 77r.50 Solution. First step. Let 6 6 y = sin x. 4w 4wr y = sin x. 240 -.8 - 2400 4 r.86 Second step. Assuming 3 3 values of x differing by 37r 3 7 2700 - 1.00 - 270 1.00 30~, we calculate the corre- 2 2 sponding values of y from 5 7r 5 r the table on p. 9. In tabu- 3 lating the results it will be 11 33 117r.50 noticed that the angles are. 330 expressed both in degree 3600 27 0 -360~ -27r 0 measure and in circular 96 PLANE TRIGONOMETRY measure. It is most convenient to use the degree measure of an angle when looking up its function, while in plotting it is necessary to use its circular measure. Third step. In plotting the points we must use the circular measure of the angles for abscissas. The most convenient way of doing this is to lay off distances 7r = 3.1416 to the right and left of the origin and then divide each of these into six equal parts. Then when x = O, y = O; x =, y =.50 =AB; 6 7l' x= -, y=.86=CD; 3 x = -, y = 1.00 =EF; etc. 2 7' Also when X = —, y =-.50 = GH; etc. 6.- 2 V. X of sin x for values of x between - 2 r and 2 7r. It is called the sine curve or sinusoid. Discussion. (a) Since sin (x 2 7r) = sin x, it follows that y - sin x = sin (x + 2 7r), that is, the graph is unchanged if we replace x by x 2 7r. This means, however, that every point is moved a distance 2 7r to the right or left. Hence the arc PNMLO may be moved parallel to XX' until P falls at O, N at F, M at I, etc., that is, into the position OFIJK, and it will be a part of the curve in its new position. In the case of the sine curve it is then only necessary to plot points, say, from x f- to x = w, giving the arc or double undulation ML OFI. The sine curve consists of an indefinite number of such arcs extending to the right and left. (b) From the graph we see that the maximum value of sin x (= y) is 1(=EF= QN, etc.) and the minimum value is -1(= SJ= RL, etc.), while x can take on any value whatever. (c) Since the graph crosses the axis of x an infinite number of times, we see that the equation sin x = 0 has an infinite number of real roots, namely, x = 0, ~ 2 7r, ~ 4 r, etc. GRAPHICAL REPRESENTATION 97 54. Periodicity of the trigonometric functions. From the graph of sin x in the above example we saw that as the angle increased from 0 to 2 7r radians, the sine first increased from 0 to 1, then decreased from 1 to - 1, and finally increased from - 1 to 0. As the angle increased from 2 7r radians to 4 7r radians, the sine again went through the same series of changes, and so on. Thus the sine goes through all its changes while the angle changes 2 7r radians in value. This is expressed by saying that the period of the sine is 2 7r. Similarly, the cosine, secant, or cosecant passes through all its changes while the angle changes 2 wr radians. The tangent or cotangent, however, passes through all its changes while the angle changes by 7r radians. Hence, the period of the sine, cosine, secant, or cosecant is 27r radians; while the period of the tangent or cotangent is 7r radians. As each trigonometric function again and again passes through the same series of values as the angle increases or decreases uniformly, we call them periodic functions. 55. Graphs of trigonometric functions plotted by means of the unit circle. The following example will illustrate how we may plot the graph of a trigonometric function without using any table of numerical values of the function for different angles such as given on p. 9. Ex. 1. Plot the graph of sinx. Solution. Let y = sin x. Draw a unit circle. Divide the circumference of the circle into any number of equal parts (12 in C.J,^., G.s.in of..arA,; * tadimte. n the sine of he ne the same... a i I i,!l,.sine of arcAJ. OJ etc.: It is evident that if we take the lengths of the arcs as the abscissas and the this case). At length s of division drop perpendiculars to the horizontal diameter. Then the sine o f the angle AOB, or, what amounts to the same thing, sine of arcAB = QB, sine of arc AE = NE, sine of arcAJ = OJ, etc. It is evident that if we take the lengths of the arcs as the abscissas and the corresponding lengths of the perpendiculars as the ordinates of points in a plane, these points will lie on the graph of sin x. If we choose the same scale as in 98 PLANE TRIGONOMETRY Ex. 1, p. 96, the two graphs could be made to coincide, but in this example the unit of length chosen is larger. The main features of the two graphs of sin x are the same, however, the discussion being the same for both. When IN IN CIRCLE GRAPH x = arc zero = zero, x = arcAB = OA, x= arcAC = OC, x = arcAD = OE, x = arcAE = OG, x =arcAF = 0I, x = arcAG = OK, x = arcAH = OL, IN IN CIRCLE GRAPH y = zero = zero; y= QB =AB; y = PC = CD; y = OD = EF; y=NE= G-H; y = MF =IJ; y = zero = zero; y = MH = LM, etc. EXAMPLES 1. Plot the graphs of the following functions: (a) x + 2. (b) 3x - 6. (c) 2 + 1. (d) X2. (e) X3. (f) 1 x-2 x-2 (h) x+l' x + 1 (i) 2Z. (j) logio x. (k) 2 x2 - 4. (1) 8 - 2. (m) 6 + 5x + x2. (n) x2 -3x +2. (o) x2-4 x + 3. (p) x3 - 4 x. (q) 3 - 2x +1. (r) X3-7 z + 6. (s) x3- 5x- 12. (t) X - 1. (u) X5 - 2. 2. Plot the graph of cos x. Solution. Let y = cos x. The cosine curve is found to be as follows::'\ Yx O To plot the graph of cos x by means of the unit circle we may use the circle on p. 97. Taking the abscissas as arcs zero, AB, AC, AD, etc., and the corresponding ordinates as OA, OQ, OP, zero, etc., respectively, we will get points lying on the cosine curve. 3. Plot the graph of tan x. Solution. The tangent curve is shown on next page. GRAPHICAL REPRESENTATION 99 To construct the tangent curve from the unit circle shown, we have IN IN IN IN CIRCLE GRAPH CIRCLE GRAPH When x = arc zero = zero, y = zero = zero; x = arc AB = OA, y =AM = AB; x =arc AC = OC, y=AN=CD; =arcAD = OD, y= =; =0; x = arc A = OE, y =A Q =EF, etc. 4. Plot the graph of sec x. Solution. The secant curve is given below. I,, YA I i I U -1; II r I II r III II II II I i ti!, -1! I I I I I II Using the unit circle, we have IN IN IN IN CIRCLE GRAPH CIRCLE GRAPH When x = arc zero = zero, y = OA = OA; x = arcAB = OB, y = OM= BC; x=arcAC = OD, y= ON= DE; x = arcAD = OF, y = = o; x = arcAE = OG, y = OQ GH, etc. 100 PLANE TRIGONOMETRY 5. Plot the cotangent curve. 6. Plot the cosecant curve. 7. Draw graphs of (a) sin x + cos x, (b) cos x - sin x, (c) sin 2 x, (d) tan 2 x, (e) sin x cos x. CHAPTER VII SOLUTION OF OBLIQUE TRIANGLES 56. Relations between the sides and angles of a triangle. One of the principal uses of Trigonometry lies in its application to the solution of triangles. That is, having given three elements of a triangle (sides and angles) at least one of which must be a side, to find the others. In Plane Geometry the student has already been taught how to solve triangles graphically. That is, it has been shown how to construct a triangle, having given CASE I. Two angles and one side. CASE II. Two sides and an opposite angle. CASE III. Two sides and the included angle. CASE IV. Three sides. From such a construction of the required triangle the parts not given may be found by actual measurement with a graduated ruler and a protractor. On account of the limitations of the observer and the imperfections of the instruments used, however, the results from such measurements will, in general, be only more or less rough approximations. After having constructed the triangle from the given parts by geometric methods, it will be seen that Trigonometry teaches us how to find the unknown parts of the triangle to any degree of accuracy desired, and the two methods may then serve as checks on each other. The student should always bear in mind, when solving triangles, the two following geometrical properties which are common to all triangles: (70) The sum of the three angles equals 180~. (71) The greater side lies opposite the greater angle, and conversely. The trigonometric solution of oblique triangles depends upon the application of three laws, —the law of sines, the law of cosines, and the law of tangents, to the derivation of which we now turn our attention. 101 102 PLANE TRIGONOMETRY 57. Law of sines. The sides of a triangle are proportional to the sines of the opposite angles. Proof. Fig. 1 represents a triangle all of whose angles are acute, while Fig. 2 represents a triangle, one angle of which is obtuse (as A). FIG. 1 FIG. 2 Draw the perpendicular CD(= h) on AB or AB produced. From either figure, using the right triangle A CD, (A) sinA= -- b In Fig. 2, sinA= sin (1800 - A)= sin CAD = h. I b ] Also, using the right triangle BCD, (B) Dividing (A) by (B) gives h sinB = -- a sin A a sin B b or, by alternation in proportion, (C) a b sin A sinB Similarly, by drawing perpendiculars from A and B we get (D) (E) b c =- - C' and sinB sin C c a si =n -A respectively. sin C sinA Writing (C), (D), (E) as a single statement, we get the law of sines. a b c (72) A- B = si sin A sin B= sin C SOLUTION OF OBLIQUE TRIANGLES 103 Each of these equal ratios has a simple geometrical meaning, as may be shown if the law of sines is proved as follows: Circumscribe a circle about the triangle ABC as shown in the figure, and draw the radii OB, OC. Denote the radius of the circle by R. Draw OM perpendicular to BC. Since the inscribed angle A is measured by one half of the arc BC and the central angle BOC is measured by the whole arc BC, it follows that the angle BOC =2A, or, angle BOM = A. A Then BM= R sinBOM= R sinA, by (7), p. 11 and a = 2BM = 2R sinA, a or, 2R = a \ sin A/ b In like manner it may be shown that 0 b c 2R1b= and 2R c. \l sin B sin C Hence, by equating the results, we get 2R= a = b c sinA sinB sin C The ratio of any side of a triangle. to the sine of the opposite angle is numerically equal to the diameter of the circumscribed circle. It is evident that a triangle may be solved by the aid of the law of sines if two of the three known elements are a side and its opposite angle. The case of two angles and the included side being given, may also be brought under this head, since by (70), p. 101, we may find the third angle which lies opposite the given side. Ex. 1. Given A = 65~, B= 400, a = 50 ft.; -solve the triangle. Solution. Construct the triangle. Since two angles are given we get the third angle at once from (70), p. 101. Thus, C = 180~ - (A + B) = 180~ - 105~ = 75~. Since we know the side a and its opposite angle A we may use the law of sines, but we must be careful to choose such ratios in (72) that only one unknown quantity is involved. Thus, to find the side b use C a b sin A sin B Clearing of fractions and solving for the only Co/ \ unknown quantity b, we get A 650 40c ^S b asinB A B i - Ac=?. sin A Substituting the numerical values of sin A and sin B from the table on p. 9, and a = 50 ft., we get.n n.Aa4 b = 0 ". = 35.46 ft. '0.9063 104 PLANE TRIGONOMETRY Similarly, to find the side c, use a c sin A sin C Clearing of fractions and solving for c, we get a sin C 50 x 0.9659 5 c...= 53.29 ft. sin A 0.9063 By measurements on the figure we now check the results to see that there are no large errors. Since we now know all the sides and angles of the triangle, the triangle is said to be solved. 58. The ambiguous case. When two sides and an angle opposite one of them are given, the solution of the triangle depends on the law of sines. We must first find the unknown angle which lies opposite one of the given sides. But when an angle is determined by its sine, it admits of two values which are supplements of each other; hence either value of that angle may be taken unless one is excluded by the conditions of the problem. C Let a and b be the given sides and A (opposite the side a) the given angle. b/ \a If a> b, then by Geometry A >B, and B must be acute whatever be the value A, ___ \B ~of A, for a triangle can have only one c obtuse angle. Hence there is one, and only one, triangle that will satisfy the given conditions. If a = b, then by Geometry A= B, both A c and B must be acute, and the required triangle is isosceles. b a If a <b, then by Geometry A < B, and A must be acute in order that the triangle shall A be possible; and when A is acute it is evident c from the figure that the two triangleAs ACB and ACB' will satisfy C the given conditions provided a is greater than the perpendicular CP; that is, provided b</| \c a > b sinA. a a The angles ABC and AB'C are A B\ supplementary (since Z B'BC = Z BB'C); they are, in fact, the supplementary angles obtained (using the law of sines) from the formula bs b sin A sinB= ---- a SOLUTION OF OBLIQUE TRIANGLES 105 That is, we get the corresponding acute value B from a table of sines, and the supplementary obtuse value as follows: B'= 180~-B. If, however, a = b sin A = CP, then sin B = 1, B = 90~, and the triangle required is a right triangle. If a < b sinA (that is, greater than CP), then sinB >1, and the triangle is impossible. These results may be stated in compact form as follows: Two solutions: If A is acute and the value of a lies between b and b sin A. No solution: If A is acute and a < b sin A, or f A is obtuse and a < b or a =b. One solution: In all other cases. The number of solutions can usually be determined by inspection on constructing the triangle. In case of doubt find the value of b sil A and test as above. Ex. 1. Given a = 21, b = 32, A = 115~; find the remaining parts. Solution. In this case a < b and A > 90~; hence the triangle is impossible and there is no solution. Ex. 2. Given a = 32, b = 86, A = 30~; find the remaining parts. Solution. Here b sinA =86 x - 43; hence a < b sinA, and there is no solution. Ex. 3. Given a = 40, b = 30, A = 75~; find the remaining parts. Solution. Since a > b and A is acute there is one solution only. By the law of sines, C a b K sin A sin B _bsinA 30 x.9659 A/ AD or, sin B = bsin A 0 x. 659 a 4O.-. sin B =.7244, or, B = 46.4~, the only admissible 75 B value of B. c=? Then C = 180 - (A + B) = 180 - 121.4~ = 58.6~. To find C, we get, by the law of sines, c a sin C sin A a sin C 40 x.8535 orc -- A - - = 35.3. orv isin A.9659 Check the results by measurements on the figure. 106 PLANE TRIGONOMETRY Ex. 4. Solve the triangle, having given b = 15, a = 12, A = 52~. Solution. Here bsinA = 15 x.7880 = 11.82; hence, since A is acute and a lies between b and b sin A, there are two solutions. That is, there are two triangles, A CB1 and A CB2, which satisfy the given conditions. By the law of sines, a b sin A sin B1 ob sinA 15 x.7880 or, sin B1 a..9850. a 12 This gives B1 = 80.07~, and the supplementary angle B2 = 180 - B1 = 99.93~. Let us first solve completely the triangle AB1C. C = 180~ - (A + B1) = 47.93~. Bythe law of sines, a = 1c sin A sin C1 C or, a sin C1 12 x.7423 c1= =- = 11.3. sinA.7880 Now, solving the triangle AB2C, C2 = 180~ - (A + B2) = 28.07~. By the law of sines, a= sin A sin C2 a sin C2 12 x.4706 or, c2 = 7.2. sin A.7880 A A A — k —c-2-> K --- C( ---- ^ The solutions then are: For triangle AB1C For triangle AB2C B1= 80.07~, 2 = 99. 93~, C = 47.93~, C2 = 28.07~, cl = 11.3. c2 = 7.2. Check the results by measurements on the figure. In the ambiguous case care should be taken to properly combine the calculated sides and angles. EXAMPLES 1. Find the number of solutions in the following triangles, having given: (a) a = 80, b = 100, A = 30~. Ans. Two. (b) a = 50, b = 100, A = 30~. One. (c) a = 40, b = 100, A = 30~. None. (d) a = 13, b = 11, A = 69~. One. (e) a = 70, b = 75, A = 60~. Two. (f) a = 134, b = 84, B = 52~. None. (g) a = 200, b = 100, A = 30. One. 2. Solve the triangle, having given a = 50, A = 65~, B = 40~. Ans. C = 75~, b = 35.46, c = 53.29. 3. Solve the triangle, having given b = 7.07, A = 30~, C = 105~. Ans. B = 45~, a = 5, c =9.66. SOLUTION OF OBLIQUE TRIANGLES 107 4. Solve the triangle, having given c 9.56, A = 45~, B = 60~ Ans. C=75~, a = 7, b=8.57. 5. Solve the triangle when c = 60, A =50~, B = 75~. Ans. C=55~, b=70.7, a=56.1. 6. Solve the triangle when a = 550, A = 10~ 12', B = 46~ 36'.?ns. C = 123~ 12', b = 2257.4, c = 2600.2. 7. Solve the triangle when a = 18, b = 20, A = 55.4~. Ans. B1 = 66.2~, C1 = 58.4~, cl = 18.6; B2 = 113.8~, C2 = 10.8~, c2 = 4.1. 8. Solve the triangle when a = 3 2, b = 2 3, A = 60~. Ans. C =75~, B -45~, c= 4.73. 9. Solve the triangle when b = 19, c = 18, C = 15~ 49'. Ans. B1 = 16~ 42', Al = 147~ 28', al = 35.5; B2 = 163~ 18', A2 = 5.3', a2 = 1.04. 10. Solve the triangle when a = 119, b = 97, A = 50~. Ans. B = 38.6~, C = 91.4~, c = 155.3. 11. Solve the triangle when a = 120, b = 80, A = 60~. Ans. B = 35.3~, C = 84.7~, c = 137.9. 12. It is required to find the horizontal distance from a point A to an inaccessible point B on the opposite bank of a river. We measure off any convenient horizontal distance as A C, and then measure the angles CAB and A CB. Let AC=283 feet, angle CAB=38~, and angle ACB=66.3~. Solve the triangle ABC for the side AB. Ans. 267.4 ft. 13. A railroad embankment stands on a horizontal plane and it is required to find the distance from a point A in the /// plane to the top B of the embankment. Select a point C at B the foot of the embankment lying in the same vertical plane as A and B, and measure the distances A C and CB, and the angle BAC. Let AC = 48.5 ft., BC = 84 ft., and angle BAC =21.5~. Solve the triangle for the side AB. Ans. 127.2 ft. 14. A tree A is observed from two points B and C, 270 ft. apart, on a straight road. The angle BCA is 55~ and the angle CBA = 65~. Find the distance from the tree to the nearer point B. Ans. 255.4 ft. 15. To determine the distance of a hostile fort A from a place B, a line BC and the angles ABC and BCA were measured and found to be 1006.62 yd., 44~, and 70~ respectively. Find the distance AB. Ans. 1035.5 yd. 16. A triangular lot has two sides of lengths 140.5 ft. and 70.6 ft., and the angle opposite the former is 400. Find the length of a fence around it. Ans. 353.9 ft., or 529.6 ft. 17. Two buoys are 64.2 yd. apart, and a boat is 74.1 yd. from the nearer buoy. The angle between the lines from the buoys to the boat is 27.3~. How far is the boat from the further buoy? Ans. 120.3 ft. 108 PLANE TRIGONOMETRY 18. Prove the following for any triangle: (a) a = b cos C + c cosB, b = a cos C + c cosA, c = a cosB + b cosA. b2 sin C + c2 sin B (b) /bc sin B sin C = -— + — b +c sinA + 2sinB sill C ) a + 2b c sin2 A - m sin2 B sin2 C (d) a2 - mb2 c2 (e) asin(B - C) + bsin(C-A) + csin(A - B) = 0. 19. If R is the radius of the circumscribed circle, prove the following for any triangle [s = - (a + b + c)]: (a) R (sin A + sin B + sin C) = s. (b) bc = 4 R2 (cosA + cos B cos C). 1 1 1 1 4R (c) - s-c s /s-a s-b s( )(s-b)(s-c) 20. Show that in any triangle a+ b cos -(A -B) c sin.- C 59. Law of cosines. In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides into the cosine of their included angle. Proof. Suppose we want to find the side a in terms of the other two sides b and c and their included angle A. C C b a a A c B D c B FIG. 1 FIG. 2 When the angle A is acute (as in Fig. 1) we have, from Geometry, 2 2 CB =AC +AB - 2AB X AD, [The square of the side opposite an acute angle equals the sumof the squares of the other two sides minus twice the product of one of thbse sides into the projection of the other upon it._ or, a2 b2 c2- 2 cAD. But AD= b cosA. (8), p. 11 Hence a2 = b2 + c2 - be cos A. SOLUTION OF OBLIQUE TRIANGLES 109 When the angle A is obtuse (as in Fig. 2) we have, from Geometry, - 2 2 2 CB =AC2 AB2 + 2AB X AD, rThe square of the side opposite an obtuse angle equals the sum] of the squares of the other two sides plus twice the product of one of those sides into the projection of the other upon it._ or, a2 = b2 + c2 + 2 cAD. But AD = b cosDAC (8), p. 11 =b cos (180~ -A) - b cos A. Hence in any case (73) a2= b2 + c2- 2 bccosA. Similarly, we may find (74) b2 = a2 + c2 - 2ac cosB. (75) c2= a2+ b2 - 2 ab cosC* Observe that if A = 90~, then cos A = 0, and (73) becomes a2 = b2 + c2, which is the known relation between the sides of a right triangle where A is the right angle. Solving (73), (74), (75) for the cosines of the angles, we get (76) cos A = 2 C2 bc a2 + C2 - b2 (77) cosB = 2 ac a2 + b2 - C2 (78) cosC = 2a These formulas are useful in finding the angles of a triangle, having given its sides. Formulas (73), (74), (75) may be used for finding the third side of a triangle when two sides and the included angle are given. The other angles may then be found either by the law of sines or by formulas (76), (77), (78). * Since a and A, b and B, c and C stand for any side of a triangle and the opposite angle, from any formula expressing a general relation between these parts another formula may be deduced by changing the letters in cyclical order. Thus, in (73) by changing a to b, b to c, c to a, and A to B we obtain (74); and in (74) by changing b to c, c to a, a to b, and B to C we get (75). This is a great help in memorizing some sets of formulas. 110 PLANE TRIGONOMETRY Ex. 1. Having given A = 47~, b = 8, c = 10; solve the triangle. pC ~ ~ ~ Solution. To find the side a use (73). a2 = b2 + c2 - 2 bccosA 6// * \z%. 64 + 100 - 2 x 8 x 10 x.6820 54.88. 4~/ o~7 \~.7. a = -/54.88 = 7.408. A B C = 1o To find the angles C and B use the law of sines. bsinA 8 x.7314 sin B= --.7898... B =52.20. a 7.408 c sinA 10 x.7314 sin C =. = 1.9873..'. C = 80.8~. a 7.408 To check our work we note the fact that A + B + C = 47~ + 52.2~ + 80.8~= 180~. Ex. 2. Having given a = 7, b = 3, c = 5; solve the triangle. Solution. Using formulas (76), (77), (78) in order to find the angles, we get b2 + C2-a2 32 + 52 72 1 120 cosA- ---.5000.. A=1200. 2bc 2.3.5 2 a2 + c2 - b2 72 + 52 - 32 13 cosB a.19286..B = 21.80~. 2 ac 2 *.7.5 14 a2 + b2 - c2 72 + 32 - 52 11 cos C = =.7857. C = 38.2~. 2 ab 2 ~ 7 ~ 3 14 Check: A +B+C=120~ + 21.8 + 38.2 = 180~. EXAMPLES 1. Having given a = 30, b = 54, C = 46~; solve the triangle. Ans. A = 33.1~, B= 100.9~, c = 39.56. 2. Having given A = 60~, b = 8, c = 5; find a and the cosines of the angles B and C. Ans. 7,, -4 3. Having given a = 33, c = 30, B = 35.4~; find A and C. Ans. A = 80.7~, C = 63.9~. 4. Having given a = 4, b = 7, c = 10; solve the triangle. Ans. A = 18.2~, B = 33.1, C = 128.7~. 5. Having given a = 21, b = 24, c = 27; solve the triangle. Ans. A = 48.2~, B = 58.4~, C = 73.4~. 6. Having given a = 2, b = 3, c = 4; find the cosines of the angles A, B, C. Ans. 7 11 -, TUI - 4 7. Having given a = 77.99, b = 83.39, C = 72~ 15'; solve the triangle. Ans. A = 51~ 15', B = 56~ 30', c = 95.24. 8. If two sides of a triangle are 10 and 11 and the included angle is 50~, find the third side. Ans. 8.92. SOLUTION OF OBLIQUE TRIANGLES Ill 9. The two diagonals of a parallelogram are 10 and 12 and they form an angle of 49.3~; find the sides. Ans. 10 and 4.68. 10. In order to find the distance between two objects, A and B, separated by a pond, a station C was chosen, and the distances CA = 426 yd., CB = 322.4 yd., together with the angle A CB = 68.7~, were measured. Find the distance from A to B. ~C -' ~~~~~Ans. 430.85 yd. 11. A ladder 52 ft. long is set 20 ft. from the foot / I ' I of an inclined buttress, and /, I, I reaches 46 ft. up its face. Find the inclination of the face of the buttress. ~/////'/''//''/ Ans. 95.-9. 12. Under what visual angle is an object 7 ft. long seen by an observer whose eye is 5 ft. from one end of the object and 8 ft. from the other end? Ans. 60~. 13. Two stations, A and B, on opposite sides of a mountain, are both visible from a third station C. The distance AC = 11.5 mi., BC = 9.4 mi., and angle ACB = 59.5~. Find the distance between A and B. Ans. 10.5 mi. 14. Prove the following for any triangle: (a) a(b2+ c2)cosA + b(c2+ a2)cosB+ c(a2+ b2)cos C= 3abc. b + c cosB + cosC a 1 - cosA (c) a + b + c = (b + c)cosA + (c + a) cosB+ (a + b)cos C. cos A cos B cos C a2 + b2 + c2 (d) —+ + + --- a b c 2abe (e) a2 + b2 + c2 = 2 (ab cos C + be cosA + ca cos B).. 60. Law of tangents. The sum of any two sides of a triangle is to their difference as the tangent of half the sum of their opposite angles is to the tangent of half their difference. Proof. By the law of sines, a b sinA sinB and by division and composition in proportion, a a- b sin A + sinB (A) a -b sinA-sinB But from (66), p. 74, sin A + sin B tan V (A + B) sinA - sinB tan (A -B) 112 PLANE TRIGONOMETRY Hence equating (A) and (B), we get (79) a +b tan (A+ B) * (79) 2 a-b tan (A - B) a + c tan - (A + C) Similarly, we get t Aa C-C tan -(A - C) b + c tan (B+ C) t b- c tan - (B -C) When two sides and the included angle are given, as a, b, C, the law of tangents may be employed in finding the two unknown angles A and B.t Since a + b, a - b, A + B (= 180~- C), and therefore also tan (A +-B), are known, we clear (79) of fractions and solve for the unknown quantity tan (A -B). This gives (80) tan (A -B) = a tan (A + B). a+b We shall illustrate the process by means of an example. Ex. 1. Having given a = 872.5, b = 632.7, C = 80~; solve the triangle. Solution. a + b =1505.2, a-b =239.8, A + B 180-C = 100~, and (A + B)= 50~. From (79), since tan - (A + B) tan 50~ = 1.1918, a-b 239.8 tan - (A - B) - tan (A + B) = 2 x 1.1918 =.1899. a + b 1505.2.. (A -B) - 10.6~. i 80 Adding this result to - (A + B) = 50~ gives A = 60.6~. ^^^^*^^^? Subtracting the result from I (A + B) = 50~ gives B = 39.4~. To find the side c, use the law of sines. Thus, a sin C 872.5 x.9848 C = = 986.2. sinA.8712 We will now derive formulas for solving triangles having three sides given, which arel more convenient than (76), (77), (78), p. 109. * If b > a, then B > A, making a - b and A - B negative. The formula still holds true, but b+a. tan (B+A) to avoid negative quantities it is better to write the formula in form b- = ta (B-A) t These may also be found by changing the letters in cyclical order (see footnote, p. 109). $ When logarithms are used in solving triangles, having given two sides and the included angle, the law of tangents, which involves products, is to be preferred to the law of cosines, which involves sums. SOLUTION OF OBLIQUE TRIANGLES 113 61. Trigonometric functions of the half angles of a triangle in terms of its sides. Denote half the sum of the sides of a triangle (i.e. half the perimeter) by s. Then (A) 2s=a+b+c. Subtracting 2 c from both sides, 2s - 2 c = a +b + c-2c, or, (B) 2(s -c)= a + b - c. Similarly, (C) 2(s -b)= a- b + c, (D) 2 (s - a)=- a + b + c. In (49a), (49b), p. 70, replace 2x by A, and, what amounts to the same thing, x by - A. This gives (E) 2 sin2 A = 1-cosA, (F) 2 cos2 -A = 1 + cos A. b2 + C2 _ a2 But from (76), p. 109, cos A = -; hence (E) becomes 2 b ee (E) becomes b2 + c2 _ a2 (G) 2 sin2 A = b - )2 2 be 2 be - b2 - c2 + a2 2bc a_ 2 _ (b2 - 2 b + C2) 2 be a - (b - C)2 2 be (a + b - c) (a - b + c) 2 be [a2 - (b - c)2 being the product of the sum and difference of a and b - c.] _ 2 (s - c)2(s-b) by (B), (C) 2 be.'.=(s - b),(s ). 2 ~bc 114 PLANE TRIGONOMETRY Similarly, (F) becomes b2 + C2 _ a2 2cos2 I 1 A =1 + 2 be - b2 + c2- a2 2bc (b + C)2 - a2 2 be (b + c + a) (b + c - a)' 2 bc 2s2 (s - a) 2 be (82). cos A = (- ). bc Since tan sin A Since tanA= cos- we get, by substitution from (81) and (82), Ccos A (83) tan 1A = s b) (s -c) ) S (S - a) Since any angle of a triangle must be less than 180~, -A must be less than 90~ and all the functions of A must be positive. Hence only the positive signs of the radicals in (81), (82), (83) have been taken. Similarly, we may get sin B = (S- a)(s - c) sin C - a)(s -b) s (s - b) s (s - c) cos B = - cos 2C = h 2 ac 2 ab tan B- (S - a) (s - c) tan( - a) (s - b) 2 N (S - 6) 2 S (s - C) There is then a choice of three different formulas for finding the value of each angle, If half the angle is very near 0~, the formula for the cosine will not give a very accurate result, because the cosines of angles near 0~ differ little in value; and the same holds true of the formula for the sine when half the angle is very near 90~. Hence in the first case the formula for the sine, in the second that for the cosine, should be used. In general, however, the formula for the tangent is to be preferred. * Also found by changing the letters in cyclical order. SOLUTION OF OBLIQUE TRIANGLES 115 When two angles, as A and B, have been found, the third angle, C, may be found by the relation A +B + C = 180~, but it is best to compute all the angles from the formulas, so that we use the sum of the angles as a test of the accuracy of the results. It is customary to use a second form of (83), found as follows: tan - A = X(S - b)(s - c) tan I A 2s (s - a) (s - a)(s - ) (s - c) s(s - a)2 Multiplying both numerator and denominator of the fraction under the radical by s - a.] -1 (s-a) (s-b) (s-c) s - C S Denoting the radical part of the expression by r, (s- a) (s- b) (s- c) (84) r a) (s-(s —c) and we get r =, (85) tan 1A = -- Similarly, 2 s - ad (86) tan B = -b r (87) tan C SBy proving one of the last three formulas geometrically it may be shown that r is the radius of the inscribed circle. C Proof. Since angle NA 0 = A, L (A) tan- A = N 2 AN If s denotes half the perimeter, we have 2s =AN+ NB+BL+ LC+ CM+MA. A 1B But NB = BL, CM = LC, MA = AN; therefore 2s = 2AN 2 BL + 2LC, or, s = AN + (BL + LC)= AN + a. This gives AN = s - a. NO Substituting in (A), tan - A = sComparing this result with (85) and (84) shows that NO ==- - a) (s - b) (s - c) NO=r==Cs * When logarithms are used in solving triangles, having given the three sides, formulas (84), (85), (86), (87), which involve products, are more convenient than the law of cosines, which involves sums. 116 PLANE TRIGONOMETRY Ex. 1. Solve the triangle whose sides are 13, 14, 15. Solution. Let a = 13, b = 14, c = 15. Then 2s = a + b + c = 42, or, s = 21. Also, s- a = 8, s- b = 7, s - c = 6. From (84), r= (s-a)(s- b) (s- c) 7. 6 = 4. r 4 1 From (85), tan A = 4 =.5000. s-a 8 2-.. A = 26.56~, or A = 53.12~. r 4 From (86), tan I B =.5714. s-b 7.. B = 29.74~, or B = 59.48~. r 4 2 From (87), tan C = = =.6667. s —c 6 3. C. C = 33.69~, or C = 67.38~. Check: A + B + C = 53.12~ + 59.48~ + 67.38~ = 179.98~.* EXAMPLES 1. Solve Examples 1, 3, 8, p. 110, using the law of tangents. 2. Solve Examples 4, 5, 6, p. 110, using formulas (84), (85), (86), (87), p. 115. 3. Prove the following for any triangle: (a) (a + b) sin - C = c cos I (A - B). b+c-a (b) tan - B tan I C = b + (c) b cos2 I C + c cos2 - B = s. (d) (b + c - a) tan A = (c + a - b)tan B. (e) c2 = (a + b)2 sin2 I C + (a - b)2 cos2 - C. (f) c (cosA + cosB) = 2 (a + b) sin2 - C. cos2 2 A a (s - a) (g) (g) cos2 B b(s - b) C B (h) bsin2-+ c sin2- = s- a. 2 2 (i) a cos - B cos - C c-sc A = s. A A 2 ___________ (j) sinA = 2 sin - cos- = - Vs (s - a) (s - b) (s - c). 2 2 bc * The error.020 arises from the fact that we used a four-place table. If we had used a table giving the first five significant figures of the tangent,.the error would have been less; if a six-place table, still less, etc. For ordinary purposes, however, the results we get, using a four-place table, are sufficiently accurate. SOLUTION OF OBLIQUE TRIANGLES 117 4. If R and r denote the radii of the circumscribed and inscribed circles respectively, prove the following for any triangle: a sin - B sin i C 1 1 (a) r= 2 2 (C) -- --- cos A bc ca ab 2 rR abc esn abc (b) R abe (d) R =1 abc 4 s (s - a) (s- b)(s - c) 2 in A sin B sin C (e) abcr = 4 R (s - a) (s - b) (s - c). 62. Formulas for finding the area of an oblique triangle. CASE I. When two sides and the included angle are known. Let b, c, and A be known. Take c as the base. Denote the altitude by h and the area by S. Then, by Geometry, S- ch. But b = b sin A (from (7), p. 11); hence (88) S = 1 bc sinA. Similarly, S = ac sin B = 2 ab sin C. C C b h a h a A \B A The area of a triangle equals half the product of any two sides multiplied by the sine of the included angle. Ex. 1. Find the area of a triangle, having given b = 20 in., c = 15 in., A = 60~. Solution. Substituting in (88), 1 1 /3 = besinA = x 20 15x 5 = 75V3 sq. in. Ans. 2 2 2 CASE II. When the three sides are known. sin A = 2 sin I A cos A (51), p. 72 -2 (s - b)(s -).s(s-a) - i (81), (82), be be pp. 113, 114 b= _s (s-a) (s- b) (s - ). Substituting this value of sin A in (88), we get (89) S = s (s -a) (s- b) (s-c). Ex. 2. Having given a = 13, b = 14, c = 15; find the area. Solution. S = (a+b+c)=21, s-a=8, s-b=7, s-c=6. Substituting in (89), S = /s (s - a) (s - b) (s - c) = 21 x 8 x7x6 = 84. Ans. 118 PLANE TRIGONOMETRY CASE III. Problems which do not fall under Cases I or II directly may be solved by Case I, if we first find an additional side or angle by the law of sines. Ex. 3. Given a = 10 V3, b = 10, A = 120~; find the area of the triangle. Solution. This does not come directly under either Case I or Case II, but, by the law of sines, b sin A 10 x -\/3 1 sinB = 2n 1 a 10o 2 Therefore B = 30~ and C = 180~ - (A + B) = 30~. Since we now have the twro sides a and b and the included angle C, the problem comes under Case I, and we get S= ab sill C = x 10V3x 10 x1 = 25V3. Ans. EXAMPLES 1. Find the areas of the following triangles, having given (a) a =40, b = 13, c = 37. Ans. 240. (b) b = 8, c = 5, A = 60. 17.32. (c) b = 10, c = 40, A = 75~. 193.18. (d) a = 10, b = 12, C = 600. 30 V3. (e) a = 40, c = 60, B = 30~. 600. (f) a = 7, c=5 v, B = 1350. 17-. (g) b = 149, A = 70~ 42', B = 39~ 18'. 15,447.7. (h) a= 5, b =6, c = 7. 66. (i) a = 409, b = 169, c = 510. 30,600. (j) a =140.5, b = 170.6, A = 40~. 11,981 or 2345.8. (k) c = 8, B = 100.1~, C = 31.10. 46. (1) a = 7, c = 3, A = 60~. 10.4. 2. Show that the area of a parallelogram equals the product of any two adjacent sides multiplied by the sine of the included angle. 3. Find a formula for the area of an isosceles trapezoid in terms of the parallel sides and an acute angle. 4. Show that the area of a quadrilateral equals one half the product of its diagonals into the sine of their included angle. 5. The base of an isosceles triangle is 20, and its area is 100 - </3; find its angles. 6. Prove the following for any triangle: abe 2abc A B (a) S = (e) S= b (cos - cos - cos - 4R a 42 b +6 c 2 2 2 (b) S = rs. a2 b2 (c) S = Rr(sinA + sin B+ sin C). (f) =4 sin 2 B +4 sin2A (d) S = a2sinB sin C cs A. CHAPTER VIII THEORY AND USE OF LOGARITHMS 63. Need of logarithms* in Trigonometry. Many of the problems arising in Trigonometry involve computations of considerable length. Since the labor connected with extensive and complicated calculations may be greatly lessened by the use of logarithms, it is advantageous for us to use them in much of our trigonometric work. Especially is this true of the calculations connected with the solution of triangles. We shall now give the fundamental principles of logarithms and explain the use of logarithmic tables. Definition of a logarithm. The power to which a given number called the base must be raised to equal a second number is called the logarithm of the second number. Thus, if (A) b = N, (exponential form) then x = the logarithm of N to the base b. This statement is written in abbreviated form as follows: (B) x = log, N. (logarithmic form) (A) and (B) are then simply two different ways of expressing the same relation between b, x, and N. (A) is called the exponential form. (B) is called the logarithmic form. The fact that a logarithm is an exponent may be emphasized by writing (A) in the form (base)l~g = number. For example, the following relations in exponential form, namely, 32= 9, 25= 32, (Q)3 = 1, x =, are written respectively in the logarithmic form 2 = log39, 5 =log232, 3 = log-, y = log^x; * Logarithms were invented by John Napier (1550-1617), Baron of Merchiston in Scotland, and described by him in 1614. 119 120 PLANE TRIGONOMETRY where 2, 5, 3, y are the logarithms (exponents), 3, 2, -, x are the bases, and 9, 32, 1, z are the numbers respectively. Similarly, the relations 1 25 = - 25= 5, 10-3=.-001 -=10 — 1000 82 = X 64=4 b~ =-=1 are written in logarithmic form as follows: = log25 5, - 3 = g.001, log 4, 0 = logb1. EXAMPLES 1. In the following name the logarithm (exponent), the base, and the number, and write each in logarithmic form: 23 = 8, 42 = 16, 52 = 25, 33 = 27, 34 = 81. Solution. In the first one, 3 = logarithm, 2 = base, 8 = number; hence log2 8 = 3. Ans. 2. Express the following equations in logarithmic form: (~)2 = 9, 125 = 5, 2-4 = -6, 10-2 -.01, p = q. 3. Express the following equations in the exponential form: log464 = 3, log749 = 2, log216 = 3, log.Ol0001 - 4, log42 = 2, loga = 1, logal= 0, logba = c. 4. When the base is 2, what are the logarithms of the numbers 1, 2, -, 4, l, 8, 64, 128? 5. When the base is 5, what are the logarithms of the numbers 1, 5, 25, 125, 1 1 1 9 5W 25 625 ' 6. When the base is 10, what are the logarithms of the numbers 1, 10, 100, 1000, 10,000,.1,.01,.001,.0001? 7. When the base is 4 and the logarithms are 0, 1, 2, 3, - 1, - 2, -, what are the numbers? 8. What must be the bases when the following equations are true: log 64 = 2? log 121 = 2? log 625 = 4? log - =-2? 9. When the base is 10, between what integers do the logarithms of the following numbers lie: 83, 251, 1793? Solution. Since loglolO = 1 and logO 100 = 2, and 83 is a number lying between 10 and 100, it follows that log1o83 = a number lying between 1 and 2. Similarly, loglo251 = a number lying between 2 and 3, loglo 1793 = a number lying between 3 and 4. THEORY AND USE OF LOGARITHMS 121 10. Verify the following: (a) log1o 1000 + loglo 100 + loglo 10 + loglo = 6. (b) logioIo + logloT- - log10o o 0.= (c) log1o.001 - log10.l0 - logo1.l=- 2. (d) log28 - 3 log 2 + log2 1 = 2. (e) 2 logaa + 2 loga- + loga 1= 0. a (f) 2 log42 + - log24 - log22 = 1. (g) log33 + log03 - log 81 = - 5. (h) 3 log273 - 3 log 27 + log93 = (i) 4 log164 + 2 log4 -1'l + - 1og216 0. (j) 2 logs 64 - log749 + I log5 - = 1. (k) log 64 + log464 + log264 = 11. (1) log5 25 - log5 125 + 2 log5 5 = 1. (m) 2 log366 - logo36 + log 6 - 3. 64. Properties of logarithms. Since a logarithm is simply a new name for an exponent, it follows that the properties of logarithms must be found from the laws in Algebra governing exponents. Theorem I. The logarithm of the product of two factors equals the sum of the logarithms of the two factors. Proof. Let the two factors be 11 and N, and let x and y be their logarithms to the common base b. Then (A) logb -= x, and log N =y. Writing these in the exponential form, (B) bx = M, and b = TN. Multiplying together the corresponding members of equations (B), b+ Y =- 7IN. Writing this in the logarithmic form gives logb MN- = + y = logb M + logb N. from (A) By successive applications this theorem may evidently be extended to the product of any number of factors as follows: logb iMNPQ = logbM. NPQ = logbM + logbNPQ Th. I = logb J + logb N + logb PQ = logb 3 + logb N + log, P + log, Q. 122 PLANE TRIGONOMETRY Theorem II. The logarithm of the quotient of two numbers is equal to the logarithm of the dividend minus the logarithm of the divisor. Proof. As in Theorem I, let (A) logb M= x, and logbN = y. Writing these in the exponential form, (B) b= M, and b = N. Dividing the corresponding members of equations (B), we get M bxz-y J N Writing this in logarithmic form gives logb N x - Y = log M- logb N. from (A) Theorem III. The logarithm of the pth power of a number is equal to p times the logarithm of the number. Proof. Let logb N= zx. Then bx = N. Raising both sides to the pth power, bPX = NP. Writing this in logarithmic form gives log1 NP = px = p logb N. Theorem IV. The logarithm of the rth root of a number is equal to the logarithm of the number divided by r. Proof. Let logb N= x. Then bx = N. Extracting the rth root of both sides, x 1 b _ N. Writing this in logarithmic form gives x log, N log N - =- - =logb N. r' r r' From the preceding four theorems it follows that if we use the logarithms of numbers instead of the numbers themselves, then the operations of multiplication, division, raising to powers, and extracting roots are replaced by those of addition, subtraction, multiplication, and division respectively. THEORY AND USE OF LOGARITHMS 123 Ex. 1. Find the value of logioOl001. Solution. logio V.001 = logo-.001 Th. IV -=2 logi oo0 = - (-3) = 3-3. Ans. 5,3 (C + d).2 Ex. 2. Write logb ( + d) in expanded form. c2 5a3 ( d) 2/ 1 a3 (c + d)T Solution. logb \ v(C ) logb Th. IV c2 5 c2 = I logb a~ + logb (c + d) - logb c2 Th. I, II = 31ogba + logb(c + d)-2logbc. Th. III, IV When no base is indicated we mean that the same base is to be used throughout. Thus, the relation log 3 d) = c {3 loga + log(c + d) -2 logc} holds true for any number used as the base. For the sake of convenience we shall call the left-hand member of an equation like the last one the contracted form of the logarithmic expression, and the right-hand member the expanded form. Ex. 3. Write 3 log (x + 1) + 3 log(x - 1) + - log x - 2 log (x2 + 1) in the contracted form. Solution. 3 log (x + 1) + 3 log (x - 1) + - log x - 2 log (x2 + 1) = log (x + 1)3 + log(x - 1)3 + log x - log(x2 + 1)2 (X + 1)3 (X - 1)3 X \ VaX2 - 1)3 = log (x + 1)(x -- 1)3 log vx(x2- 1) Ans. (X2 + 1)2 (X2 + 1)2 Another form of the answer is found as follows: V (/ 2 - 1)3 (X2 - 1)6\2 1 x(x2- 1)6 log og og = - log (x2 + 1)2 \(x2+1)4/ 2 (x2 + 1)4 EXAMPLES 1. Verify the following: (a) logio v/10O + loglo /.Ol =. (e) log1 2 +log3 ()2 - -. (b) loglo (.1)4 -log. loglog2 (.5)3 - log4 /16 = —. (c) loglo 1/X = + loglo 1 = 0. (g) log -/1 log 121 = l3-. (d) loglo 10 - loglo (.01)2 = 34. (h) logs (2)5 + log7 (1 )1- = 1. 2. Write the following logarithmic expressions in expanded form: * (a) log - (s - b) (s - c) (e) log x ( - s - a z.(z + x) (b) logab sin C (f) log /p2 (1 -q) z Vw(1 + q) (c) logP(1 + r). (g) log (m+n)s2 m - n (1 + s),,a3b2C 41/a2(b - c)\3 (d) log-. (h) log 9 (4 To v a -b/ * To verify your results, reduce them back to the original form. 124 PLANE TRIGONOMETRY 3. Write the following logarithmic expressions in contracted form: (a) 2 log x + logy - 3 log z. (b) 31og(1 - x) - 21og(2 + x) + logc. (c) 5 log (x - 1) log z - log (x + 2) + log c. (d) log y - - log (y2 + 4) + log c. (e) l21og(x - 1) + 3 og(x + 1) + -l og- 2log(za2 + 1). 65. Common* system of logarithms. Any positive number except unity may be taken as the base, and to every particular base chosen there corresponds a set or system of logarithms. In the common system the base is 10, being the one most convenient to use with our decimal system of numbers. In what follows the base is usually omitted when writing expressions in the logarithmic form, the base 10 being always understood. Thus log10100 = 2 is written log 100 =2, etc. The logarithm of a given number in the common system is then the answer to the question: TWhat power of 10 will equal the given number? The following table indicates what numbers have integers for logarithms in the common system. Exponential Form Logar-ithmic Form Since 104 = 10,000 we have log 10,000 = 4 103 =1000 log 1000 = 3 102 = 100 log 100 = 2 10' =10 log 10 =1 10~ =1 log = 0 10-1=.1 log.1 =-1 10-2 =.01 log.01 =-2 10-3 =.001 log.001 =-3 10-4=.0001 log.0001 =-4 etc., etc. Assuming that as a number increases its logarithm also increases, we see that a number between 100 and 1000 has a logarithm lying between 2 and 3. Similarly, the logarithm of a number between.1 and.01 has a logarithm lying between - 1 and - 2. In fact the logarithm of any number not an exact power of 10 consists, in general, of a whole-ntmber part and a decimal part. * Also called the Briggs System, from Henry Briggs (1556-1631), professor at Gresham College, London, and later at Oxford. He modified the new invention of logarithms so as to make it convenient for practical use. THEORY AND USE OF LOGARITHMS 125 Thus, since 4587 is a number lying between 103 and 104, we have log 4587 = 3 + a decimal. Similarly, since.0067 is a number lying between 10-3 and 10-2, log.0067 = - (2 + a decimal) =- 2 - a decimal. For practical reasons the logarithm of a number is always written in such a form that the decimal part is positive. When the logarithm as a whole is negative, the decimal part may be made positive by adding plus unity to it. Then, so as not to change the value of the logarithm, we add minus unity to the whole part. Thus in the last example, log.0067 = (- 2) + (- a decimal) = (- 1 - 2)+ (1- a decimal) - 3 + a new decimal. To emphasize the fact that only the whole part of a logarithm is negative, the minus sign is usually written over the whole part. For example, log.004712 =- 2.3268 =-2 -.3268 = (- - 2) + (1 -.3268) =3.6732. The whole-number part of a logarithm is called the characteristic of the logarithm. The decimal part of a logarithm is called the mantissa of the logarithm. Thus if log 357 = 2.5527 and log.004712 = 3.6732, 2 and - 3 are the characteristics and.5527 and.6732 the mantissas. From the previous explanations and by inspection of the table on the opposite page we get the following: 66. Rules for determining the characteristic of a logarithm. The characteristic of a number greater than unity is positive, and one less than the number of digits in the number to the left of the decimal point. The characteristic of a number less than unity is negative, and is one greater numerically than the number of zeros between the decimal point and the first significant figure of the number. Ex. Write down the characteristics of the logarithms of the numbers 27,683, 456.2, 9.67, 436,000, 26,.04,.0000612,.7963,.8,.0012. Ains. 4, 2, 0, 5, 1, -2, - 5, -1, - 1, - 3. 126 PLANE TRIGONOMETRY Theorem V. Numbers with the same significant part * (and which therefore differ only in the position of the decimal point) have the same mantissa. Proof. Consider, for example, the numbers 54.37 and 5437. Let 10x = 54.37. If we multiply both members of this equation by 100 (= 102), we have 102. 10x = 10x+2 = 5437 or, x + 2 = log 5437. Hence the logarithm of one number differs from that of the other merely in its whole part (characteristic). Thus, if log 47,120 = 4.6732, then log 47.12 = 1.6732, and log.004712 = 3.6732. Special care is necessary in dealing with logarithms because of the fact that the mantissa is always positive, while the characteristic may be either positive or negative. When the characteristic is negative it is best for practical reasons to add 10 to it and write - 10 after the logarithm, thus giving the logarithm a new form without change of value. Thus, if (A) log.0249 = 2.3962, we add 10 to - 2, giving 8 in the place of the characteristic, and counteract this by writing - 10 after the logarithm; that is (B) log.0249 = 8.3962 - 10. In case we wish to divide a logarithm having a negative characteristic by an integer (as is sometimes required in applying Theorem IV, p. 122), it is convenient to add and subtract 10 times that integer. Thus in case we wish to divide such a logarithm by 2, we add and subtract 20; if by 3, we add and subtract 30; and so on. Suppose we want to divide the logarithm of.0249, which is 2.3962, by 3. We would then add and subtract 30, so that (C) log.0249 = 28.3962 - 30, a form more convenient than (A) or (B) when we wish to divide the logarithm by 3. Thus, ~ log.0249 = 1 (28.3962 - 30)= 9.4654 - 10. * The significant part of a number consists of those figures which remain when we ignore all initial and final zeros. Thus, the significant part of 24,000 is 24; of 6.050 is 605; of.00907 is 907; of.00081070 is 8107. THEORY AND USE OF LOGARITHMS 127 This result may be written in form (A) by adding the 9 in front to the - 10 at the end, giving - 1 = 1 as the characteristic. Hence log.0249 = 1.4654. Another method for dividing a logarithm which has a negative characteristic will now be illustrated. Suppose we wish to divide 2.3962 (=-2 + 0.3962) by 2. We get at once 21- 2 + 0.3962 - 1 + 0.1981 = 1.1981. In case we wish to divide by 3 (as in the above example), we first add and subtract 1 in order to make the negative characteristic exactly divisible by 3. Thus, 3- 3 - 1.3962 - 1 + 0.4654 = 1.4654. The following examples will illustrate the best methods for performing the four fundamental operations of Arithmetic on logarithms. CASE I. Addition of logarithms. (a) To add two logarithms having positive characteristics, as 3.2659 and 1.9866. 3.2659 1.9866 5.2525 This is in no way different from ordinary addition. (b) To add two logarithms, one having a negative characteristic, as 4.2560 and 2.8711. 4.2560 or, 6.2560 - 10 2.8711 2.8711 1.1271 9.1271 - 10 i.e. 1.1271 Since the mantissas (decimal parts) are always positive, the carrying figure 1 from the tenth's place is positive. Hence in adding the first way, 1- 4 + 2 = -1=1 will be the characteristic of the sum. (c) To add two logarithms having negative characteristics, as 2.4069 and 1.9842. 2.4069 or, 8.4069 - 10 1.9842 9.9842 - 10 2.3911 18.3911 -20 i.e. 2.3911 128 PLANE TRIGONOMETRY CASE II. Subtraction of logarithms. (a) To subtract logarithms having positive characteristics. From 5.6233 From 2.4673 or, 12.4673 - 10 take 3.8890 take 3.7851 3.7851 1.7343 2.6822 8.6822 - 10 i.e. 2.6822 In the first example we have ordinary subtraction. In the second we subtract a greater logarithm from a smaller one and the result as a whole is negative. (b) To subtract logarithms having negative characteristics. From 2.1163 or, 12.1163 - 10 take 3.4492 7.4492 - 10 4.6671 4.6671 From 1.6899 or, 9.6899 - 10 take 1.9083 1.9083 3.7816 7.7816 - 10 i.e. 3.7816 From 2.1853 or, 18.1853 -20 take 1.7442 9.7442 -10 2.4411 8.4411 - 10 i.e. 2.4411 CASE III. Multiplication of logarithms by numbers. Multiply 0.6842 Multiply 2.7012 or, 8.7012 - 10 by 5 by 3 3 3.4210 4.1036 26.1036 -30 i.e. 4.1036 In the second example the carrying figure from tenth's place is + 2. Adding this + 2 to - 2 x 3 gives 2 - 6 =- 4 =4 for the characteristic. CASE IV. Division of logarithms by nunbers. (a) Divide 3.8530 by 2. 213.8530 1.9265 (b) Divide 2.2411 by 3. Here we first add and then we subtract 30, writing the logarithm in the form 28.2411-30. 3 |28.2411 - 30 9.4137 - 10 i.e. 1.4137 67. Tables of logarithms. The common system (having the base 10) of logarithms is the one used in practical computations. For the convenience of the calculator the common logarithms of numbers up to a certain number of significant figures have been computed and arranged in tabulated forms called logarithmic tables. The common system has two great advantages. THEORY AND USE OF LOGARITHMS 129 (A) The characteristic of the logarithm of a number may be written down on mere inspection by following the rules on p. 125. Hence, as a rule, only the mantissas of the logarithms of numbers are printed in the tables. (B) The logarithms of numbers having the same significant part have the same mantissa (Th. V, p. 126). Hence a change in the position of the decimal point in a number affects the characteristic alone, and it is sufficient to tabulate the mantissas * of integers only. Thus, log 3104 = 3.4920, log 31.04 = 1.4920, log.03104 = 2.4920, log 310,400 = 5.4920; in fact, the mantissa of any number whatever having 3104 as its significant part will have.4920 as the mantissa of its logarithm. Table I, pp. 2, 3,t gives immediately the mantissas of the logarithms of all numbers whose first significant figure is 1 and whose significant part consists of four or fewer digits; and on pp. 4, 5 are found the mantissas of the logarithms of all numbers whose first significant figure is greater than 1 and whose significant part consists of three or fewer digits. 68. To find the logarithms of numbers from Table I, pp. 2-5. When the first significant figure of the number is 1, and there are four or fewer digits in its significant part, follow RULE I. First step. Determine the characteristic by inspection, using the rule on p. 125. Second step. Find in the vertical column N, Table I, 2pp. 2, 3, the first three significant figures of the number. The mantissa required is in the same horizontal row with these figures and in the vertical column having the fourth significant figure at the top (and bottom). Ex. 1. Find log 1387. Solution. First step. From the rule on p. 125 we see that the characteristic will be + 3, that is, one less than the number of digits (four) to the left of the decimal point. Second step. On p. 2, Table I, we find 138 in column N. The required mantissa will be found in the same horizontal row with 138 and in the vertical column which has 7 at the top. This gives the mantissa.1421. Therefore log 1387 = 3.1421. Ans. * In order to save space the decimal point in front of each mantissa is usually omitted in the tables. f The tables referred to in this book are Granville's Four-Place Tables of Logarithms (Ginn & Company). 130 PLANE TRIGONOMETRY If the significant part of the number consists of less than four digits, annex zeros until you do have four digits. Ex. 2. Find log 17. Solution. First step. By the rule on p. 125 the characteristic is found to be 1. Second step. To find the mantissa of 17 we look up the mantissa of 1700. On p. 3, Table I, we locate 170 in column N. The required mantissa is found in the same horizontal row with 170, and in the vertical column having 0 at the top. This gives the mantissa.2304. Therefore log 17 = 1.2304. Ans. Ex. 3. Find log.00152. Solution. First step. By the rule on p. 125 we find that the characteristic is - 3, that is, negative, and one greater numerically than the number of zeros (two) immediately after the decimal point. Second step. Locate 152 in column N, Table I, p. 3. In the same horizontal row with 152 and in the vertical column with 0 at the top we find the required mantissa.1818. Therefore log.00152 = 3.1818 = 7.1818 - 10. Ans. To find the logarithm of a number when the first significant figure of the number is greater than I and there are three or fewer digits in its significant part, follow RULE II. First step. Determine the characteristic by rule on p. 125. Second step. Find in the vertical column N, Table I, pp. 4, 5, the first two significant figures of the number. The mantissa required is in the horizontal row with these figures and in the vertical column having the third significant figure at the top (and bottom). Ex. 4. Find log 5.63. Solution. First step. The characteristic here is zero. Second step. On p. 4, Table I, we locate 56 in column N. In the horizontal row with 56 and in the vertical column with 3 at the top we find the required mantissa.7505. Therefore log5.63 = 0.7505. Ans. If the significant part of the number consists of less than three digits, annex zeros until you do have three digits. Ex. 5. Find log460,000. Solution. First step. The characteristic is 5. Second step. On p. 4, Table I, we locate 46 in column N. In the horizontal row with 46 and in the vertical column with 0 at the top we find the required mantissa.6628. Therefore log 460,000 = 5.6628. Ans. THEORY AND USE OF LOGARITHMS 131 Ex. 6. Find log.08. Solution. First step. The characteristic is - 2. Second step. Using 800, we find that the mantissa is.9031. Therefore log.08 = 2.9031 = 8.9031- 10. Ans. Ex. 7. Find (a) log1872, (b) log5, (c) log.7, (d) log20,000, (e) log 1.808, (f) log.000032, (g) log.01011, (h) log9.95, (i) log 17.35, (j) log.1289, (k) log2500, (1) log 1.002. Ans. (a) 3.2723, (b) 0.6990, (c) 1.8451, (d) 4.3010, (e) 0.2572, (f) 5.5051, (g) 2.0048, (h) 0.9978, (i) 1.2393, (j) 1.1103, (k) 3.3979, (1) 0.0009. When the first significant figure of a number is 1 and the number of digits in its significant part is greater than 4, its mantissa cannot be found in Table I; nor can the mantissa of a number be found when its first significant figure is greater than 1 and the number of digits in its significant part be greater than 3. By interpolation,* however, we may, in the first case, find the mantissa of a number having a fifth significant figure; and in the second case, of a number having a fourth significant figure. In this book no attempt is made to find the logarithms of numbers with more significant figures, since our four-place tables are in general accurate only to that extent. We shall now illustrate the process of interpolation by means of examples. Ex. 8. Find log 2445. Solution. By rule on p. 125 the characteristic is found to be 3. The required mantissa is not found in our table. But by Rule II, p. 130, log 2450 =3.3892 and log 2440 3.3874 Difference in logarithms =.0018 Since 2445 lies between 2440 and 2450, it is clear that its logarithm must lie between 3.3874 and 3.3892. Because 2445 is just halfway between 2440 and 2450 we assume that its logarithm is halfway between the two logarithms.t We then take half (or.5) of their difference,.0018 (called the tabular difference), and add this to log 2440 = 3.3874. This gives log 2445 = 3.3874 +.5 x.0018 = 3.3883. If we had to find log 2442, we should take not half the difference, but.2 of the difference between the logarithms of 2440 and 2445, since 2442 is not halfway between them but two tenths of the way. * Illustrated by examples on pp. 16-19 in the case of trigonometric functions. t In this process of interpolation we have assumed and used the principle that the increase of the logarithm is proportional to the increase of the number. This principle is not strictly true, though for numbers whose first significant figure is greater than 1 the error is so small as not to appear in the fourth decimal place of the mantissa. For numbers whose first significant figure is 1 this error would often appear, and for this reason Table I, pp. 2, 3, gives the mantissas of all such numbers exact to four decimal places. 132 PLANE TRIGONOMETRY In order to save work in interpolating, when looking up the logarithms of numbers whose mantissas are not found in the table, each tabular difference occurring in the table has been multiplied by.1,.2,.3, ',.9, and the results are printed in the large right-hand column with " Prop. Parts (propor- DIFFERENCE tional parts) at the top. Thus, on p. 4, Table I, the EXTRA 22 --- DIGIT first section in the Prop. Parts column shows the 22 21 products obtained when multiplying the tabular i 2.2 2.1 differences 22 and 21* by.1,.2,.3,...,.9. Thus, 2 4.4 4.2 3 6.6 6.3.1 x 22= 2.2.1x21 = 2.1 4 8.8 8.4.2 x 22 = 4.4.2x 21 = 4.2 5 11.0 10.5.3 x 22 = 6.6.3 x 21= 6.3 6 13.2 12.6.4 x 22= 8.8.4 x 21 8.4 7 15.4 14.7 8 17.6 16.8.5 x 22 = 11.0.5 x 21 10.5 19.8 18.9 etc. etc. Hence To find the logarithm of a number whose mantissa is not found in the table, t use RULE III. First step. Find the logarithm of the number, using only the first three (or four) digits of its significant part when looking up the mantissa.T Second step. Subtract the mantissa just found from the next greater mantissa in the table to find the corresponding tabular difference. Third step. In the Prop. Parts column locate the block corresponding to the tabular difference found. Under this difference and opposite the extra digit ~ of the number will be found the proportional part of the tabular difference which should be added to the extreme right of the logarithm found in the first step. The steum will be the logarithm of the given number. Ex. 9. Find log 28.64. Solution. Since the mantissa of 2864 is not found in our table, this example comes under Rule III, the extra digit being 4. First step. log 28.60 = 1.4564 Rule II Second step. log 28.70 = 1.4579 Rule II Tabular difference = 15 11 * These are really.0022 and.0021, it being customary to drop the decimal point. t That is, a number whose logarithm cannot be found by Rule I or Rule II, because its significant part contains too many digits. i When the first significant figure is 1, use the first four digits, following Rule I; when the first significant figure is greater than 1, use the first three digits, following Rule II. ~ In finding log 4836, for instance, 6 is called the extra digit, or, in finding log 14,835 the extra digit is 5. 11 The tabular difference =.0015, but the decimal point is usually omitted in practice. THEORY AND USE OF LOGARITHMS 133 Third step. About halfway down the Prop. Parts column on p. 4 we find the block giving the proportional parts corresponding to the tabular difference 15. Under 15 and opposite the extra digit 4 of our number we find 6.0. Then log 28.60 = 1.4564 6 Prop. Part log28.64 = 1.4570. Ans. Ex. 10. Find log.12548. Solution. Since the mantissa of 12,548 is not found in our table, this example comes under Rule III, the extra digit being 8. First step. log. 12540 = 1.0983 Second step. log. 12550 = 1.0986 Tabular difference = 3 Third step. In the Prop. Parts column on p. 2 we find the block giving the proportional parts corresponding to the tabular difference 3. Under 3 and opposite the extra digit 8 we find 2.4(= 2). Then log.12540 = 1.0983 2 Prop. Part log.12548 = 1.0985. Ans. Ex. 11. Verify the following: (a) log 4583 = 3.6612. (e) log 1000.7 -3.0003. (b) log 16.426 = 1.2155. (f) log 724,200 -5.8598. (c) log.09688 = 2.9862. (g) log 9.496 0.9775. (d) log.10108 = 2.9862. (h) log.0004586 = 4.6614. 69. To find the number corresponding to a given logarithm, use RULE IV. On pp2. 2-5, Table I, look for the mantissa of the given logarithm. If the mantissa is found exactly in the table, the first significant figures of the corresponding number are found in the same row under the N column, while the last figure is at the top of the column in which the mantissa was found. Noting what the characteristic in the given logarithm is, place the decimal point so as to agree with the ruie on p. 125. In case the mantissa of the given logarithm is not found exactly in the table we must take instead the following steps: First step. Locate the given mantissa between two mantissas in the tables. Second step. Write down the number corresponding to the lesser of the two mantissas. This will give the first three (or four) significant figures of the required number. Third step. Find the tabular difference between the two mantissas from the table, and also the difference between the lesser of the two and the given mantissa. 134 PLANE TRIGONOMETRY Fourth step. Under the Prop. Parts column find the block corresponding to the tabular difference found. Under this tabular difference pick out the proportional part nearest the difference found between the lesser mantissa and the given mantissa, and to the left of it will be found the last (extra) figure of the number, which figure we now annex. Fifth step. Noting what the characteristic of the given logarithm is, place the decimal point so as to agree with the rule on p. 125. Ex. 12. Find the number whose logarithm is 2.1892. Solution. The problem may also be stated thus: find x, having given log x = 2.1892. On p. 3, Table I, we find this mantissa,.1892 exactly, in the same horizontal row with 154 in the N column and in the vertical column with 6 at the top. Hence the first four significant figures of the required number are 1546. Since the characteristic is 2, we place the decimal point so that there will be three digits to the left of the decimal point, that is, we place it between 4 and 6. Hence x 154.6. Ans. Ex. 13. Find the number whose logarithm is 4.8409. Solution. That is, given log x = 4.8409, to find x. Since the mantissa.8409 is not found exactly in our table, we follow the last part of Rule IV. First step. The given mantissa,.8409, is found to lie between.8407 and.8414 on p. 4, Table I. Second step. The number corresponding to the lesser one, that is, to.8407, is 693. Third step. The tabular difference between.8407 and.8414 is 7, and the difference between.8407 and the given mantissa.8409 is 2. Fourth step. In the Prop. Parts column under the block corresponding to the tabular difference 7, we find that the proportional part 2.1 is nearest to 2 in value. Immediately to the left of 2.1 we find 3, the (extra) figure to be annexed to the number 693 found in the second step. Hence the first four significant figures of the required number are 6933. Fifth step. Since the characteristic of the given logarithm is 4, we annex one zero and place the decimal point after it in order to have five digits of the number to the left of the decimal point. Hence x = 69,330. Ans. Ex. 14. Find the numbers whose logarithms are (a) 1.8055, (b) 1.4487, (c) 0.2164, (d) 2.9487, (e) 2.0529, (f) 5.2668, (g) 3.9774, (h) 4.0010, (i) 8.4430 - 10,* (j) 9.4975 - 10. Ans. (a) 63.9, (b).281, (c) 1.646, (d) 888.6, (e).011295, (f) 184,850, (g) 9493, (h).00010023, (i).02773, (j).3144. * By (A), (B), p. 126, 8.4430 - 10 = 2.4430. THEORY AND USE OF LOGARITHMS 135 70. The use of logarithms in computations. The following examples will illustrate how logarithms are used in actual calculations. Ex. 1. Calculate 243 x 13.49, using logarithms. Solution. Denoting the product by x, we may write x = 243 x 13.49. Taking the logarithms of both sides, we get log x = log 243 + log 13.49. Th. I, p. 121 Looking up the logarithms of the numbers, log 243 = 2.3856 Rule II, p. 130 log13.49 = 1.1300 Rule I, p. 129 Adding, logx = 3.5156 By Rule IV, p. 133, x 3278. Ans. 1375 x.06423 Ex. 2. Calculate 1375 x 423 76,420. Lt - 1375 x.06423 Solutioz. Let x = -— 7 ---- 76,420 ~ Then log x = log 1375 + log.06423 - log 76,420 Th. I, p. 121, and Th. II, p. 122 log 1375 = 3.1383 Rule I, p. 129 log.06423 8.8077 - 10 Rule III, p. 132 Adding, 11.9460- 10 log 76,420 = 4.8832 Rule III, p. 132 Subtracting, logx 7.0628 - 10 logx= 3.0628. Bv Rule IV. D. 133. x =.0011555. Ans. or Ex. 3. Calculate (5.664)3. Solution. Let Then 1( Multiplying by 3, By Rule IV, p. 133, Ex. 4. Calculate -.7182. Solution. Let Then ~~~~Dividing by 3, Dividing by 3, x = (5.664)3. log x - 3 log 5.664. og 5.664 = 0.7531 3 logx = 2.2593 x =181.67. Ans. x = -.7182 = (.7182). log x = I log.7182. og.7182 = 1.8562 = 29.8562 - 30. 3129.8562 - 30 logx= 9.9521-10 = 1.9521. x.8956. Ans. Th. III, p. 122 Rule III, p. 132 Th. IV, p. 122 Rule III, p. 132 (b), Case IV, p. 128 By Rule IV, p. 133, 136 PLANE TRIGONOMETRY 3 7194 x 87 Ex. 5. Calculate 8 98,080,000 87 (7194) x 87 2 Solution. Let x = 87 7194)= x87 98,080,000 98,080,000 Then logx = [- log 7194 + log 87 - log 98,080,000], log7194 3.8569 Dividing by 2, Adding, or, Subtracting, or, Dividing by 3, 213.8569 - log7194 = 1.9285 log 87 = 1.9395 3.8680 13.8680 - 10 log 98,080,000 = 7.9916 5.8764 - 10 25.8764 - 30 3 25.8764 - 30 logx = 8.6255 - 10 = 2.6255... x =.04222. Ans. (a), Case II, p. 128 (b), Case IV, p. 128 8 x 62.73 x.052 Ex. 6. Calculate 8 6273 x.052 56 x 8.793 8 x 62.73 x.052 Solution. Let x = 56 x 8.793 Then logx = [log8 + log 62.73 + log.052] - [log 56 + log 8.793]. log 8= 0.9031 log 62.73 = 1.7975 log.052= 8.7160-10 log numerator = 11.4166 - 10 log denominator =2.6924 logx= 8.7242-10 =2.7242..x. x=.05299. Ans.* log 56= 1.7482 log 8.793 = 0.9442 log denominator = 2.6924 * Instead of looking up the logarithms at once when we write down log 8, log 62.73, etc., it is better to write down an outline or skeleton of the computation before using the tables at all. Thus, for above example, log 8= 0. log 62.73= 1. log.052= 8. - 10 log numerator = log denominator = log x = log56= 1. log 8.793 = 0. log denominator = It saves time to look up all the logarithms at once, and, besides, the student is not so apt to forget to put down the characteristics. THEORY AND USE OF LOGARITHMS 137 71. Cologarithms. The logarithm of the reciprocal of a number is called its cologarithm (abbreviated colog). Hence if N is any positive number, colog N = log = log 1 - log N Th. II, p. 122 = 0 -log N = - log N. That is, the cologarithm of a number equals minus the logarithm of the number, the minus sign affecting the entire logarithm, both characteristic and mantissa. In order to avoid a negative mantissa in the cologarithm, it is customary to subtract the logarithm of the number from 10 - 10. Thus, taking 25 as the number, colog 25 = log -1 = log 1 - log 25. But log 1 = 0, or, what amounts to the same thing, log 1 = 10.0000 - 10. Also, log 25 = 1.3979 colog 25= 8.6021- 10 Since dividing by a number is the same as multiplying by the reciprocal of the number, it is evident that when we are calculating by means of logarithms we may either subtract the logarithm of a divisor or add its cologarithm. When a computation is to be made in which several factors occur in the denominator of a fraction, it is more convenient to add the cologarithms of the factors than to subtract their logarithms. Hence RULE V. Instead of subtracting the logarithm of a divisor, we may add its cologarithm. The cologarithm of any number is found by subtracting its logarithm from 10.0000 -- 10. Ex. 1. Find colog52.63. Solution. 10.0000 - 10 log 52.63 = 1.7212 colog 52.63 = 8.2788 - 10. Ans. Rule V Ex. 2. Find colog.016548. Solution. 10.0000 - 10 log.016548 = 8.2187 - 10 colog.016548 = 1.7813. Ans. Rule V Thus we see that the cologarithm may be obtained from the logarithm by subtracting the last significant figure of the mantissa from 10 and each of the others from 9. 138 PLANE TRIGONOMETRY In order to show how the use of cologarithms exhibits the written work in more compact form, let us calculate the expression in Ex. 6, namely, 8 x 62.73 x.052 56 x 8.793 Solution. Using cologarithms, log x = log 8 + log 62.73 + log.052 + colog 56 + colog 8.793. log8= 0.9031 log62.73= 1.7975 log.052 = 8.7160' — i colog 56 = 8.2518 -10 colog 8.793 = 9.0558 - 10 log = 28.7242 -30 =2.7242... x =.05299. Ans. since log 56= 1.7482 since log 8.793 = 0.9442 Calculate the following expressions, using logarithms: 3. 9.238 x.9152. Ans. 4. 336.8 - 7984. 5. (.07396)5. 15.008 x.0843.06376 x 4.248 7. V2. 8. /65. 9. /.02305. 10..03296 7.962 8.454..04218..000002213. 4.671. 1.414. 1.495..2846..1606..08726) 11. /.08726\ns..1321 12. (538.2 x.0005969)1. 16 -401.8 52.37 52.37.5010..8678..631..6443..7035. - 7.672. 17. (-2563) x.03442 714.8 x (-.511) 18. 121.6 x (- 9.025) (- 48.3).x 3662 x (-.0856) Ans..2415. -.0725. 72. Change of base in logarithms. We have seen how the logarithm of a number to the base 10 may be found in our tables. It is sometimes necessary to find the logarithm of a number to a base different from 10. For the sake of generality let us assume that the logarithms of numbers to the base a have been computed. We wish to find the logarithm of a number, as N, to a new base b; that is, we seek to express logbN in terms of logarithms to the base a. Suppose that is, logbN = x, b= 1N.. * From the definition of a logarithm, p. 119, it is evident that a negative number can have no logarithm. If negative numbers do occur in a computation, they should be treated as if they were positive, and the sign of the result determined by the rules for signs in Algebra, irrespective of the logarithmic work. Thus, in Example 16 above, we calculate the value of 401.8., 52.37 and write a minus sign before the result. THEORY AND USE OF LOGARITHMS 139 Taking the logarithms of both sides of this equation to the base a, we get loga b = loga N, or, x logb = logaN. Th. III, p. 122 logaN Solving, x log log b But logbN = x. By hypothesis. (90).1. log N log N log, b Theorem VI. The logarithm of a number to the new base b equals the logarithm of the same number to the original base a, divided by the logarithm of b to the base a. This formula is also written in the form logbN = M- log,, N, where M1 = — is called the modulus of the new system with logab respect to the original one. This number M does not depend on the particular number N, but only on the two bases a and b. In actual computations a = 10, since the tables we use are computed to the base 10. Ex. Find log321. Solution. Here N 21, b = 3, a = 10. Substituting in (90), lo-lo 21 1.3222 log321 - - =2.771. Ans. loglo 3.4771 EXAMPLES 1. Verify the following: (a) log27 =2.807. (e) log 8 = 0.9464. (i) log 10 =2.096. (b) log34 = 1.262. (f) logs 5 = 0.7743. (j) log5100 = 2.86. (c) log4 9 = 1.585. (g) log7 14 = 1.356. (k) logs.1= - 2.096. (d) logs 7 = 1.209. (h) logs 102 = 2.873. (1) log5.01= - 2.86. 2. Find the logarithm of y7 in the system of which 0.5 is the base. 3. Find the base of the system in which the logarithm of 8 is 2. 4. Prove logb a ~ logab = 1. 5. Prove logy 10 logio N If, then, we have given the logarithms of numbers to a certain base a, and we wish to find the logarithms of the same numbers to a new base b, we multiply the given logarithms by the constant multiplier (modulus) M-= 1' Thus, having given the common logarithms (base 10) of numbers, we can reduce them to the logarithms of the same numbers to the base e (= 2.718) by multiplying them by M= el = 2.3026. Sog,, e 140 PLANE TRIGONOMETRY 73. Exponential equations. These are equations in which the unknown quantities occur in the exponents. Such equations may often be solved by the use of logarithms, as illustrated in the following examples: Ex. 1. Given 81x = 10; find the value of x. Solution. Taking the logarithms of both members, log 81i = log 10, or, x lo 81 = log 10. log 10 1.0000 Solving, x = — = 0.524. Ans. Solvig' log 81 1.9085 Ex. 2. Express the solution of a2x+3bx = c in terms of logarithms. Solution. Taking the logarithms of both members, log a2 + 3 log b = logc. (2 x + 3) log a + x log b = log c. 2 x log a + 3 log a + x log b = log c. x (2 log a + log b) = log c - 3 log a. log 3 og A X Ans. 2 log a + log b Ex. 3. Solve the simultaneous equations Th. III, p. 122 Th. I, p. 121 Th. III, p. 122 (A) (B) 2x 3 = 100. x + y = 4. Solution. Taking the logarithms of both members of through by log 2, we get xlog2 + y log3 = 2 x log 2 + y log 2 = 4 log 2 Subtracting, y (log 3 - log 2) = 2 - 4 log 2 Solving, 2 - 4 log 2 2 - 1.2040 Solving, y = ---- = — log3-log2.4771-.3010.7960 y - = 4.52..1761 Substituting back in (B), we get x =-.52. (A), and multiplying (B) Th. I, III, p. 122 EXAMPLES 1. Solve the following equations: (a) 5 = 12. Ans. 1.54. (b) 7x= 25. 1.65. (c) (0.4)-x= 7. 2.12. (d) 10x-1 = 4. 1.602. (e) 4x-1 = 5+1. 13.86. (f) 4x = 40. 2.66. (g) (1.3)x = 7.2. 1 1 (h) (0.9)x = (4.7) 3. (i) 7+ 3 = 5. (j) 22x+3 _ 6x-1 =. Ans. 7.53. 0.45. - 2.1728. 9.542. THEORY AND USE OF LOGARITHMS 141 2. Solve the following simultaneous equations: (a) 4x. 3 = 8, Ans. x =.9005, (c) 2x 2y = 222, Ans. x = 13, 2x 8 = 9. y=.7565. x - y = 4. y = 9. (b) 3z. 4y = 15,552, x = 5, (d) 2x 3Y = 18, x = 1, 4x.5 = 128,000. y=3. 5x 7y = 245. y = 2. 3. Indicate the solution of the following in terms of logarithms: logA - logP (a) A = P(r + 1)x. Ans. x log log(r l) (b) a2+2x =b. x-1 _ogab \ loga log d log m - log b log n (c) ax. by = m, X= log a log d - log b log c c-d. dy=. y I log a log n- log c log m log a log d - log b log c (d) a2x-3. a3y-2 = a8, x = 5, 3x+2y=17. y= 1. 74. Use of the tables of logarithms of the trigonometric functions. On p. 9 the values of the trigonometric functions of angles from 0~ to 90~ were given in tabulated form. When we are using logarithms in calculating expressions involving these trigonometric functions it saves much labor to have the logarithms of these functions already looked up for us and arranged in tabulated form.* Two complete sets of such logarithms of the trigonometric functions are given. Table II, pp. 8-16, should be used when the given or required angle is expressed in degrees, minutes, and the decimal part of a minute; and Table III, pp. 20-37, when the given or required angle is expressed in degrees, and the decimal part of a degree.t In both tables the following directions hold true: Angles between 0~ and 45~ are in the extreme left-hand column on each page,? and the logarithm of the function of any angle will be found in the same horizontal row with it and in the vertical column with the name of the function at the top; that is, sines in the first column, tangents in the second, cotangents in the third, and cosines in the fourth, counting from left to right. * To distinguish between the two kinds of tables, that on p. 9 is called a Table of Natural Functions, while the logarithms of these functions arranged in tabulated form is called a Table of Logarithmic Functions. t The division of the degree into decimal parts, instead of using minutes and seconds, has much to recommend it theoretically, and is also regarded with favor by many expert computers. In fact, a movement towards the adoption of such a system of subdivision is not only gaining headway in France and Germany, but is making itself felt in America. $ The angles increase as we read downwards. 142 PLANE TRIGONOMETRY Angles between 45~ and 90~ are in the extreme right-hand column on each page,* and the logarithm of the function of any angle will be found in the same horizontal row with it and in the vertical column with the name of the function at the bottom; that is, cosines in the first column, cotangents in the second, tangents in the third, and sines in the fourth, counting from left to right. In order to avoid the printing of negative characteristics, the number 10 has been added to every logarithm in the first, second, and fourth columns (those having log sin, log tan, and log cos at the top). Hence in writing down any logarithm taken from these three columns - 10 should be written after it. Logarithms taken from the third column, having "log cot" at the top, should be used as printed. Thus, log sin 38~ 30' = 9.7941 - 10 = 1.7941. p. 16 log cot 0~ 10' = 2.5363 = 2.5363. p. 8 log tan 75.6~ = 0.5905 = 0.5905. p. 31 log cos 2.94~ = 9.9994 - 10 = 1.9994. p. 25 75. Use of Table II, pp. 8-I6, the given or required angle being expressed in degrees and minutes.t This table gives the logarithms of the sines, cosines, tangents, and cotangents of all angles from 0~ to 5~ and from 85~ to 90~ for each minute on pp. 8-12; and on pp. 13-16, from 5~ to 85~ at intervals of 10 minutes. The small columns headed " diff. 1'" immediately to the right of the columns headed " log sin " and " log cos" contain the differences, called tabular differences, in the logarithms of the sines and cosines corresponding to a difference of 1' in the angle. Similarly, the small column headed "com. diff. 1"' contains the tabular differences for both tangent and cotangent corresponding to a difference of 1' in the angle. It will be observed that any-tabular difference is not in the same horizontal row with a logarithm, but midway between the two particular logarithms whose difference it is. Of course that tabular difference should always be taken which corresponds to the interval in which the angle in question lies. Thus, in finding log sin 78~ 16', the tabular difference corresponding to the interval between 78~ 10' and 78~ 20' is 6.1. * The angles increase as we read upwards. t In case the given angle involves seconds, first reduce the seconds to the decimal part of a minute by dividing by 60. Thus, 880 18' 42" = 88~ 18.7', since 42" = - =.7'; 2~ 0' 16"= 2~ 0.27', since 16" = -1 =.266'... If the angle is given in degrees and the decimal parts of a degree, and it is desired to use Table II, the angle may be quickly found in degrees and minutes by making use of the Conversion Table on p. 17. THEORY AND USE OF LOGARITHnMS 143 76. To find the logarithm of a function of an angle when the angle is expressed in degrees and minutes, use RULE VI. When the given angle is found exactly in Table II, the logarithm of the given function of the angle is immediately found in the same horizontal row and in the vertical column having the given function at the top when the angle is less than 45~, or at the bottom when the angle is greater than 45~. In case the given angle is not found exactly in the table we should take the following steps: (a) Write down the logarithm of the same function of the next less angle found in the table, and also the corresponding tabular difference for 1'. (b) To find the correction necessary, multiply this tabular difference by the excess in minutes of the given angle over the angle whose logarithm was written down. (c) If sine or tangent, add t co c v th/is correction.* If cosine or cotangent, subtract j This rule, as well as the next three, assumes that the differences of the logarithms of functions are proportional to the differences of their corresponding angles. Unless the angle is very near 0~ or 90~, this is in general sufficiently exact for most practical purposes. Ex. 1. Find log tan 32~ 30'. Solution. On p. 15, Table II, we find the angle 32~ 30' exactly; hence, by Rule VI, we get immediately from the table log tan 32~ 30' = 9.8042 - 10. Ans. Ex. 2. Find log cot 88~ 17'. Solution. On p. 9, Table II, we find the angle 88~ 17' exactly; hence, by Rule VI, we get at once log cot 88~ 17' = 8.4767 - 10. Ans. Ex. 3. Find log sin 23~ 26'. Solution. The exact angle 23~ 26' is not found in Table II; but then, by Rule VI, from p. 14, log sin 23~ 20' = 9.5978 - 10 Tab. diff. = 2.9 corr. for 6' = 17 Excess = 6 log sin 23~ 26' = 9.5995 - 10. Ans. Corr. =17.4 * The sine and tangent increase as the angle increases, hence we add the correction; the cosine and cotangent, however, decrease as the angle increases, hence we subtract the correction. Of course this is true only for acute angles. 144 PLANE TRIGONOMETRY Ex. 4. Find log cos 54~ 42' 18". Solution. Since 18" is less than half a minute, we drop it, and from p. 16, Table II, by Rule VI, log cos 54~ 40' = 9.7622 - 10 Tab. diff.= 1.8 Excess = 2 corr. for 2' = 4Excess Corr. = 3.6 log cos 54~ 42' = 9.7618 - 10. Ans. i.e.= 4 Ex. 5. Find log cot 1~ 34.42'. Solution. From p. 9, Table II, by Rule VI, log cot 1 34' = 1.5630 Tab. diff. 46 corr. for.4'= 18 Excess =.4 log cot 1~ 34.4'= 1.5612. A ns. Corr. = 18.4 When the angles are given in the table at intervals of 10', it is only necessary to take our angle to the nearest minute, while if the angles are given for every minute, we take our angle to the nearest tenth of a minute. Thus, in Ex. 4, we find cos 54~ 42', dropping the seconds; and in Ex. 5 we find log cot 1~ 34.4', dropping the final 2. Ex. 6. Verify the following: (a) log tan 35~ 50' = 9.8586 - 10. (g) log cos 27~ 28' = 9.9480 - 10. (b) log sin 61~ 58' = 9.9458 - 10. (h) log cot 510 49' = 9.8957 - 10. (c) log tan 82~ 3' 20"=.8550. (i) log sin 85~ 57' - 9.9989 - 10. (d) log cos 44~ 32' 50"= 9.8528 -10. (j) log cot 45~ 0' 13" = 0.0000. (e) log tan 1~ 53.2' = 8.5177 - 10. (k) log sin 120~ 24.3' = 9.9358 - 10. (f) log tan 87~ 15.6' = 1.3201. (1) log tan 243~ 42' 15" = 0.3060. 77. To find the acute angle in degrees and minutes which corresponds to a given logarithmic function, use RULE VII. When the given logarithmic function is found exactly in Table II, then the corresponding angle is immediately found in the same horizontal row, to the left if the given function is written at the top of the column, and to the right if at the bottom. In case the given logarithmic function is not found exactly in the table we should take the following steps: (a) Write down the angle corresponding to the next less logarithm of the same function found in the table, and also the corresponding tabular difference for 1'. (b) To find the necessary correction in minutes divide this tabular difference into the excess of the given logarithmic function over the one written down. (c) If sine or tangent, add ti orrction. If cosine or cotangent, subtract * See footnote, p. 143. THEORY AND USE OF LOGARITHMS 145 In searching the table for the given logarithm, attention must be paid to the fact that the functions are found in different columns according as the angle is less or greater than 45~. If, for example, the logarithmic sine is found in the column with "log sin" at the top, the degrees and minutes must be taken from the left-hand column, but if it is found in the column with "log sin" at the bottom, the degrees and minutes must be taken from the right-hand column. Similarly, for the other functions. Thus, if the logarithmic cosine is given, we look for it in two columns on each page, the' one having "log cos" at the top and also the one having "log cos" at the bottom. Ex. 7. Find the angle whose log tan = 9.6946 - 10. Solution. This problem may also be stated as follows: having given log tan x = 9.6946 - 10; to find the angle x. Looking up and down the columns having "log tan" at top or bottom, we find 9.6946 exactly on p. 15, Table II, in the column with " log tan " at top. The corresponding angle is then found in the same horizontal row to the left and is x = 26~ 20'. Ex. 8. Find the angle whose log sin = 9.6652 - 10. Solution. That is, having given log sin x = 9.6652 - 10; to find the angle x. Looking up and down the columns having "log sin" at top or bottom, we do not find 9.6652 exactly; but (Rule VII) the next less logarithm in such a column is found on p. 15, Table II, to be 9.6644, which corresponds to the angle 27~ 30', and the corresponding tabular difference for 1' is 2.4. Hence log sin x = 9.6652 - 10 Tab. diffr. 1' Excess Corr. log sin 27~ 30'= 9.6644 - 10 2! 3 72 excess = 8 8 Since the function involved is the sine, we add this correction, giving x = 27~ 30' + 3'= 27~ 33'. Ans. Ex. 9. Find the angle whose log cos = 9.3705 - 10. Solution. That is, having given log cos = 9.3705 - 10; to find the angle x. Looking up and down the columns having "log cos" at top or bottom, we do not find 9.3705 exactly; but (Rule VII) the next less logarithm in such a column is found on p. 13, Table II, to be 9.3682, which corresponds to the angle 76~ 30', and the corresponding tabular difference for 1' is 5.2. Hence log cosx = 9.3705 - 10 Tab. diff. 1' Excessl Corr. log cos 76~ 30' = 9.3682 - 10 5.2 123.0 4 208 excess = 23 22 Since the function involved is the cosine, we subtract this correction, giving x = 76~ 30' - 4'= 76~ 26'. Ans. 146 PLANE TRIGONOMETRY Ex. 10. Given log tan x = 8.7570 - 10; find x. Solution. By Rule VII the next less logarithmic tangent is found on p. 11, Table II. log tanx = 8.7570 - 10 Tab. diff.1' Excess Corr. 22 5.0.2 log tan 3~ 16' = 8.7565 - 10 44 excess = 5 6 Hence x = 3 16' +.2' = 30 16.2'. Ans. Ex. 11. Given cotx = (1.01)5; find x. Solution. Taking the logarithms of both sides, log cot x = log 1.01. Th. III, p. 122 But log 1.01 = 0.0043 and, multiplying by 5, 5 log cot x = 0.0215; to find x. By Rule VII the next less logarithmic cotangent is found on p. 16, Table II. log cot x = 0. 0215 Tab. diff. ' I Excess Corr. log cot 43~ 40' = 0.0020 2.6 13.0 5 excess = 13 13 Hence x = 43~ 40' - 5' = 43~ 35'. Ans. Ex. 12. Verify the following: (a) If log sin x = 9.5443 - 10, then x = 20~ 30'. (b) If log cosx = 9.7531 - 10, then x = 55~ 30'. (c) If log tan x = 9.9570 - 10, then x =.42~ 10'. (d) If log cot x = 1.0034, then x = 5~ 40'. (e) If log sin x = 8.0435 - 10, then x = 0~ 38'. (f) If log cos x = 8.7918 - 10, then x = 86~ 27'. (g) If log tan x 9.5261 - 10, then x = 18~ 34'. (h) If log cotx = 0.6380, then x = 12~ 58'. (i) If log sin x = 9.9995 - 10, then x = 870 16'. (j) If log cosx = 8.2881 - 10, then x = 88~ 53.3'. (k) If log tan x = 2.1642, then x = 89~ 36.4'. (1) If log tan x = 7.9732 - 10, then x = 0~ 32.3'. (m) If log sin x = 9.8500 - 10, then x = 45~ 4'. (n) If log cos x = 9.9000 - 10, then x = 37~ 25'. (o) If log tan x = 0.0035, then x = 45~ 14'. (p) If log cot x = 1.0000, then x = 5~ 43'. (q) If log cot x = 3.9732, then x = 890 27.7'. * When there are several angles corresponding to the given logarithmic function, we choose the middle one. THEORY AND USE OF LOGARITHMS 147 EXAMPLES Use logarithms when making the calculations in the following examples: 1. Given 184 sin3x = (12.03)2 cos 57~ 20'; find x. Solution. First we solve for sin x, giving sx (12.03)2 cos 57~ 20' sill X = \ - - 184 Taking the logarithms of both sides, log sin x = ~ [2 log 12.03 + log cos 57~ 20' + colog 184]. 2 log 12.03 = 2.1606 since log 12.03 = 1.0803 log cos 57~ 20' = 9.7322 - 10 colog 184 = 7.7352 - 10 since log 184 = 2.2648 19.6280 - 20 3 129.6280 - 30 log sinx = 9.8760-10.. x = 48~ 44'. Ans. 2. Given cos x =(.9854)}; find x. Ans. 77~ 37'. 4.236 cos 52~ 19' 3. Calculate 42 cs 2.659. 13.087 sin 48~ 5' 4. Given 1.5 cot 82 = x2 sin 12~ 15'; find x..9968. Hint. First solve for x, giving /1.5 cot 820 sin 120 15' 5. Given 50 tan x =.2584; find x. 0~ 49'. sin 24~ 13' cot 58~ 2' 6. Calculate *.8426. cos 33~ 17' tan 19~ 58' 7. Calculate v/cos 10~ 5' tan 73~ 11'. 1.805. 8. Calculate (sin 330 18')3 v/cot 71 20'.04422. 8. Calcu-late.04422. 10. 658 tan 63~ 54' 9. Given 3 cot x =.7; find x. 72~ 45'. 10. Given silz = (.9361)10; find x. 31~ 6'. 11. Given 2.3 tanx = (1.002)125; find x. 290 24'. 78. Use of Table III, pp. 20-37, the given or required angle being expressed in degrees and the decimal part of a degree.* This table gives, on pp. 20-29, the logarithms of the sines, cosines, tangents, and cotangents of all angles from 0~ to 5~, and from 85~ to 90~ for every hundredth part of a degree; and on pp. 30-37 from 5~ to 85~ for every tenth of a degree. The tabular differences between the logarithms given in the table are given in the same manner as were the tabular differences in Table II, and the general arrangement is the same. *In case the angle is given in degrees, minutes, and seconds, and it is desired to use Table III, we may quickly reduce the angle to degrees and the decimal part of a degree by using the Conversion Table on p. 17. 148 PLANE TRIGONOMETRY 79. To find the logarithm of the function of an angle when the angle is expressed in degrees and the decimal part of a degree, use RULE VIII. When the given angle is found exactly in Table III, the logarithm of the given function of the angle is immediately found in the same horizontal row and in the vertical column having the given function at the top when the angle is less than 450, or at the bottom when the angle is greater than 45~. In case the given angle is not found exactly in the table we should take the following steps: (a) Write down the logarithm of the same function of the next less angle * found in the table and note the tabular difference which follows. (b) In the Prop. Parts column locate the block corresponding to this tabular difference. Under this difference and opposite the extra digit of the given angle will be found the proportional part of the tabular difference (that is, the correction). (c) If sine or tangent, addis cor v this correction.t If cosine or cotangent, subtract Ex. 1. Find log sin 27.4~. Solution. On p. 34, Table III, we find the angle 27.4~ exactly; hence, by Rule VIII, we get at once log sin 27.4~ = 9.6629 - 10. Ans. Ex. 2. Find log cot 3.17~. Solution. On p. 26, Table III, we find the angle 3.17~ exactly; hence, by Rule VIII, we get immediately from the table log cot 3.17~ = 1.2566. Ans. Ex. 3. Find log tan 61.87~. Solution. The exact angle 61.87~ is not found in our tables. But then, by Rule VIII, the next less angle is 61.8~, the extra digit of the given angle being 7, and we have, from p. 34, Table III, log tan 61.8~ = 10.2707 - 10. The tabular difference between log tan 61.8~ and log tan 61.9~ is 18. In the Prop. Parts column under 18 and opposite the extra digit 7 we find the proportional part 12.6 (= 13). Then log tan 61.80~ = 0.2707 13 Prop. Part. log tan 61.87~ = 0.2720. Ans. * This " next less angle " will not contain the last (extra) digit of the given angle. t See footnote, p. 143. THEORY AND USE OF LOGARITHMS 149 Ex. 4. Find log cot 2.1580. Solution. The exact angle 2.158~ is not found in our tables. But then, by Rule VIII, the next less angle is 2.15~, the extra digit of the given angle being 8, and we have, from p. 24, Table III, log cot 2.150 = 1.4255. The tabular difference between log cot 2.15~ and log cot 2.16~ is 20. In the Prop. Parts column under 20 and opposite the extra digit 8 we find the proportional part 16. Then log cot 2.150~ = 1.4255 16 Prop. Part. log cot 2.158~ = 1.4239. Ans. Ex. 5. Verify the following: (a) log tan 37.6~ = 9.8865 - 10. (g) log tan 88.564~ = 1.6009. (b) log sin 63.87 = 9.9532 - 10. (h) log cos 20.03~ = 9.9729 - 10. (c) log cot 1.111~ = 1.7123. (i) log sin 89.97~ = 0.0000. (d) log sin 0.335~= 7.7669 -10. (j) log cot 34.84 = 0.1574. (e) log cos 45.68 = 9.8443 - 10. (k) log sin 155.42~ = 9.6191 - 10. (f) log tan 3.867~ = 8.8299 - 10. (1) log tan 196.85 = 9.4813 - 10. 80. To find the acute angle in degrees and decimal parts of a degree which corresponds to a given logarithmic function, use RULE IX. When the given logarithmic function is found exactly in Table III, then the corresponding angle is immediately found in the same horizontal row; to the left, if the given function is written at top of the column, and to the right if written at the bottom. In case the given logarithmic function is not found exactly in the table we should take the following steps: (a) Locate the given logarithm between two of the logarithms of the same function given in the tables. (b) The lesser angle of the two angles corresponding to these logarithms will be the required angle complete except for the last digit. Write this angle down with the corresponding logarithmic function. (c) Find the difference between the logarithm just written down and the given logarithm, also noting the corresponding tabular difference in.the table. (d) In the Prop. Parts column, under this tabular difference, pick out the proportional part nearest the difference found in (c), and to the left of it will be found the last (extra) digit of the required angle, which wAe now annex. 150 PLANE TRIGONOMETRY Ex. 6. Having given log tan x = 9.5364 - 10; to find the angle x. Solution. Looking up and down the columns having "log tan" at top or bottom, we do not find 9.5364 exactly. But then, by Rule IX, we locate it between 9.5345 and 9.5370, on p. 32, Table III. Except for the last digit the required angle will be the lesser of the two corresponding angles, that is, 18.9~. Then log tan 18.9~ = 9.5345 -10 log tan x = 9.5364 -10 19 = difference. The corresponding tabular difference being 25, we find in the Prop. Parts column that 20 is the proportional part under 25 which is nearest 19. To the left of 20 is the last (extra) digit 8 of the required angle. Hence x = 18.98~. Ans. Ex. 7. Having given log cos x = 8.6820 - 10; find x. Solution. On p. 25, Table III, we locate 8.6820 between 8.6810 and 8.6826. Except for the last digit, the required angle must be the lesser of the two corresponding angles, that is, 87.240. Then log cos 87.240 = 8.6826 - 10 log cos x = 8.6820 - 10 6 = difference. The corresponding tabular difference being 16, we find in the Prop. Parts column that 6.4 is the proportional part under 16 which is nearest 6. To the left of 6.4 is the last (extra) digit 4 of the required angle. Hence x = 87.244~. Ans. Ex. 8. Verify the following: (a) If logsinx = 9.6371 - 10, then x = 25.7~. (b) If log cos = 9.9873 - 10, then x = 76.2~. (c) If log tan x = 8.9186 - 10, then x = 4.74~. (d) If log cot = 1.1597, then x = 3.96~. (e) If log sin x = 9.5052 - 10, then x = 18.67~. (f) If log cosx = 9.9629 - 10, then x = 23.35~. (g) If log tan x = 9.8380 - 10, then x = 34.55~. (h) If log cot x = 9.3361 - 10, then x = 77.77~. (i) If log sinx = 8.6852 - 10, then x = 2.776~ (j) If log cos x = 9.9995 - 10, then x = 2.74~. (k) If log tan x = 7.2642 - 10, then x = 0.105~. (1) If log cot x = 1.7900, then x = 0.929~. (m) If log sinx = 9.5350 - 10, then x = 20.05~. (n) If log cos x = 9.8000 - 10, then x = 50.88~. (o) If log tan x = 0.0035, then x = 45.23~. (p) If log cot x = 2.0000, then x = 0.573~. (q) If log sin x = 0.0000, then x = 90~. (r) If log tanxa = 0.0000, then x = 45~. THEORY AND USE OF LOGARITHMS 151 EXAMPLES Use logarithms when making the calculations in the following examples: 1. Given tanx = (1.018)12; find x. Solution. Taking the logarithms of both sides, But and, multiplying by 12, log tan x = 12 log 1.018. log 1.018 = 0.0077 12 log tan x = 0.0924 Th. III, p. 122 On p. 36 we locate 0.0924 between 0.0916 and 0.0932. Then log tan 51.0~ = 0.0916 log tan x = 0.0924 8 = difference. The tabular difference is 16. In the Prop. Parts column under 16 we find 8.0 exactly. To the left of 8.0 we find the last digit 5 of the required angle. Hence x = 51.05~. Ans. 2. Given 56.4 tan3x = (18.65)5 cos 69.8~; find x. Solution. First we solve for tan x, giving tan x = k/(18.65)5 cos 69.8~ Taking the logarithms of both sides, log tan x = - [5 log 18.65 + log cos 69.8~+ colog 56.4]. 5 log 18.65 = 6.3535 since log 18.65 = 1.2707 log cos 69.8~ = 9.5382 - 10 colog 56.4 = 8.2487 - 10 24.1404 - 20 5154.1404 - 50 log tan x = 10.8281 - 10.. x =81.55~. Ans. 3. Given cos x = 9681; find x. 26.52 tan 33.860 4. Calculate 100.85 cot 88. 963 ~ 5. Given X3 sin 48.06~ = x3 cos 2.143~; find x. since log 56.4 = 1.7513 Ans. 10.25~. 9.745. 1.0885. Hint. First solve for x, giving 3\/3 sin 48.06~ A= cos 2.143~ 6. Given 5 cot x = /.4083; find x. -\/83 cos 52.82~ 7. Given sinx =;(3 find x. 8.Calculate 361 tan 87.5~ sin 9.53~. 8. Calculate V/361 tan 87i.60 sin 9.530 81.56~. 0.557~. 37. 152 PLANE TRIGONOMETRY 81. Use of logarithms in the solution of right triangles. Since the solutions of right triangles involve the calculation of products and quotients, time and labor may be saved by using logarithms in the computations. From p. 7 we have the following: General directions for solving right triangles. First step. Draw a figure as accurately as possible representing the triangle in question. Second step. TWhen one acute angle is known, subtract it from 90~ to get the other acute angle. Third step. To find an unknown part, select from (1) to (6), p. 2, a formula involving the unknown part and two known parts, and then solve for the unknown part.* Fourth step. Check the values found by seeing whether they satisfy relations different from those already employed in the third step. A convenient numerical check is the relation a2 = b2= (c + b) (c - b).t Large errors may be detected by measurement. For reference purposes we give the following formulas from p. 8 and p. 11. ab Area of a right triangle = -. (7) Side opposite an acute angle = hypotenuse X sine of the angle. (8) Side adjacent an acute angle = hypotenuse X cosine of the angle. (9) Side opposite an acute angle = adjacent side X tangent of the angle. It is best to compute the required parts of any triangle as far as possible from the given parts, so that an error made in determining one part will not affect the computation of the other parts. This also includes formulas (7), (8), (9), on p. 11. t When we want the hypotenuse, the other two sides being given, this formula is not well adapted to logarithmic computation, since c= /a2 + b2, and we have a summation under the radical that cannot be performed by the use of our logarithmic tables. If, however, we have the hypotenuse c and one side (as b) given to find the other side a, then a= Vc2 —b2= /(c- b)(c+ b), and we have a product under the radical. The factors c- b and c+ b of this product are easily calculated by inspection, and then we can use logarithms advantageously. Thus log a= I [log (c - b) + log (c + b)]. $ In case a or b is not given, or both a and b are not given, we first find what we need from the known parts, as when solving the triangle, so that we can use the above formula for finding the area. THEORY AND USE OF LOGARITHMS 153 In trigonometric computations it sometimes happens that the unknown quantity may be determined in more than one way. When choosing the method to be employed it is important to keep in mind the following suggestions: (a) An angle is best determined from a trigonometric function which changes rapidly, that is, one having large tabular differences, as the tangent or cotangent. (b) When a number is to be found (as the side of a triangle) from a relation involving a given angle, it is best to employ a trigonometric function of the angle which changes slowly, as the sine or cosine. As was pointed out on pp. 13, 14, the solution of isosceles triangles and regular polygons depends on the solution of right triangles. The following examples will illustrate the best plan to follow in solving right triangles by the aid of logarithms. Ex. 1. Solve the right triangle if A = 48~17', c = 324. Also find the area. Solution. First step. Draw a figure of the triangle indicating the known and unknown parts. Second step. B = 900 - A = 41~ 43'. Third step. To find a use a = c sinA. by (7), p. 11 Taking the logarithms of both sides, log a = log c + log sin A. Hence, from Tables I and II,* / logc= 2.5105, log sinA = 9.8730 -10 log a = 12.3835 -10 = 2.3835. /48~17.. a = 241.8. b=? To find b use b= c cosA. by (8), p. 11 Taking the logarithms of both sides, log b = log c + log cosA. Hence, from Tables I and II, log c= 2.5105 log cosA 9.8231 - 10 log b = 12.3336-10 = 2.3336.'. b = 215.6. * If we wish to use Table III instead of Table II, we reduce 17' to the decimal of a degree. Thus, A = 480 17'= 48.28~. 154 PLANE TRIGONOMETRY Fourth step. To check these results numerically, let us see if a, b, c satisfy the equation a2 = c2- b2 = (c+ b) (c - b), or, using logarithms, 2 log a = log (c + b) + log (c - b), that is, log a = - [log (c + b) + log (c - b)]. Here c + b = 539.6 and c - b = 108.4. log (c + b) = 2.7321 log(c - b) = 2.0350 2 loga = 4.7671 log a = 2.3835. Since this value of log a is the same as that obtained above, the answers are probably correct. To find the area use formula ab Area = -a 2 log area = log a + log b - log 2. log a = 2.3835 log b = 2.3336 4.7171 log2 = 0.3010 log area = 4.4161.-. area = 26,070. Ex. 2. Solve the right triangle, having given b = 15.12, c = 30.81. Solution. Here we first find an acute angle; to find A use A b cosA = -. C log cos A = log b - log c. logb = 11.1796 -10 logc= 1.4887 log cosA = 9.6909- 10.-. A = 60~ 36'. Hence B = 90~ - A = 29~ 24'. To find a we may use a = b tanA. log a = log b + log tan A. logb = 1.1796. log tanA = 0.2491 log a = 1.4287.'. a =26.84. from Table II, p. 15 by (9), p. 11 (2), p. 2 THEORY AND USE OF LOGARITHMS 155 To check the work numerically, take a2 = (c + b) (c - b), or, log a = - [log (c + b) + log (c - b)]. Here c + b = 45.93 and c - b = 15.69. log (c + b) = 1.6621 log(c - b) = 1. 1956 2 log a = 2.8578 log a = 1.4288. This we see agrees substantially with the above result. Ex. 3. Solve the right triangle, having given B = 2.325~, a = 1875.3. Solution. A = 90 - B = 87.675~. sinA -=-. by (1), p. 2 c Solving for the unknown side c, a sin A log c = log a - log sinA. Hence, from Tables I and III,* l log a = 13.2731 - 10 log sin A = 9.9996 -10 logc= 3.2735 A C.. c =1877. b? tanA =- by (3), p. 2 b Solving for the unknown side b, a tan A log b = log a - log tan A. log a = 13.2731 - 10 log tan A = 11.3915-10 logb= 1.8816.. b = 76.13. To check the work we may use formulas a2 = (c + b) (c - b), or, b = c sinB, by (7), p. 11 since neither one was used in the above calculations. * If we wish to use Table II instead of Table III, we reduce 2.325~ to degrees and minutes. Thus, B= 2.3250= 2~ 19.5'. 156 PLANE TRIGONOMETRY EXAMPLES Solve the following right triangles (C = 90~), using logarithmic Tables I and II.* No. GIVEN PARTS REQUIRED PARTS 1 A = 43~ 30' c = 11.2 B = 46~ 30' a = 7.709 b = 8.124 2 B = 68~ 50' a = 729.3 A = 21~ 10' b 1883.5 c= 2019.5 3 B =62~ 56' b = 47.7 A = 27~ 4' a =24.37 c = 53.56 4 a =.624 c=.91 A = 43~ 18' B = 46~ 42' b =.6623 5 A = 720 7' a = 83.4 B = 17~ 53' b 26.91 c = 87.64 6 b= 2.887 c= 5.11 B= 34~ 24' A=55~36' a 4.216 7 A = 52~ 41' b = 4247 B = 37~ 19' a = 5571 c = 7007 8 a = 101 b = 116 A =41 2' B = 48~ 58' c = 153.8 9 A = 43~ 22' a = 158.3 B = 46~ 38' b = 167.6 c = 230.5 10 a = 204.2 c = 275.3 A = 47~ 53' B = 42~ 7' b = 184.7 11 B = 10~ 51' c =.7264 A = 79~ 9' a =.7133 b =.1367 12 a = 638.5 b = 501.2 A = 51~ 53' B = 38~ 7' c = 811.7 13 b =.02497 c =.04792 A = 58~ 36' B = 31~ 24' a =.0409 14 B = 2 19' 30" a =1875.3 A = 870 40' 30" b = 76.13 c = 1877 15 B = 210 33' 51" a =.8211 A = 68~ 26' 9" b =.3246 c =.8826 16 A = 7400' 18" c= 275.62 B = 15~ 59' 42" a=264.9 b =75.9 17 B 340 14' 37" b = 120.22 A = 55~ 45' 23" a =176.57 c = 213.6 18 a =10.107 b = 17.303 A = 30~ 18' B = 59~ 42' c = 20.04 19 a=24.67 b =33.02 A = 36~ 46' B = 53~ 14' c = 41.22 20 A =78~ 17' a=203.8 B = 11~ 43' b = 42.27 c = 208.15 21. Find areas of the first five of the above triangles. Ans. (1) 31.32; (2) 686,900; (3) 581.3; (4).2067; (5) 1122.5. Solve the following isosceles triangles where A, B, C are the angles and a, b, c the sides opposite respectively, a and b being the equal sides. 22. Given A = 68~ 57', b = 35.09. Ans. C = 42~ 6', c = 25.21. 23. Given B = 27~ 8', c = 3.088. Ans. C = 125~ 44', a = 1.735. 24. Given C = 80~ 47', b = 2103. Ans. A = 49~ 36.5', c = 2726. 25. Given a = 79.24, c = 106.62. Ans. A = 47~ 43', C = 84~ 34'. 26. Given C = 151~ 28', c = 95.47. Ans. A = 14~ 16', a = 49.25. 27. One side of a regular octagon is 24 ft.; find its area and the radii of the inscribed and circumscribed circles. Ans. Area = 2782, r = 28.97, R = 31.36. * For the sake of clearness and simplicity, one set of triangle examples is given which are adapted to practice in using Table II, the given and required angles being expressed in degrees and minutes; and another set is given on p. 157 for practice in the use of Table III, the given and required angles being expressed in degrees and the decimal part of a degree. There is no reason why the student should not work out the examples in the first set, using Table III, and those in the second set, using Table II, if he so desires, except that it may involve a trifle more labor. This extra work of reducing minutes to the decimal part of a degree, or the reverse, may be reduced to a minimum by making use of the Conversion Tables on p. 17. It is possible, however, that an answer thus obtained may differ from the one given here by one unit in the last decimal place. This practice of giving one set of triangle examples for each of the Tables II and III will be followed throughout this book when solving triangles. THEORY AND USE OF LOGARITHMS 157 Solve the following right triangles(C = 90~), using logarithmic Tables I and III. No. GIVEN PARTS REQUIRED PARTS 28 a = 5 b = 2 A = 68.2~ B = 21.8 c = 5.385 29 B = 32.17~ c =.02728 A = 57.83~ a =.02309 b =.01452 30 A =58.65~ c = 35.73 B = 31.35~ a = 30.51 b =.18.59 31 A =22.23 b = 13.242 B = 67.77 a = 5.413 c = 14.31 32 b =.02497 c =.04792 A = 58.6~ B =31.4~ a =.0409 33 a = 273 b = 418 A = 33.15~ B 56.85~ c = 499.3 34 B =23.15~ b =75.48 A = 66.85~ a =176.5 c = 191.9 35 A =31.75~ a =48.04 B = 58.25~ b = 77.64 c = 91.29 36 b = 512 c = 900 A = 55.32 B = 34.68~ a = 740.2 37 a =52 c 60 A = 60.06~ B 29.94~ b = 29.94 38 A = 2.49~ a =.83 B = 87.51 b = 19.085 c = 19.107 39 A = 88.426~ b 9 B = 1.574~ a =327.5 c = 327.6 40 B = 4.963~ b =.07 A = 85.037~ a =.8062 c =.8092 41 B = 85.475~ c = 80 A = 4.525~ a = 6.313 b = 79.74 42 a = 100.87 b = 2 A = 88.844 B = 1.136~ c = 100.9 43. Find the areas of the first five of the above triangles. Ans. (28) 5; (29).0001677; (30) 283.6; (31) 35.83; (32).00051. 44. The perimeter of a regular polygon of 11 sides is 23.47 ft. Find the radius of the circumscribed circle. Ans. 3.79 ft. 45. Two stations are 3 mi. apart on a plain. The angle of depression of one from a balloon directly over the other is observed to be 8~ 15'. How high is the balloon? Ans..435 mi. 46. A rock on the bank of a river is 130 ft. above the water level. From a point just opposite the rock on the other bank of the river the angle of elevation of the rock is 14~ 30' 21". Find the width of the river. Ans. 502.5 ft. 47. A rope 38 ft. long tied to the top of a tree 29 ft. high just reaches the level ground. Find the angle the rope makes with the tree. Ans. 40~ 15'. 48. A man 5 ft. 10 in. high stands at a distance of 4 ft. 7 in. from a lamp-post, and casts a shadow 18 ft. long. Find the height of the lamp-post. Ans. 7.32 ft. 49. The shadow of a vertical cliff 113 ft. high just reaches a boat on the sea 93 ft. from its base. Find the altitude of the sun. Ans. 50~ 33'. 50. The top of a tree broken by the wind strikes the ground 15 ft. from the foot of the tree and makes an angle of 42~ 28' with the ground. Find the original height of the tree. Ans. 34.07 ft. 51. A building is 121 ft. high. From a point directly across the street its angle of elevation is 65~ 3'. Find the width of the street. Ans. 56.3 ft. 52. Given that the sun's distance from the earth is 92,000,000 mi., and the angle it subtends from the earth is 32'. Find diameter of the sun. Ans. About 856,400 mi. 53. Given that the radius of the earth is 3963 mi., and that it subtends an angle of 57' at the moon. Find the distance of the moon from the earth. Ans. About 239,100 mi. 158 PLANE TRIGONOMETRY 54. The radius of a circle is 12,732, and the length of a chord is 18,321. Find the angle the chord subtends at the center. Ans. 92~ 2'. 55. If the radius of a circle is 10 in., what is the length of a chord which subtends an angle of 77~ 17' 40" at the center? Ans. 12.488 in. 56. The angle between the legs of a pair of dividers is 43~, and the legs are 7 in. long. Find the distance between the points. Ans. 5.13 in. 82. Use of logarithms in the solution of oblique triangles. As has already been pointed out, formulas involving principally products, quotients, powers, and roots are well adapted to logarithmic computation; while in the case of formulas involving in the main sums and differences, the labor-saving advantages of logarithmic computation are not so marked. Thus, in solving oblique triangles, the law of sines a b c sinA sinB sin C and the law of tangents a-b (A-B)tan ( (A + B), are well adapted to the use of logarithms, while this is not the case with the law of cosines, namely, a2= b2+ c2-2becosA. In solving oblique triangles by logarithmic computation, it is convenient to classify the problems as follows: CASE I. When two angles and a side are given. CASE II. When two sides and the angle opposite one of them are given (ambiguous case). CASE III. When two sides and included angle are given. CASE IV. When all three sides are given. CASE I. When two angles and a side are given. First step. To find the third angle, subtract the sum of the two given angles from 180~. Second step. To find an unknown side, choose a pair of ratios from the law of sines a b c sin A sin B sin C uhich involve only one unknown part, and solve for that part. Check: See if the sides found satisfy the law of tangents. THEORY AND USE OF LOGARITHMS 159 Ex. 1. Having given b = 20, A = 104~, B = 19~; solve the triangle. Solution. Drawing a figure of the triangle on which we indicate the known and unknown parts, we see that the problem comes under Case I. First step. C = 180~ - (A + B) = 180~ - 123~ = 57~. a, b Second step. Solving = -- for a we get sin A sin B b sin A a = -—, sin B or, log a = log b + log sin A - log sin B. logb= 1.3010 B logsinA= 9.9869-10* 11.2879 - 10 log sinB = 9.5126-10 loga= 1.7753 \\\ \.-> a = 59.61. Solving b - for c, we get sin B sin C 104 b sin C A C c= -—, b=20 sin B or, log c = log b + log sin C - log sin B. logb= 1.3010 log sin C = 9.9236 - 10 11.2246 - 10 log sin B = 9.5126 - 10 logc= 1.7120 c = 51.52. Check: a + c =111.13, a-c = 8.09; A + C =161, A - C = 47~; (A + C) = 80~ 30', (A - C) = 23~ 30'. a-c Here, tan l (A - C) = tan (A + C), a+c or, log tan (A - C) = log (a - c) + log tan I (A + C) -log (a + c). log(a-c)= 0.9079 log tan - (A + C) = 10.7764 - 10 11.6843 - 10 log(a + c) = 2.0458 log tan - (A- C)= 9.6385 - 10... (A -C) = 230 31', which substantially agrees with the above results. * sin A = sin 104~ = sin (180~ - 104~)= sin 760. Hence log sin 104~ = log sin 76~ = 9.9869 - 10. 160 PLANE TRIGONOMETRY EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. GIVEN PARTS REQUIRED PARTS 1 a=10 A=38~ B=77~10' C=64~50' b=15.837 c=14.703 2 a=795 A =79~59' B = 4441' C=55~20' b = 567.6 c=664 3 b=.8037 B=52~20' C=101~40' A=26~ a=.445 c=.9942 4 c=.032 A=36~8' B=44~27' C=99~25' a=.01913 b=.02272 5 b=29.01 A=87~40' C =33~15' B=59~5' a=33.78 c=18.54 6 a=804 A=99~55' B=45~1' C=35~4' b=577.3 c=468.9 7 a=400 A=54~28' C=60~ B=65~32' b=447.4 c=425.7 8 c=161 A=35~15' C_123~39' B=21~6' a=111.6 b=69.62 9 a=5.42 B=42~17.3' C=82028.4' A=55014.3' b=4.439 c=6.542 10 b=2056 A =6352.8' B=70~ C=4607.2' a=1965.5 c=1578 11 a=7.86 B=3202'52" C=43025'26" A-104031'42" b=4.309 c=5.583 12 b=8 A=80~ B=2015'46" C=97044'14" a=199.53 c=200.73 Solve the following oblique triangles, using logarithmic Tables I and III. No. GIVEN PARTS REQUIRED PARTS 13 a =500 A =10.2 B = 46.6 C = 123.2 b = 2052 c = 2363 14 a =45 A =36.8 C = 62 B = 81.2 b = 74.25 c = 66.33 15 b =.085 B =95.6~ C = 24.2~ A = 60.2~ a =.0741 c =.035 16 b=5685 B=48.63~ C = 83.26~ A= 48.11~ a = 5640 c =7523 17 c=7 A=59.58~ C=60~ B=60.42~ a = 6.971 b=7.03 18 c=.0059 B=75~ C=36.87~ A =68.13~ a =.00916 b =.0095 19 a=76.08 B=126~ C=12.44~ A =41.56~ b = 92.8 c =24.7 20 a =22 A =3.486~ B =73~ C = 103.514~ b = 346 c =351.8 21 b = 8000 A = 24.5~ B = 86.495~ C = 69.005~ a = 3324 c = 7483 22 b = 129.38 A = 19.42~ C = 64~ B = 96.58~ a = 43.29 c = 117.05 23 c=95 A =2.086~ B =112~ C = 65.914~ a = 3.788 b =96.5 24 b = 132.6 A = 1~ C =75~ B =104~ a = 2.385 c = 131.98 25. A ship S can be, seen from each of two points A and B on the shore. By measurement AB = 800 ft., angle SAB = 67~ 43', and angle SBA = 74~ 21'. Find the distance of the ship from A. Ans. 1253 ft. 26. Two observers 5 mi. apart on a plain, and facing each other, find that the angles of elevation of a balloon in the same vertical plane with themselves are 55~ and 58~ respectively. Find the distances of the balloon from the observers. Ans. 4.607 mi.; 4.45 mi. 27. One diagonal of a parallelogram is 11.237, and it makes the angles 19~0 1' and 42~ 54' with the sides. Find the sides. Ans. 4.15 and 8.67. 28. To determine the distance of a hostile fort A from a place B, a line BC and the angles ABC and BCA were measured and found to be 322.6 yd., 60~ 34', 56~ 10' respectively. Find the distance AB. Ans. 300 yd. THEORY AND USE OF LOGARITHMS 161 29. From points A and B at the bow and stern of a ship respectively, the foremast, C, of another ship is observed. The points A and B are 300 ft. apart, and the angles ABC and BA C are found to be 65.460 and 112.85~ respectively. What is the distance between the points A and C of the two ships? Ans. 92.54 ft. 30. A lighthouse was observed from a ship to bear N. 34~ E.; after the ship sailed due south 3 mi. it bore N. 23~ E. Find the distance from the lighthouse to the ship in each position. Ans. 6.143 mi. and 8.792 mi. 31. In a trapezoid the parallel sides are 15 and 7, and the angles one of them makes with the nonparallel sides are 70~ and 40~. Find the nonparallel sides. Ans. 8 and 5.47. CASE II. When two sides and the angle opposite one of them are given, as a, b, A (ambiguous case *). First step. Using the law of sines as in Case, calculate log sinB. If log sinB = 0, sinB — 1, B = 90~; it is a right triangle. If log sinB > 0, sin B > 1 (impossible); there is no solution. If log sinB< 0 and b < a, only the acute value of B found from the table can be used; there is one solution.t If log sin B < 0 and b > a, the acute value of Bfound from the table and also its stuplement, should be used; and there are two solutions. $ Second step. Find C (one or two values according as we have one or two values of B) from C = 180~- (A + B). Third step. Find c (one or two values), using law of sines. Check: Use law of tangents. Ex. 1. Having given a = 36, b = 80, A = 28~; solve the triangle. Solution. In attempting to draw a figure of the triangle, the construction appears impossible. To verify this, let us find log sin B in order to apply our tests. a b First step. Solving - =A iB for sin B sinA sin B b sin A sin B = —, a or, log sin B = log b + log sin A - log a. logb = 1.9031 log sin A= 9.6716 - 10 11.5747 - 10 log a = 1.5563 log sin B = 10.0184 - 10 = 0.0184. Since log sin B > 0, sin B > 1 (which is impossible), and there is no solution. * In this connection the student should read over ~ 58, pp. 104, 105. t For if b <a, B must be less than A, and hence B must be acute. I Since b > a, A must be acute, and hence B may be either acute or obtuse. 162 PLANE TRIGONOMETRY Ex. 2. Having given a = 7.42, b = 3.39, A = 105~; solve the triangle. Solution. Draw figure. First step. From law of sines, B sin B b si nA a or, log sin B = log b + log sin A - log a. logb = 0.5302 ^c,\, log sinA = 9.9849 - 10* " \~ \<~ ~ 10.5151- 10 loga 0.8704 \05~ \ log sinB 9.6447 -10 105 ~__< A =3.39 C.'. B = 26011'. Using Table II Since log sin B < 0 and b < a, there is only one solution. Second step. C = 180~ - (A + B) = 180 - 131~ 11' = 48~ 49'. Third step. By law of sines, a sin C sin A or, log c = log a + log sin C - log sin A. loga= 0.8704 log sinC = 9.8766- 10 10.7470 - 10 log sinA = 9.9849 - 10 logc= 0.7621.. c= 5.783. Check: Use law of tangents. c-b tan (C -B) = tan -( C + B), or, log tan - (C - B) = log (c - b) + log tan - (C + B) - log (c + b). Substituting, we find that this equation is satisfied. Ex. 3. Given a = 732, b = 1015, A = 40~; solve the triangle. Solution. It appears from the construction of the triangle that there are two solutions. First step. By law of sines, bsinA C b, sinB?, a or, log sill B = log b + log sin A- log a. / logb= 3.0065 logsinA = 9.8081-10/ = 12.8146 - 10 A B1 log a= 2.8645 - log sin B = 9.9501-10 C * Sin A = sin 105~ = sin (180~ -1050) = sin 750. Hence log sin A = log sin 750 = 9.9849 -10. THEORY AND USE OF LOGARITHMS 163 Since log sin B < 0 and b > a, we have two solutions, which test verifies our construction. From Table II we find the first value of B to be B1 = 63~ 3'. Hence the second value of B is B2 = 180- B1 = 116~ 57'. Second step. C1 = 180~ - (A + B1) = 180~ - 103~ 3' = 76~ 57'; C2 = 180~ - (A + B2) = 180~ - 156~ 57' = 23~ 3'. Third step. From law of sines, a sin C1 sin A or, log ce = log a + log sin C1 - log sin A. loga= 2.8645 log sin C1 = 9.9886- 10 12.8531 - 10 log sinA = 9.8081 -10 logcl = 3.0450.~. c =1109.3. a sin C2 In the same manner, from Ca = — sin A we get C2 = 445.9. c-b Check: Use tan - (C - B) = btan I (C + B) for both solutions. EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. GIVEN PARTS REQUIRED PARTS 1 a= 50 c=66 A = 123~ 11' Impossible 2 a=5.08 b=3.59 A=63050' B=39021' C=76049' c=5.511 3 a=62.2 b=74.8 A=27018' B =33028' C1=119~14' c1=118.32 B2= 146~ 32' C2 =6~ 10' C2= 14.567 4 b=.2337 c=.1982 B=109~ A=1741' C= 53019' a=.07508 5 a=107 c=171 C= 3153' A=19018' B= 128049' b=252.2 6 b =3069 c=1223 C= 55~ 52' Impossible 7 b=5.161 c=6.84 B=4403' A1=68047' C1=67010' a1=6.92 A2=2307' C2=112050' a2=2.913 8 a=8.656 c=10 A=59057' B=3003' C=90~ b=5.009 9 a=214.56 b=284.79 B=104020' A=46053' C=28~47' c=141.5 10 a=32.16 c=27.08 C=52024' A1=70012' B1=57024' b1=28.79 A2= 109048' B2= 170 48' b2= 10.45 11 b=811.3 c=606.4 B=12605'20" A=16044'40" C=37~10' a=289.2 164 PLANE TRIGONOMETRY Solve the following oblique triangles, using logarithmic Tables I and III. No. GIVEN PARTS REQUIRED PARTS 12 a =840 b = 485 A = 21.5 B = 12.21 C = 146.29~ c = 1272 13 a =72.63 b = 117.48 A = 80~ Impossible 14 a = 177 b = 216 A = 35.6~ B1 = 45.27 C1 = 99.13 c1 = 300.3 B2= 134.73 C2 = 9.67~ c2 = 51.09 15 b =9.399 c= 9.197 B = 120.4~ A = 2.02 C = 57.58~ a =.3841 16 b =.048 c =.0621 B = 57.62~ Impossible 17 b = 19 c = 18 C= 15.8~ A = 147.5~ B1 = 16.7~ a, = 35.52 A2 = 0.9~ B2 = 163.3~ a2 =1.0385 18 a =55.55 c= 66.66 C = 77.7 A =54.5~ B=47.8~ b=50.54 19 a =34 c = 22 C = 30.35~ A = 51.37~ B1 98.28~ b1= 43.07 A2= 128.63~ B2=21.02 b2 = 15.613 20 a=528 b = 252 A = 124.6~ B=23.14~ C 32.26 c= 342.3 21 b = 91.06 = 77.04 B =51.12 A =87.69~ =41.19~ a =116.88 22 a=17,060 b=14,050 B=40~ A = 51.32 C1= 88.68~ c= 21,850 A2=128.68~ C2=11.32~ C2=4290 23. One side of a parallelogram is 35, a diagonal is 63, and the angle between the diagonals is 21~ 37'. Find the other diagonal. Ans. 124.62. 24. The distance from B to C is 145 ft., from A to C is 178 ft., and the angle ABC is 41~ 10'. Find the distance from A to B. Ans. 259.4 ft. 25. Two buoys are 2789 ft. apart, and a boat is 4325 ft. from the nearer buoy. The angle between the lines from the buoys to the boat is 16~ 13'. How far is the boat from the further buoy? Ans. 6667 ft. CASE III. When two sides and the included angle are given, as a, b, C.* First step. Calculate a +b, a-b; also (A +B) from A +B =180~-C. Second step. From tlaw of tangents, tan (A -B)= b tan (A +B), we find I (A -B). Adding this result to I (A + B) gives A, and subtracting it gives B. Third step. To find side c use law of sines; for instance, a sin C C —. sin A Check: Check by law of sines,t that is, see if log a - log sin A= log b - log sinB = log c - log sin C. * In case any other two sides and included angle are given, simply change the cyclic order of the letters throughout. Thus, if b, c, A are given, use tan (B-C) = -tan 1(B + C), etc. 2 b+c 2 a b c f From law of sines, sin A sin B sin C THEORY AND USE OF LOGARITHMS 165 Ex. 1. Having given a = 540, b = 420, C = 52~ 6'; solve the triangle, using logarithms from Tables I and II. Solution. Drawing a figure of the triangle on which we indicate the known and unknown parts, we see that the problem comes under Ca a ase II, since two sides and the included angle are given. C First step. /.\ a = 540 540 180~ b = 420 420 C= 52~ 6' I + b =960 a - b=120 A + B = 127~ 54'.. 2(A + B)= 63~ 57'. Fy _____ 526) Second step. tan (A - B) =a- tan -(A+B) A =420 B 2 a+b 2 log tan - (A - B) = log (a - b) + log tan (A + B) - log (a + b). log(a-b) = 2.0792 log tan - (A + B) = 10.3108 - 10 12.3900 - 10 log(a + b) = 2.9823 log tan (A - B) = 9.4077 - 10... (A - B) =14~21' or - (A - (A Adding, Third step. log log Check: By law of sines, log a = 12.7324 - 10 log sinA = 9.9909 - 10 2.7415 + B) = 63~ 57' - B) = 14 21' A = 78~ 18'. 63~ 57' 14~ 21' Subtracting, B = 49~ 36'. a sin C c a c =. From - sin A sin C sin A log c = log a + log sin C - log sin A. loga= 2.7324,sinC= 9.8971-10 12.6295 - 10 sinA = 9.9909 -10 logc= 2.6386.'. c = 435.1. log b= 12.6232 - 10 logsin B = 9.8817 -10 2.7415 log c = 12.6386 -10 log sin C= 9.8971 -10 2.7415 Ex. 2. Having given a = 167, c = 82, B = 98~; solve the triangle, using logarithms from Tables I and III. Solution. First step. a =167 c= 82 a + c = 249 167 82 a-c= 85 180~ B = 98~ A+ C= 82~... F(A+C)= 410. 166 PLANE TRIGONOMETRY 1 1 Second step. tan - ta (A + C), ta ( C, 2 a + c 2 or, log tan I (A - C) = log (a - c) + log taln - (A + C) - log (a + c)o log(a-c)= 1.9294 log tan I (A + C) = 9.9392 -10 11.8686 - 10 log(a + c) = 2.3962 log tan I (A- C) = 9.4724 - 10.. (A - C) = 16.530 -(A + C) 41.00~ 41.000 (A- C)= 16.530 16.530 A = 57.53~. Subtracting, C = 24.47~. Adding, Third step. a sin B bsin sin A b a from = sin B sin A log b = log a + log sin B - log sin A. B A? C loga= 2.2227 log sinB = 9.9958 -10 * 12.2185 - 10 log sinA = 9.9262 -10 logb = 2.2923.'. b = 196. Check: By law of sines, log a = 12.2227 - 10 log sinA = 9.9262 - 10 2.2965 which substantially agree. log b = 12.2923 - 10 log sin B = 9.9958 -10 2.2965 logc = 11.9138 -10 log sin C= 9.6172 - 10 2.2966 EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. GIVEN PARTS REQUIRED PARTS 1 a=27 c=15 B =46~ A= 100 57' C=33 3' b = 19.78 2 a=486 b=347 C=51036' A=830 15' B=45 9' c=383.5 3 b=2.302 c=3.567 A=62 B-=39~ 16' C=78~ 44' a=3.211 4 a=77.99 b=83.39 C=72~ 16' A=51 14' B=56~30' c=95.24 5 a=0.917 b=0.312 C=33~ 7.2' A=1320 17' B=14~ 35' c=.6769 6 a=.3 b=.363 C=124~56' A=24~41.8' B=30~22.2' c=.5886 7 b=1192.1 c=356.3 A=26~16' B=143~29' C=10~15' a=886.2 8 a=7.4 c=11.439 B=82~26' A=35~2' C=62~32' b=12.777 9 a=53.27 b=41.61 C=78~33' A=59~16.5' B=42~10.5' c=60.74 10 b=.02668 c=.05092 A=115047' B=210 1' C=430 1.3' a=.06697 11 a=51.38 c=67.94 B=79~12'36" A=40052.7' C=59 54.7' b=77.12 12 b= /5 c-=V3 A =350 53' B=930 28' C=500 39' a=1.313 sin = sin 9 = sin n (180 - 98~)= sin 82~... log sin 98~ = losinin 82= 9.9958- 10. THEORY AND USE OF LOGARITHMS 167 Solve the following oblique triangles, using logarithmic Tables I and III. No. GIVEN PARTS REQUIRED PARTS 13 a = 17 b = 12 C = 59.3~ A = 77.2~ B = 43.5~ c = 14.99 14 a = 55.14 b = 33.09 C =30.4~ A = 117.4 B = 32.2~ c = 31.44 15 b = 101 c= 158 A 37.38~ B= 38.26~ C = 104.37~ a 99.04 16 a =101 b =29 C =32.18~ A = 136.4~ B=11.42 c = 78 17 c=45 b =29 A =42.8~ B=39.72 C =97.48~ a = 30.84 18 a =.085 c=.0042 B 56.5~ A = 121.07~ C = 2.43 b =.08258 19 b =.9486 c =.8852 A = 84.6 B =49.88 C = 45.52~ a = 1.235 20 b=6 c=9 A =88.9~ B=34.03~ C =57.07~ a =10.72 21 a =12 b = 19 C = 5.24~ A = 8.84 B = 78.54~ c = 7.132 22 a =42,930 c = 73,480 B =24.8 A = 27.56~ C=127.64 b =38,920 23. In order to find the distance between two objects, A and B, separated by a swamp, a station C was chosen, and the distances CA = 3825 yd., CB = 3476 yd., together with the angle ACB= 620 31', were measured. What is the distance AB? Ans. 3800 yd. 24. Two trains start at the same time from the same station and move along straight tracks that form an angle of 30~, one train at the rate of 30 mi. an hour, the other at the rate of 40 mi. an hour. How far apart are the trains at the end of half an hour? Ans. 10.27 mi. 25. In a parallelogram the two diagonals are 5 and 6 and form an angle of 490 18'. Find the sides. Ans. 5.002 and 2.338. 26. Two trees A and B are on opposite sides of a pond. The distance of A from a point C is 297.6 ft., the distance of B from C is 864.4 ft., and the angle ACB is 87.72~. Find the distance AB. Ans. 903 ft. 27. Two stations A and B on opposite sides of a mountain are both visible from a third station C. The distances AC, BC, and the angle A CB were measured and found to be 11.5 mi., 9.4 mi., and 59~ 31' respectively. Find the distance from A to B. Ans. 10.533 mi. 28. From a point 3 mi. from one end of an island and 7 mi. from the other end the island subtends an angle of 33~ 55.8'. Find the length of the island. Ans. 4.812 mi. 29. The sides of a parallelogram are 172.43 and 101.31, and the angle included by them is 61~ 16'. Find the two diagonals. Ans. 152.93 and 238.5. 30. Two yachts start at the same time from the same point, and sail, one due north at the rate of 10.44 mi. an hour, and the other due northeast at the rate of 7.71 mi. an hour. How far apart are they at the end of 40 minutes? Ans. 4.936 mi. CASE IV. When all three sides a, b, c are given. First step. Calculate s = - (a + b + c), s -, s - b, s - c. Second step. Find log r from (s - )(s - b)(s - c). (84) to (87), p. 115 S 168 PLANE TRIGONOMETRY Third step. Find angles A, B, C fromn r' r r tan -A =, tan -B = - tan - C = 2 s s-a s-b s —c Check: See if A + B + C = 180~. Ex. 1. Having given a = 51, b = 65, c = 20; solve the triangle. Solution. Drawing a figure of the triangle on which we indicate the known and unknown parts, we see that since the three sides are given, the problem comes under Case IV. First step. a= 51 Hence b = 65 s = 68 s = 68 s = 68 c 20 a 51 b =65 c = 20 2s=136 s - a 17 s-b= 3 s-c=-48 s 68. Second step. or, r- (s - a) (s - b) (s - c) ogr [log(s - a) + log(s - b) + log(s - c) - logs]. logr =- [log (s - a) + log(s - b) + log (s - c)- log s]. From the table of logarithms, B ICG.b= A b SC log (s - a) =1.2304 log(s - b) = 0.4771 log(s - c) = 1.6812 3.3887 logs = 1.8325 211.5562 logr = 0.7781 r Third step. From the formula tan l A -, 8s-a log tan i A = log r - log (s - a). log r = 10.7781 -10 log(s - a) = 1.2304 log tan - A = 9.5477-10 A = 19~ 27', or, A = 38~ 54'. r From the formula tan = B - 2 s-b log tan - A = log r - log (s - b). logr = 10.7781-10 log(s - b) = 0.4771 log tan I B = 10.3010 - 10 B = 63~ 26', B = 126~ 52'. using Table II using Table II * If we use Table III instead, we get ' A= 19.44~, B= 63.43~, and A= 38.88~, B = 126.86~, Check: A + B + C= 179.98~. C= 7.12~, C= 14.24~. THEORY AND USE OF LOGARITHMS From the formula taln C = --- s-c 169 log tan I C = log r - log (s - c). logr = 10.7781 - 10 log(s - c)= 1.6812 log tan- C = 9.0969-10 C = 70 8'. C =14~ 16'. A = 38~ 54' B = 126~ 52' C = 14~ 16' A B + C = 180~ 2' Using Table II Check: EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. GIVEN PARTS REQUIRED PARTS 1 a =2 b=3 c = 4 A = 28 58' B = 46~ 34' C = 104~ 28' 2 a=2.5 b=2.79 c=2.33 A = 57~ 36' B = 70~ 28' C =51~54' 3 a = 5.6 b = 4.3 c =4.9 A = 74~ 40' B =47~ 46' C =57~ 34' 4 a=111 b = 145 c=40 A = 2720' B = 143~ 8' C=9~ 32' 5 a = 79.3 b = 94.2 c = 66.9 A= 55 56' B = 79~ 44' C = 44~ 20' 6 a = 321 b = 361 c = 402 A = 49~ 24' B = 58~ 38' C = 71~ 58' 7 a =.641 b =.529 c =.702 A = 60 50' B =46~ 6' C =73~ 2' 8 a =3.019 b =6.731 c = 4.228 A = 18~ 12' B = 135~ 50' C = 25~ 56' 9 a =.8706 b =.0916 c =.7902 A = 149~ 50' B = 3~ 2' C = 27~ 10' 10 a = 73 b = 82 c = 91 A = 49~ 34' B = 58 46' C = 71~ 38' 11 a =1.9 b =3.4 c = 4.9 A = 16~ 26' B =30 26' C = 133~ 10' 12 a =.21 b =.26 c =.31 A = 42~ 6' B= 56~ 6' C = 81~ 48' 13 a = 513.4 b = 726.8 c = 931.3 A = 33 16' B = 50 56' C = 95~ 48' 14 a= 5 b5= 6 c= 7 A = 51~ 52' B=59~ 32' C = 68~ 34' Solve the following oblique triangles, using logarithmic Tables I and III. No. GIVEN PARTS REQUIRED PARTS 15 a=4 b = 7 c=6 A =34.78 B = 86.42 C = 58.82~ 16 a = 43 b = 50 c = 57 A = 46.82~ B = 57.98~ C = 75.18~ 17 a =.23 b =.26 c =.198 A = 58.44~ B = 74.38~ C = 47.18~ 18 a =61.3 b = 84.7 'c = 47.6 A = 45.2~ B = 101.38 C = 33.44~ 19 a =.0291 b =.0184 c =.0358 A = 54.06~ B = 30.8 C = 95.16~ 20 a= 705 b = 562 c = 639 A = 71.56~ B = 49.14~ C = 59.32~ 21 a = 56 b = 43 c = 49 A = 74.68~ B = 47.78~ C = 57.56~ 22 a = 301.9 b = 673.1 c = 422.8 A = 18.2 B = 135.86~ C = 25.94~ 23 a = 2.51 b = 2.79 c = 2.33 = 57.88~ B = 70.3~ C = 51.84~ 24 a = 80 b = 90 c = 100 A = 49.46~ B = 58.76~ C = 71.78~ 170 PLANE TRIGONOMETRY 25. The sides of a triangular field are 7 rd., 11 rd., and 9.6 rd. Find the angle opposite the longest side. Ans. 81~ 24'. 26. A pole 13 ft. long is placed 6 ft. from the base of an embankment, and reaches 8 ft. up its face. Find the slope of the embankment. Ans. 44~ 2'. 27. Under what visual angle is an object 7 ft. long seen when the eye of the observer is 5 ft. from one end of the object and 8 ft. from the other end? Ans. 60~. 28. The distances between three cities, A, B, and C, are as follows: AB = 165 mi., AC = 72 mi., and BC= 185 mi. B is due east from A. In what direction is C from A? Ans. 7~ 19.5' N. of W. 29. Three towns, A, B, and C, are connected by straight roads. AB = 4 mi., BC = 5 mi., AC = 7 mi. Find the angle made by the roads AB and BC. Ans. 101.54~. 30. The distances of two islands from a buoy are 3 and 4 mi. respectively. If the islands are 2 mi. apart, find the angle subtended by the islands at the buoy. Ans. 28.96~. 31. A point P is 13,581 ft. from one end of a wall 12,342 ft. long, and 10,025 ft. from the other end. What angle does the wall subtend at the point P? Ans. 60.86~. 83. Use of logarithms in finding the area of an oblique triangle. From ~ 62, p. 117, we have the following three cases. CASE I. When two sides and the included angle are given, use one of the formulas ab sin C bc sin A ac sin B (88) S= ----, S =- -—, S = —, 2 2 2 where S = area of the triangle. Ex. 1. Given a = 25.6, b = 38.2, C = 41~ 56'; find the area of the triangle. absin C Solution. S = 2 log S = log a + log b + log sin C - log 2. loga= 1.4082 logb= 1.5821 log sin C = 9.8249 - 10 12.8152 - 10 log2= 0.3010 logS = 12.5142 - 10 = 2.5142..S. S= 326.8. Ans. CASE II. When the three sides are given, use formula (89) S = Vs(s- a) (s- b) (s-c), where S = area of the triangle, and s = I(- a + b +). THEORY AND USE OF LOGARITHMS 171 Ex. 2. Find the area of a triangle, having given a =12.53, b = 24.9, c =18.91. Solution. a= 12.53 Hence b = 24.9 s = 28.17 s = 28.17 s = 28.17 c= 18.91 a= 12.53 b= 24.9 c= 18.91 2s= 56.34 s-a = 15.64 s-b= 3.27 s-c= 9.26 s = 28.17. S = /s (s - a) (s - b) (s - c). logS = - [log s + log(s- a) + log(s- b)+ log(s -c)]. log s = 1.4498 log(s - a) = 1.1942 log (s - b) = 0.5145 log(s - c) = 0.9666 2 4.1251 log S= 2.0626.-. S = 115.5. Ans. CASE III. Area problems which do not fall directly under Cases I or II may be solved by Case I if we first find an additional side or angle by the law of sines. Ex. 3. Given A = 34~ 22', B = 66~ 11', c = 78.35; find area of triangle. Solution. This does not now come directly under either Case I or Case II. But C = 180~ - (A + B) = 180~ - 100~ 33' = 79~ 27'. And, by law of sines, c sin A sin C log a = log c + log sin A - log sin C. logc= 1.8941 logsinA = 9.7517 -10 11.6458 - 10 logsin C = 9.9926-10 loga= 1.6532 Now it comes under Case I. ac sin B 2 logS = loga + logc + logsinB - log2. loga= 1.6532 logc= 1.8941 log sinB = 9.9614 -10 13.5087 - 10 log2 = 0.3010 log S = 13.2077 -10 = 3.2077 -. S = 1613.3. Ans. 172 PLANE TRIGONOMETRY EXAMPLES Find the areas of the following oblique triangles, using Tables I and II for the first ten and Tables I and III for the rest. No. GIVEN PARTS AREA 1 a = 38 c = 61.2 B = 67~ 56' 1078 2 b = 2.07 A = 70 B = 36~ 23' 3.257 3 b= 116.1 c=100 A = 118~ 16' 5113 4 a = 3.123 A = 53 11' B = 13~ 57' 1.344 5 b =.439 A =76~ 38' C = 40~ 35'.0686 6 a =.3228 c =.9082 B = 60~ 16'.1273 7 c = 80.25 B = 100~ 5' C = 31~ 44' 4506 8 a =.010168 b =.018225 C = 11~ 18.4'.000018155 9 a = 18.063 A = 96~ 30' B = 35~ 70.55 10 b = 142.8 c = 89.6 a = 95 4174 11 a= 100 B = 60.25 C = 54.5~ 3891 12 a = 145 b = 178 B = 41.17~ 12,383 13 a = 886 b = 747 C = 71.9~ 314,600 14 a =266 b = 352 C = 73~ 44,770 15 a = 960 b = 720 C = 25.67~ 149,730 16 a = 79 b = 94 c = 67 2604 17 a = 23.1 b = 19.7. c = 25.2 215.9 18 a= 5.82 b =6 c = 4.26 11.733 19. The sides of a field ABCD are AB = 37 rd., BC = 63 rd., and DA = 20 rd., and the diagonals AC and BD are 75 rd. and 42rd., respectively. Required the area of the field. 1570 sq. rd. 20. In a field ABCD the sides AB, BC, CD, and DA are 155 rd., 236 rd., 252 rd., and 105 rd., respectively, and the length from A to C is 311 rd. Find the area of the field. 29,800 sq. rd. 21. The area of a triangle is one acre; two of its sides are 127 yd. and 150 yd. Find the angle between them. 30~ 32'. 22. Given the area of a triangle = 12. Find the radius of the inscribed circle if a = 60 and B = 40~ 35.2'. 84. Measurement of land areas. The following examples illustrate the nature of the measurements made by surveyors in determining land areas, and the usual method employed for calculating the area from the data found. The Gunter's chain is 4 rd., or 66 ft., in length. An acre equals 10 sq. chains, or 160 sq. rd. EXAMPLES 1. A surveyor starting from a point A runs N. 27~ E. 10 chains to B, thence N.E. by E. 8 chains to C, thence S. 5~ W. 24 chains to D, thence N. 40~ 44' W. 13.94 chains to A. Calculate the area of the field ABCD. Solution. Draw an accurate figure of the field. Through the extreme westerly point of the field draw a north-and-south line. From the figure, area THEORY AND USE OF LOGARITHMS 173 ABCD = area trapezoid* GCDE - (area trapezoid GCBF + area triangle FBA + area triangle ADE) = 13.9 acres. Ans. 2. A surveyor measures S. 50~ 25' E. 6.04 chains, thence N S. 58~ 10'W. 4.15 chains, thence N. 28~ 12'W. 5.1 chains, thence to the starting point. Determine the direction and distance of the starting point from the last station, and G -C find area of the field inclosed. 8 Ans. N. 39~ 42' E. 2 chains; 1.66 acres.,-___ 3. One side of a field runs N. 83~ 30' W. 10.5 chains, the second side S. 22~ 15' W. 11.67 chains, the third side N. 71~ 45' E. 12.9 chains, the fourth side completes the 27c0 circuit of the field. Find the direction and length of the 24 fourth side, and calculate the area of the field. A Ans. N. 25 1' E. 6.15 chains; 7.9 acres. 4. From station No. 1 to station No. 2 is S. 7~20' W. 4.57 chains, thence to station No. 3 S. 61~ 55' W. 7.06 chains, thence to station No. 4 N. 3~ 10' E. 5.06 chains, thence to station No. 5 N. 33~ 50' E. 325 chains, thence ------ to station No. 1. Find the direction and distance of station No. 1 from No. 5, and calculate the area of the field inclosed. Ans. E. 1~ 15'N. 4.7 chains; 3.55 acres. 85. Parallel sailing. When a vessel sails due east or due west, that is, always travels on the same parallel of latitude, it is called parallel sailing. The distance sailed is the departure,t and it is expressed in (North pole) P between A and B. The latitudes of A and B are the same, i.e. arc EA = angle EOA = are QB = angle QOB. The difference in longitude * From Geometry the area of a trapezoid equals one half the sum of the parallel sides times the altitude. Thus, area GCDE = (GC+ ED) GE. t The departure between two meridians is the arc of a parallel of latitude comprehended between those meridians. It diminishes as the distance from the equator increases. $ A geographical mile or knot is the length of an arc of one minute on a great circle of the earth. 174 PLANE TRIGONOMETRY of A and B = arc EQ. The relation between latitude, departure, and difference in longitude may be found as follows: By Geometry, arc A B DA DA are = O = -- A = cos OAD - cos A OE = cos latitude. arc EQ OE OA... arc A B = arEQ cos latitude, or, departure (90) Diff. long. = depa cos latitude EXAMPLES 1. A ship whose position is lat. 25~ 20' N., long. 36~ 10' W. sails due west 140 knots. Find the longitude of the place reached. Solution. Here departure = 140, and latitude = 25~ 20' N. Substituting in above formula (90), 140 diff. long. = 240 cos 25~ 20' log 140 = 12.1461- 10 log cos 25~ 20' = 9.9561 - 10 log diff. long. = 2.1900 diff. long. = 154.9' = 2~ 34.9'. Hence longitude of place reached = 36~ 10' + 2~ 34.9' = 38~ 44.9'W. Ans. 2. A ship in lat. 42~ 16' N., long. 72~ 16' W., sails due east a distance of 149 geographical miles. What is the position of the point reached? Ans. Long. 68~ 55' W. 3. A vessel in lat. 44~ 49' S., long. 119~ 42' E., sails due west until it reaches long. 117~ 16' E. Find the departure. Ans. 103.6 knots. 4. A ship in lat. 36~ 48' N., long. 56~ 15' W., sails due east 226 mi. Find the longitude of the place reached. Ans. Long. 51~ 33' W. 5. A vessel in lat. 48~ 54' N., long. 10~ 55' W., sails due west until it is in long. 150 12' W. Find the number of knots sailed. Ans. 168.9 knots. 86. Plane sailing. When a ship sails in such a manner as to cross (North Pole) successive meridians at the same angle, p it is said to sail on a rhumb line. This Ij^^^"^^ ~angle is called the course, and the distance i \ ~\\\\ B between two places is measured on a \ \ \ A\ rhumb line. Thus, in the figure, if a C o ship travels from A to B on a rhumb line, I -^ \ K arc AB = distance, D -E \ \ angle CAB = course, t --- A ~c — c —; are CB = departure, I ~\ V^~ ~ arc A C = difference in latitude \^~ X -q~t0 ~ ~between A and B. THEORY AND USE OF LOGARITHMS 175 An approximate relation between the quantities involved is obtained by regarding the surface of the earth as a plane surface, that is, regarding A CB as a plane right triangle, the angle A CB being the right angle. This right triangle is called the triangle of plane sailing. From this plane right triangle we get Depart. CB =AB sinA, and C B AC=ABcosA; or, (91) Departure = distance X sin course, and (92) Diff. lat. = distance X cos course. If AB is long, the error caused by neglecting the curvature of the earth will be too great to make these results of any value. In that case AB may be divided into parts, such as AE, EG, GI, IB (figure on p. 174), which are so small that the curvature of the earth may be neglected. EXAMPLES 1. A ship sails from lat. 8~ 45' S., on a course N. 36~ E. 345 geographical mi. Find the latitude reached and the departure made. Solution. Here distance = 345 and course = 36~..-. departure = 345 sin 36~. diff. lat. = 345 cos 36~. log 345 = 2.5378 log 345 = 2.5378 log sin 36~ = 9.7692 -10 log cos 36~ = 9.9080 -10 log departure = 2.3070 log diff. lat. = 2.4458.-. departure = 202.8 mi. Ans. diff. lat. = 279. 1' 4~ 39. 1'. As the ship is sailing in a northerly direction she will have reached latitude 8~ 45'- 4~ 39.1' = 4 5.9' S. Ans. 2. A ship sails from lat. 32~ 18' N., on a course between N. and W., a distance of 344 mi., and a departure of 103 mi. Find the course and the latitude reached. Ans. Course N. 17~ 25' W., lat. 37~ 46' N. 3. A ship sails from lat. 43~ 45' S., on a course N. by E. 2345 mi. Find the latitude reached and the departure made. Ans. Lat. 5~ 25' S., departure = 457.5 mi. 4. A ship sails on a course between S. and E. 244 mi., leaving lat. 2~ 52' S., and reaching lat. 5~ 8' S. Find the course and the departure. Ans. Course S. 56~ 8' E., departure = 202.6 mi. 87. Middle latitude sailing. Here we take the departure between two places to be measured on that parallel of latitude which lies halfway between the parallels of the two places. Thus, in the figure on p. 174, the departure between A and B is LM, measured on a parallel of latitude midway between the parallels of A and B. 176 PLANE TRIGONOMETRY This will be sufficiently accurate for ordinary purposes if the run is not of great length nor too far away from the equator. The middle latitude is then the mean of the latitudes of A and B. The formula (90) on p. 174 will then become departure (93) Diff. long. departure cos mid. lat. EXAMPLES 1. A ship in lat. 42~ 30' N., long. 58~ 51' W., sails S. 33~ 45' E. 300 knots. Find the latitude and longitude of the position reached. Solution. We know the latitude of the starting point A. To get the latitude of the final position B, we first find diff. in lat. from (92). This gives A diff. lat. = 300 cos 33~ 45'. log 300 = 2.4771 0 t log cos 33~ 45'= 9.9198 - 10 S~ O^- a\<, ^log diff. lat. = 2.3969 diff. lat. = 249.4' = 4~ 9.4'. Since the ship sails in a southerly direction, she will C Depart. B have reached latitude=42030'-409.4'= 3820.6'N. Ans. To get the longitude of B we must first calculate the departure and middle latitude for substitution in (93). From (91) departure = 300 sin 33~ 45'. log 300= 2.4771 log sin 33~ 45' = 9.7448 - 10 log departure = 2.2219 departure = 166.7'. Middle latitude = 2 (42 30' + 38~ 20.6') = 40 25.3'. 166.7 Substituting in (93), diff. long. = 6 cos 400 25.3' log 166.7= 12.2219 - 10 log cos 40~ 25.3'= 9.8815 - 10 log diff. long. = 2.3404 diff. long. = 219' = 3~ 39'. Since the ship sails in an easterly direction, she will have reached longitude =58~ 51' - 3~ 39' = 55~ 12' W. Ans. 2. A vessel in lat. 260 15'N., long. 610 43' W., sails N. W. 253 knots. Find the latitude and longitude of the position reached. Ans. Lat. 29~ 13.9' N.; long. 65~ 5.1' W. 3. A ship leaves lat. 31014' N., long. 42~ 19' W., and sails E.N.E. 325 mi. Find the position reached. Ans. Lat. 33~ 18.4' N.; long. 36~ 24' W. 4. Leaving lat. 42~30'N., long. 58~0 51'W., a battleship sails S.E. byS. 300 mi. Find the place reached. Ans. Lat. 38~ 21' N.; long. 55~ 12' W. THEORY AND USE OF LOGARITHMS 1T7 5. A ship sails from a position lat. 49~ 56' N., long. 15 16' W., to another lat. 47~ 18' N., long. 20~ 10' W. Find the course and distance. Ans. Course, S. 50~ 53' W.; distance = 250.5 mi. Hint. The difference in latitude and the difference in longitude are known, also the middle latitude. 6. A torpedo boat in lat. 37~ N., long. 32~ 16' W., steams N. 36~ 56' W., and reaches lat. 41~ N. Find the distance steamed and the longitude of the position reached. Ans. Distance = 300.3 mi.; long. 36~ 8' W. 7. A ship in lat. 42~ 30' N., long. 58~ 51' W., sails S.E. until her departure is 163 mi. and her latitude 38~ 22' N. Find her course and distance and the longitude of the position reached. Ans. Course, S. 33~ 19' E.; distance = 296.7 mi.; long. 55~ 17' W. 8. A cruiser in lat. 47~ 44' N., long. 32~ 44' W., steams 171 mi. N.E. until her latitude is 50~ 2' N. Find her course and the longitude of the position reached. Ans. Course, N. 36~ 11' E.; long. 30~10' W. 9. A vessel in lat. 47~ 15' N., long. 20~ 48' W., sails S.W. 208 mi., the departure being 162 mi. Find the course and the latitude and longitude of the position reached. Ans. Course, S. 51~ 9' W.; lat. 450 4.5' N.; long. 24~ 42' W. CHAPTER IX ACUTE ANGLES NEAR 0~ OR 90~ 88. When the angle x approaches the limit zero, each of the ratios sin x tan x -,,- approaches unity as a limit, x being the circular measure x x of the angle. Proof. Let 0 be the center of a circle whose radius is unity. Let arc AP = x, and let arc AP' = x in numerical value. Draw PP', and let PT and P'T be the tangents drawn to the circle at P and P'. From Geometry (A) PQP' < PAP' < PTP'. But PQP' = PQ + QP' = 2 sin x in numerical value, PAP' = PA + AP' = 2 x in numerical value, and PTP'= PT + TP = 2 tan x in numerical value. Substituting in (A), /P 2 sinx < 2 x < 2 tan x. Dividing through by 2, we have [/: \T (B) sin < x < tanx, - A which proves that \ R=1 \ j If x be the circular measure of an acute angle, it will always lie between sin x and tan x, being greater than sin x and less than tan x. Dividing (B) through by sin x, we get x 1<. - < sin x cos x If we now let x approach the limit zero, it is seen that limit x X=O sinx must lie between the constant 1 and limit, which is also 1. x=o COS X 178 ACUTE ANGLES NEAR 0~ OR 90~ 179 Hence limit x = 1 or x=O sinx (C) limit inx = x=O X Similarly, if we divide (B) through by tan x, we get x cos < < 1. tan x As before, if x approaches zero as a limit, limit tan x x=O X must lie between the constant 1 and limit cos x, which is also 1. X=0 Hence limit - 1 or x=o tanx (D) limit tan x x=O X The limits (c) and (D) are of great importance both in pure and applied mathematics. These results may be stated as follows: When x is the circular measure of a very small angle we may repglace sin x and tan x in our calculations by x. 89. Functions of positive acute angles near 0~ and 90~. So far we have assumed that the differences in the trigonometric functions are proportional to the differences in the corresponding angles. While this is not strictly true, it is in general sufficiently exact for most practical purposes unless the angles are very near 0~ or 90~. In using logarithms we have also assumed that the differences in the logarithms of the trigonometric functions are proportional to the differences in the corresponding angles. This will give sufficiently accurate results for most purposes if we use Tables II or III in this book and confine ourselves to angles between.3~(= 18') and 89.7~ (= 89~ 42') inclusive. If, however, we have an angle between 0~ and.3~ (= 18') or one between 89.7~ (= 89~ 42') and 90~, and are looking for exact results, it is evident that the ordinary method will not do. For example, the tabular difference (Table II) between the logarithmic sine, tangent, or cotangent of 8' and the logarithm of the corresponding functions of 9' is 512, while between 9' and 10' it is 457. If we interpolate here in the usual way it is evident that 180 PLANE TRIGONOMETRY our results will be only approximately correct. In case it is desired to obtain more accurate results we may use the principle established in the last section, namely: We may replace sin x and tan x in our calculations by x when x is a very small angle and is expressed in circular measure. From a table giving the natural functions of angles, we have sin 2.2~= 0.03839 = 0.0384, tan 2.2~= 0.03842 = 0.0384. Also 2.2~= 0.0384 radians. Hence it is seen that in any calculation we may replace the sine or tangent of any angle between 0~ and 2.2~ by the circular measure of the angle without changing the first four significant figures of the result. Also since cos 87.8~= sin (90~- 87.8~) = sin 2.2~= 0.0384, cot 87.8~ = tan (90~ - 87.8~) = tan 2.2~ = 0.0384, and 2.2~= 90~- 87.8 = 0.0384 radians, we may replace the cosine or cotangent of any angle between 87.8~ and 90~ by the circular measure of the complement of that angle. We may then state the following rules: 90. Rule for finding the functions of acute angles near 0~. sin x = circular measure of x, tanx = circular measure of x, 1 cot x = -.circular measure of x cos x is found from the tables in the usual way.t * The following equivalents may be used for reducing an angle to circular measure (radians), and in other computations. 1~= - radians. 180 1~ = 0.0174533 radians, log 0.0174533 = 8.2419 - 10. 1' = 0.0002909 radians, log 0.0002909 = 6.4637 - 10. 1" = 0.0000048 radians, log0.0000048= 4.6856 - 10. 180~ -= 57.29578~ = 1 radian, log 57.29578 = 1.7581. r = 3.14159 log 7r = 0.4971. = P2 approximately. f csc x and sec x are simply the reciprocals of sin x and cos x respectively. ACUTE ANGLES NEAR 0~ OR 90~ 181 91. Rule for finding the functions of acute angles near 90~. cos x == circular meastre of the complement of x,* cot x = circular measure of the compjlement of x, tanx = circular measure of the complement of x sin x is found from the tables in the usual way.t Since any function of an angle of any magnitude whatever, positive or negative, equals some function of a positive acute angle, it is evident that the above rules, together with those on p. 57, will suffice for finding the functions of angles near ~ 90~, ~ 180~, ~ 270~, ~ 360~. Ex. 1. Find sine, tangent, and cotangent of 42'. Solution. Reducing the angle to radians, 42' = 42 x 0.0002909 radians = 0.01222 radians. Therefore sin 42' = 0.01222, tan 42' 0.01222, cot 42' = = 81.833. Ans. 0.01222 Ex. 2. Find cosine, cotangent, and tangent of 890 34.6'. Solution. The complement of our angle is 90~ - 89~ 34.6' = 25.4'. Reducing this remainder to radians, 25.4'= 25.4 x 0.0002909 radians = 0.00739 radians. Therefore cos 89~ 34.6' = 0.00739, cot 89~ 34.6' = 0.00739, tan 89~ 34.6' = = 135.32. Ans. 0.00739 When the function of a positive acute angle near 0~ or 90~ is given, to find the angle itself we reverse the process illustrated above. For instance: Ex. 3. Find the angle subtended by a man 6 ft. tall at a distance of 1225 ft. Solution. From the figure tan x = 1-25 1225 But, since the angle is very small, we may replace tan x by x, giving x = T radians = 0.0049 radians. Or, reducing the angle to minutes of arc, we get 0.0049 x = minutes of arc = 16.8'. Ans. 0.0002909 * If the angle is given in degrees, subtract it from 900 and reduce the remainder to circular measure (radians). If the angle is given in circular measure (radians), simply subtract it from - (= 1.57079). sc and se are simply the reciprocals of sin and cos respectively. f ese x and see x are simply the reciprocals of sin x and cos x respectively. 182 PLANE TRIGONOMETRY 92. Rules for finding the logarithms of the functions of angles near 0~ and 90~.* For use in logarithmic computations the rules of the last two sections may be put in the following form: If the angle is given in degrees, minutes, and seconds, it should first be reduced to degrees and the decimal part of a degree (see Conversion Table on p. 17 of Tables). Rule I. To find the logarithms of the functions of angles near 0~. log sin x ~2.2419 + log x.t log tan x~ = 2.2419 + log x. log cot x~ = 1.7581 - log x. log cos x~ is found from the tables in the usual way. Rule II. To find the logarithms of the functions of an angle near 90~. log cos x~ = 2.2419 + log (90 - x). log cot x~ = 2.2419 + log (90 - x). log tan x~ = 1.7581 - log (90 - x). log sin x~ is found from the tables in the usual way. Ex. 1. Find log tan 0.045~. Solution. As is indicated in our logarithmic tables, ordinary interpolation will not give accurate results in this case. But from the above rule, log tan 0.045~= 2.2419 + log 0.045 = 2.2419 + 2.6532.. log tan 0.045~ =4.8951. Ans. On consulting a much larger table of logarithms, this result is found to be exact to four decimal places. Interpolating in the ordinary way, we get log tan 0.045~ = 4.8924, which is correct to only two decimal places. * These rules will give results accurate to four decimal places for all angles between 0~ and 1.1~ and between 88.9~ and 90~. t Since 1 degree= 0.017453 radians, the circular measure of x degrees= 0.017453 * x radians. Hence, from p. 180, sin x~ = 0.017453' x, and log sin x~ = log 0.017453 + log x = 2.2419 + log x. $ From p. 180, cot = --, 0.017453 ~ x and log cot x~ = - log 0.017453 - log x = 1.7581 - log x. ACUTE ANGLES NEAI 0~ OR 90~ 183 Ex. 2. Find log tan 89.935~. Solution. From the above rule, log tan 89.935 = 1.7581 - log(90 - 89.935) = 1.7581 - log0.065 = 1.7581 - 2.8129... log tan 89.935~= 2.9452. Ans. If the tangent itself is desired, we look up the number in Table I corresponding to this logarithm. This gives tan 89.935~ = 881.4. 93. Consistent measurements and calculations. In the examples given so far in this book it has generally been assumed that the given data were exact. That is, if two sides and the included angle of a triangle were given, as 135 ft., 217 ft., and 25.3~ respectively, we have taken for granted that these numbers were not subject to errors made in measurement. This is in accordance with the plan followed in the problems that the student has solved in Arithmetic, Algebra, and Geometry. It should not be forgotten, however, that when we apply the principles of Trigonometry to the solution of practical problems, - engineering problems, for instance, - it is usually necessary to use data which have been found by actual measurement, and therefore are subject to error. In taking these measurements one should carefully see that they are made with about the same degree of accuracy. Thus, it would evidently be folly to measure one side of a triangle with much greater care than another, for, in combining these measurements in a calculation, the result would at best be no more accurate than the worst measurement. Similarly, the angles of a triangle should be measured with the same care as the sides. The number of significant figures in a measurement is supposed to indicate the care that was intended when the measurement was made, and any two measurements showing the same number of significant figures will, in general, show about the same relative care in measurement. If the sides of a rectangle are about 936 ft. and 8 ft., the short side should be measured to at least two decimal places. A neglected 4 in the tenths place will alter the area by 374 sq. ft. The following directions will help us to make consistent measurements and avoid unnecessary work in our calculations. 1. Let all measured lines and calculated lines show the same number of significant figures, as a rule. 2. When the lines show only one significant figure, let the angles read to the nearest 5~. 184 PLANE TRIGONOMETRY 3. When the lines show two significant figures, let the angles read to the nearest half degree. 4. When the lines show three significant figures, let the angles read to the nearest 5'. 5. When the lines show four significant figures, let the angles read to the nearest minute. EXAMPLES 1. The inclination of a railway to the horizontal is 40'. How many feet does it rise in a mile? Ans. 61.43. 2. Given that the moon's distance from the earth is 238,885 mi. and subtends an angle of 31' 8" at the earth. Find the diameter of the moon in miles. Ans. 2163.5. 3. Given that the sun's distance from the earth is 92,000,000 mi. and subtends an angle of 32' 4" at the earth. Find the sun's diameter. Ans. 858,200 mi. 4. Given that the earth's radius is 3963 mi. and subtends an angle of 57'2" at the moon. Find the distance of the moon from the earth. Ans. 238,833 mi. 5. Given that the radius of the earth is 3963 mi. and subtends an angle of 9" at the sun. Find the distance of the sun from the earth. Ans. 90,840,000 mi. 6. Assuming that the sun subtends an angle of 32' 4" at the earth, how far from the eye must a dime be held so as to just hide the sun, the diameter of a dime being 5 in.? Ans. 69.83 in. 7. Find the angle subtended by a circular target 5 ft. in diameter at the distance of half a mile. Ans. 6'30.6". MISCELLANEOUS EXAMPLES 1. A balloon is at a height of 2500 ft. above a plain and its angle of elevation at a point in the plain is 40~ 35'. How far is this point from the balloon? Ans. 3843 ft. 2. A tower standing on a horizontal plain subtends an angle of 37~ 19.5' at a point in the plain distant 369.5 ft. from the foot of the tower. Find the height of the tower. Ans. 281.8 ft. 3. The shadow of a steeple on a horizontal plain is observed to be 176.23 ft. when the elevation of the sun is 33.2~. Find the height of the tower. Ans. 115.3 ft. 4. From the top of a lighthouse 112.5 ft. high, the angles of depression of two ships, when the line joining the ships passes through the foot of the lighthouse, are 27.3~ and 20.6~ respectively. Find the distance between the ships. Ans. 81 ft. 5. From the top of a cliff the angles of depression of the top and bottom of a lighthouse 97.25 ft. high are observed to be 23~ 17' and 240 19' respectively. How much higher is the cliff than the lighthouse? Ans. 1947 ft. 6. The angle of elevation of a balloon from a station due south of it is 470 18.5', and from another station due west of the former and 671.4 ft. from it the elevation is 41 14'. Find the height of the balloon. Ans. 1000 ft. MISCELLANEOUS EXAMPLES 185 7. A ladder placed at an angle of 75~ with the street just reaches the sill of a window 27 ft. above the ground on one side of the street. On turning the ladder over without moving its foot, it is found that when it rests against a wall on the other side of the street it is at an angle of 16~ with the street. Find the breadth of the street. Ans. 34.24ft. 8. A man traveling due west along a straight road observes that when he is due south of a certain windmill the straight line drawn to a distant church tower makes an angle of 30~ with the direction of the road. A mile farther on the bearings of the windmill and church tower are N.E. and N.W. respectively. Find the distances of the tower from the windmill and from the nearest point on the road. Ans. 2.39 mi., 1.37 mi. 9. Standing at a certain point, I observe the elevation of a house to be 45~, and the sill of one of its windows, known to be 20 ft. above the ground, subtends an angle of 20~ at the same point. Find the height of the house. Ans. 54.94 ft. 10. A hill is inclined 36~ to the horizon. An observer walks 100 yd. away from the foot of the hill, and then finds that the elevation of a point halfway up the hill is 18~. Find the height of the hill. Ans. 117.58 yd. 11. Two straight roads, inclined to one another at an angle of 60~, lead from a town A to two villages B and C; B on one road distant 30 mi. from A, and C on the other road distant 15 mi. from A. Find the distance from B to C. Ans. 25.98 mi. 12. Two ships leave harbor together, one sailing N.E. at the rate of 7- mi. an hour and the other sailing north at the rate of 10 mi. an hour. Prove that the distance between the ships after an hour and a half is 10.6 mi. 13. A and B are two positions on opposite sides of a mountain; C is a point visible from A and B. From A to C and from B to C are 10 mi. and 8 mi. respectively, and the angle BCA is 60~. Prove that the distance between A and B is 9.165 mi. 14. A and B are two consecutive milestones on a straight road and C is a distant spire. The angles ABC and BAC are observed to be 120~ and 45~ respectively. Show that the distance of the spire from A is 3.346 mi. 15. If the spire C in the last example stands on a hill, and its angle of elevation at A is 15~, show that it is.866 mi. higher than A. 16. If in Example 14 there is another spire D such that the angles DBA and DAB are 45~ and 90~ respectively and the angle DAC is 45~, prove that the distance from C to D is very nearly 23 mi. 17. A and B are consecutive milestones on a straight road; C is the top of a distant mountain. At A the angle CAB is observed to be 380 19'; at B the angle CBA is observed to be 132~ 42', and the angle of elevation of C at B is 100 15'. Show that the top of the mountain is 1243.7 yd. higher than B. 18. A base line AB, 1000 ft. long, is measured along the straight bank of a river; C is an object on the opposite bank; the angles BAC and CBA are observed to be 65~ 37' and 53~ 4' respectively. Prove that the perpendicular breadth of the river at C is 829.8 ft. 186 PLANE TRIGONOMETRY 19. The altitude of a certain rock is observed to be 47~, and after walking 1000 ft. towards the rock, up a slope inclined at an angle of 32~ to the horizon, the observer finds that the altitude is 77~. Prove that the vertical height of the rock above the first point of observation is 1034 ft. 20. A privateer 10 mi. S.W. of a harbor sees a ship sail from it in a direction S. 80~ E., at a rate of 9 mi. an hour. In what direction and at what rate must the privateer sail in order to come up with the ship in 1- hr.? Ans. N. 76~ 56' E. 13.9 mi. per hour. 21. At the top of a chimney 150 ft. high, standing at one corner of a triangular yard, the angle subtended by the adjacent sides of the yard are 30~ and 45~ respectively, while that subtended by the opposite side is 30~. Show that the lengths of the sides are 150 ft., 86.6 ft., and 106.8 ft. respectively. 22. A person goes 70 yd. up a slope of 1 in 3- from the edge of a river, and observes the angle of depression of an object on the opposite bank to be 21-~. Find the breadth of the river. Ans. 442.8 yd. 23. A flagstaff h ft. high stands on the top of a tower. From a point in the plain on which the tower stands the angles of elevation of the top and bottom of the flagstaff are observed to be a and P respectively. Prove that the h tan 13 h sin 3l. cos ac height of the tower is hta ft. i.e.sin ft. tan a-tan/S sin (a-/) 24. The length of a lake subtends at a certain point an angle of 46~ 24', and the distances from this point to the two extremities of the lake are 346 and 290 ft. Find the length of the lake. Ans. 255.8 ft. 25. From the top of a cliff h ft. high the angles of depression of two ships at sea in a line with the foot of the cliff are a and P respectively. Show that the distance between the ships is h (cot - cot a) ft. 26. Two ships are a mile apart. The angular distance of the first ship from a fort on shore, as observed from the second ship, is 35~ 14' 10"; the angular distance of the second ship from the fort, observed from the first ship, is 420 11' 53". Find the distance in feet from each ship to the fort. Ans. 3121 ft., 3634 ft. 27. The angular. elevation of a tower at a place due south of it is a, and at another place due west of the first and distant d from it, the elevation is p. Prove that the height of the tower is d. d sin a sin / V/cot23 - cot2 a V/sin (a - ). sin (a + 3) 28. To find the distance of an inaccessible point C from either of two points A and B, having no instruments to measure angles. Prolong CA to a, and CB to b, and join AB, Ab, and Ba. Measure AB,.500; aA, 100; aB, 560; bB, 100; and Ab, 550. Ans. 500 and 536. 29. A man stands on the top of the wall of height h and observes the angular elevation a of the top of a telegraph post; he then descends frbm the wall and finds that the angular elevation is now /3; prove that the height of the post exceeds the height of the man by h sin - cos. sin (i - a) MISCELLANEOUS EXAMPLES 187 30. Two inaccessible points A and'B are visible from D, but no other point can be found whence both are visible. Take some point C, whence A and D can be seen, and measure CD, 200 ft.; ADC, 89~; ACD, 50~ 30'. Then take some point E, whence D and B are visible, and measure DE, 200; BDE, 54~ 30'; BED, 88~ 30'. At D measure ADB, 72~ 30'. Compute the distance AB. Ans. 345.4 ft. 31. The angle of elevation of an inaccessible tower situated on a horizontal plane is 63~ 26'; at a point 500 ft. farther from the base of the tower the elevation of its top is 32 14'. Find the height of the tower. Ans. 460.5 ft. 32. To compute the horizontal distance between two inaccessible points A and B, when no point can be found whence both can be seen. Take two points C and D, distant 200 yd., so that A can be seen from C, and B from D. From C measure CF, 200 yd. to F, whence A can be seen; and from D measure DE, 200 yd. to E, whence B can be seen. Measure AFC, 83~; ACD, 53~ 30'; ACF, 54~ 31'; BDE, 54 30'; BDC, 156~ 25'; DEB, 88~ 30'. Ans. 345.3yd. 33. A tower is situated on the bank of a river. From the opposite bank the angle of elevation of the tower is 60~ 13', and from a point 40 ft. more distant the elevation is 50~ 19'. Find the breadth of the river. Ans. 88.9 ft. 34. A ship sailing north sees two lighthouses 8 mi. apart, in a line due west; after an hour's sailing one lighthouse bears S.W. and the other S.S.W. Find the ship's rate. Ans. 13.6 mi. per hour. 35. A column in the north temperate zone is east-southeast of an observer, and at noon the extremity of its shadow is northeast of him. The shadow is 80 ft. in length, and the elevation of the column at the observer's station is 45~. Find the height of the column. Ans. 61.23 ft. 36. At a distance of 40 ft. from the foot of a tower on an inclined plane the 'tower subtends an angle of 41~ 19'; at a point 60 ft. farther away the angle subtended by the tower is 23~ 45'. Find the height of the tower. Ans. 56.5 ft. 37. A tower makes an angle of 113~ 12' with the inclined plane on which it stands; and at a distance of 89 ft. from its base, measured down the plane, the angle subtended by the tower is 23~ 27'. Find the height of the tower. Ans. 51.6 ft. 38. From the top 6f a hill the angles of depression of two objects situated in the horizontal plane of the base of the hill are 45~ and 30~; and the horizontal angle between the two objects is 30~. Show that the height of the hill is equal to the distance between the objects. 39. I observe the angular elevation of the summits of two spires which appear in a straight line to be a, and the angular depressions of their reflections in still water to be p and y. If the height of my eye above the level of the water be c, then the horizontal distance between the spires is 2 c cos2a sin (p - y) sin (p - a) sin (y - a) 40. The angular elevation of a tower due south at a place A is 30~, and at a place B, due west of A and at a distance a from it, the elevation is 18~. Show that the height of the tower is /2 - 5 + 2 188 PLANE TRIGONOMETRY 41. A boy standing c ft. behind and opposite the middle of a football goal sees that the angle of elevation of the nearer crossbar is A and the angle of elevation of the farther one is B. Show that the length of the field is c (tanA cotB - 1). 42. A valley is crossed by a horizontal bridge whose length is 1. The sides of the valley make angles A and B with the horizon. Show that the height of the bridge above the bottom of the valley is cot A + cot B 43. A tower is situated on a horizontal plane at a distance a from the base of a hill whose inclination is a. A person on the hill, looking over the tower, can just see a pond, the distance of which from the tower is b. Show that, if the distance of the observer from the foot of the hill be c, the height of the be sin a tower is a + b + c cosa 44. From a point on a hillside of constant inclination the angle of elevation of the top of an obelisk on its summit is observed to be a, and a ft. nearer to the top of the hill to be g; show that if h be the height of the obelisk, the inclination of the hill to the horizon will be C s- (a sin a sin, th sin (3p-a) CHAPTER X RECAPITULATION OF FORMULAS PLANE TRIGONOMETRY Right triangles, pp. 2-II. B (1) sin A=- (4) cscA = - o a b c (2) cos A =- (5) sec = b a a b (3) tanA =. (6) cotA = A b b a b (7) Side opposite an acute angle = hypotenuse X sine of the angle. (8) Side adjacent an acute angle = hypotenuse X cosine of the angle. (9) Side opposite an acute angle = adjacent side x tangent of the angle. Fundamental relations between the functions, p. 59. 1 1 (19) sinx = csc =- csc X sin (20) cos x = secx = sec x cos (21) tan x = cot = cot x tan x sin x cos X (22) tanx -, cot x = - Cos X sin x (23) sin2x + cos2x = 1. (24) sec2x = 1 + tan2x. (25) csc2x = 1 + cot2x. Functions of the sum and of the difference of two angles, pp. 63-69. (40) sin (x + y) = sin x cos y + cos x sin y. (41) sin (x - y) = sin x cos y - cos x sin y. (42) cos (x - y) = cos x 6os y - sin x sin y. (43) cos (x - y) = cos x cos y + sin x sin y. 189 190 PLANE TRIGONOMETRY tan x + tan y (44) tan(x + Y)= 1- - t t tan -tany (45) - tan x -tan y (45) tan (x — Y) = 1 + tan x tan y cot x cot y - 1 (46) cot ( + y)= ot y + ot cot y + cot X cot x cot y + 1 (47) cot(x - ) = cot y-cot cot y - cot x Functions of twice an angle, p. 70. (48) sin 2x = 2 sin x cos x. (49) cos 2x = cos2x - sin2x. (50) 2 tan x tan 2x = 1 - tan2x Functions of an angle in terms of functions of half x x (51) sin x = 2 sin 2 cos 2 x x (52) cos x = cos2 - - sin2 -. 2 2, the angle, p. 72. (53) x 2 tan tan x = 1- tan2 -2 Functions of half an angle, pp. 72-7: (54) sin = —z 2 2 2 x 1 1 cos x (55) cos = = ~ —2 X, a;, 1 - cos X (56) tan 2 + cos x x sin x (57) tan = 1 + cos x 2 1 + cos x x I - cos x (58) tan = - s t 2n l sinx x 1 + cos x (59) cot 1 = -cos v 2 M1 - cos x x 1 + cos x (60) cot = X-. 2 sin x x sinx (61) cot 2 = -cos x co2 1 - cosx Sums and differences of functions, p. 74. (62) sin A + sinB = 2 sin I (A + B) cos;(A -B). (63) sin A - sinB = 2 cos I (A + B) sin (A - B). (64) cos A + cos B = 2 cos z (A + B) cos - (A - B). (65) cosA -cosB =-2 sin I (A - B)sin (A -B). sin A + sin B tan I (A + B) (66) sinA - sin B tan I (A - B) RECAPITULATION OF FORMULAS 191 Law of sines, p. 102. a b c (72) (72) sin A sinb' sin C Law of cosines, p. Io9. (73) a2 = b2 + c2 - 2 bc cosA. Law of tangents, p. I 2. 9a -+ _ tan (A + B) (79) 1 a - b tan (A -B) Functions of the half angles of a triangle in terms of the sides, pp. 113-115. pp. II3-II5- S= 2 (a+ b + c). (81) sill. A = (s - )(s - ). be s ss-a) (82) cos I A = a 2cb (83) tanIA =(s (-). 2 / J - s(- - a) (84) (s- a) ( -b) (s -c) r (85) tan I A- 2 s -a r (86) tan B = b2 s -- r (87) tan C = s —c Area of a triangle, p. 117. (88) S = I besinA. (89) S = V (S - a) (s - ) (s - c). SPHERICAL TRIGONOMETRY CHAPTER I RIGHT SPHERICAL TRIANGLES 1. Correspondence between the face angles and the diedral angles of a triedral angle on the one hand, and the sides and angles of a spherical triangle on the other. Take any triedral angle O-A'B'C' and let a sphere of any radius, as OA, be described about the vertex 0 as a center. The intersections of this sphere with the faces of the B' this triangle measure the face angles A'OB', B....,,COA' of the..........A'..................... triedral angle. The angles ABC, BCA, CAB, are measured by the plane angles which also measure the diedral angles of the triedral angle; for, by Geometry, each is measured by the angle between two straight lines drawn, one in each face, perpendicular to the edge at the same point. Spherical Trigonometry treats of the trigonometric relations between the six elements (three sides and three angles) of a spherical triangle; or, what amounts to the same thing, between the face and diedral angles of the triedral angle which intercepts it, as shown in the figure. Hence we have the Theorem. From any property of triedral angles an analogous property of spherical triangles can be inferred, and vice versa. 13;Y 194 SPHERICAL TRIGONOMETRY It is evident that the face and diedral angles of the triedral angle are not altered in magnitude by varying the radius of the sphere; hence the relations between the sides and angles of a spherical triangle are independent of the length of the radius. The sides of a spherical triangle, being arcs, are usually expressed in degrees.* The length of a side (arc) may be found in terms of any linear unit from the proportion circumference of great circle: length of are:: 360~: degrees in arc. A side or an angle of a spherical triangle may have any value from 0~ to 360~, but any spherical triangle can always be made to depend on a spherical triangle having A ____ B each element less than 180~. Thus, a triangle such as ADEBC (unshaded portion of hemisphere in figure), which has a side ADEB greater than 180~, need not be considered, for its parts can be immediately found from the parts of the triangle ABC, ~D)~\ /E ~each of whose sides is less than gO0~. For arc A DEB = 360~ - arc AB, angle CAD = 180~ - angle CAB, etc. Only triangles whose elements are less than 180~ are considered in this book. 2. Properties of spherical triangles. The proofs of the following properties of spherical triangles may be found in any treatise on Spherical Geometry: (a) Either side of a spherical triangle is less than the sum of the other two sides. (b) If two sides of a spherical triangle A' are unequal, the angles opposite them are unequal, and the greater angle lies opposite A the greater side, and conversely. c b (c) The sum of the sides of a spherical triangle is less than 360~.t (d) The sum of the angles of a spher- B' ical triangle is greater than 180~ and less L a C' than 540~. T a * One of the chief differences between Plane Trigonometry and Spherical Trigonometry is that in the former the sides of triangles are expressed in linear units, while in the latter all the parts are usually expressed in units of arc, i.e. degrees, etc. t In a plane triangle the sum of the sides may have any magnitude. $ In a plane triangle the sum of the angles is always equal to 180~. RIGHT SPHERICAL TRIANGLES 195 (e) If A'B'C' is the polar triangle* of ABC, then, conversely, ABC is the polar triangle of A'B'C'. (f) In two polar triangles each angle of one is the supplement of the side lying opposite to it in the other. Applying this to the last figure, we get A = 180~ - a', B = 180 - b', C = 180 - c', A'= 180~- a, B' = 180~- b, C' = 180~- c. A spherical triangle which has one or more right angles is called a right spherical triangle. EXAMPLES 1. Find the sides of the polar triangles of the spherical triangles whose angles are as follows. Draw the figure in each case. (a) A = 70~, B = 80~, C 100~. Ans. a'= 110, b' = 100~, c'= 80~. (b) A = 56~, B =97~, C = 112~. (c) A = 68~ 14', B = 52 10', C = 98~ 44'. (d) A = 115.6~, B = 89.9~, C = 74.2~. 2. Find the angles of the polar triangles of the spherical triangles whose sides are as follows: (a) a = 94~, b = 52~, c = 100~. Ans. A' = 86~, B'= 128~, C'= 80~. (b) a - 74~ 42', b = 95~ 6', c = 66~ 25'. (c) a = 106.4, b= 64.3~, c = 51.7~. 3. If a triangle has three right angles, show that the sides of the triangle are quadrants. 4. Show that if a triangle has two right angles, the sides opposite these angles are quadrants, and the third angle is measured by the opposite side. 5. Find the lengths of the sides of the triangles in Example 2 if the radius of the sphere is 4 ft. 3. Formulas relating to right spherical triangles. From the above Examples 3 and 4, it is evident that the only kind of right spherical triangle that requires further investigation is that which contains only one right angle. In the figure shown on the next page let ABC be a right spherical triangle having only one right angle, the center of the sphere being at O. Let C be the right angle, and suppose first that each of the other elements is-less than 90~, the radius of the sphere being unity. *The polar triangle of any spherical triangle is constructed by describing arcs of great circles about the vertices.of the original triangle as poles, 196 SPHERICAL TRIGONOMETRY Pass an auxiliary plane through B perpendicular to OA, cutting OA at E and OC at D. Draw BE, BD, and DE. BE and DE are each perpendicular to OA; [If a straight line is I to a plane, it is _L to every line in the plane.] therefore angle BED = angle A. The plane BDE is perpendicular to the plane A OC'; -If a straight line is I to a plane, every plane] [passed through the line is _ to the first plane. hence BD, which is the intersection of the planes BDE and BOC, is perpendicular to the plane A OC, If two intersecting planes are each _ to a third] plane, their intersection is also I to that plane.] and therefore perpendicular to OC and DE. In triangle EOD, remembering that angle EOD = b, we have OE OD = cos b, OD or, clearing of fractions, (A) But and OE = OD. cos b. OE = cos c (= cos EOB), OD = cos a (= cos DOB). Substituting in (A), we get (1) cos c = cos a cos b. In triangle BED, remembering that angle BED = angle A, we have or, clearing of fractions, (B) But and BD = sin A, BD -BE sin A. BD = sin a (= sin DOB), BE = sin c (= sin EOB). RIGHT SPHERICAL TRIANGLES 197 Substituting in (B), we get (2) sin a = sin c sin A. Similarly, if we had passed the auxiliary plane through A perpendicular to OB, (3) sin b = sin c sin B. Again, in the triangle BED, (C) cos A = - BE DE But DE = OD sin b, from sin b = D OD OD = cos a (= cos DOB),. and BE = sin c (= sin EOB). Substituting in (C), OD sin b sin b (D) cos A= si = cos a — sn sin c sin c sin b But from (3), s = sin B. Therefore sin c (4) cos A = cos a sin B. Similarly, if we had passed the auxiliary plane through A perpendicular to OB, (5) cos B = cos b sin A. The above five formulas are fundamental; that is, from them we may derive all other relations expressing any one part of a right spherical triangle in terms of two others. For example, to find a relation between A, b, c, proceed thus: From (4), cos A = cos a sin B cos c sin b cos b sin c since cos a= from (1), and sin B= sin bfrom (3). cos b sine sin b cos c cos b sin (6).'. cos A = tan b cot c. Similarly, we may get (7) cos B = tan a cot c. (8) sin b = tan a cot A. (9) sin a = tan b cot B. (10) cos c = cot A cot B. 198 SPHERICAL TRIGONOMETRY These ten formulas are sufficient for the solution of right spherical triangles. In deriving these formulas we assumed all the elements except the right angle to be less than 90~. But the formulas hold when this assumption is not made. For instance, let us suppose that a is greater that 90~. In this case the auxiliary plane BDE will cut CO and AO produced beyond the center 0, and we have, in triangle EOD, OE (E) cos DOE (= cos b) D But OE = cos EOB - cos A OB = - cos c, and OD = cos DOB =- cos COB - cos a. Substituting in (E), we get cos c cos b -, or cos c = cos a cos b, cos a which is the same as (1). Likewise, the other formulas will hold true in this case. Similarly, they may be shown to hold true in all cases. If the two sides including the right angle are either both less or a c _ D- - C-................................................................... both greater than 90~ (that is, cos a and cos b are either both positive or both negative), then the product (F) cos a cos b will always be positive, and therefore cos c, from (1), will always be positive, that is, c will always be less than 90~. If, however, one of the sides including the right angle is less and the other is greater than 900, the product (F), and therefore also cos c, will be negative, and c will be greater than 90~. Hence we have Theorem I. If the two sides including the right angle of a right spherical triangle are both less or both greater than 90~, the hypotenuse RIGHT SPHERICAL TRIANGLES 199 is less than 90~; if one side is less and the other is greater than 90~, the hypotenuse is greater than 90~. cos A cos B From (4) and (5), sin B= —, and sin A = cos a cos b Since A and B are less than 180~, sin A and sin B must always be positive. But then cos A and cos a must have the same sign, that is, A and a are either both less than 90~ or both greater than 90~. Similarly, for B and b. Hence we have Theorem II. In a right spherical triangle an oblique angle and the side opposite are either both less or both greater than 90~. 4. Napier's rules of circular parts. The ten formulas derived in the last section express the relations between the three sides and the two oblique angles of a right spherical triangle. All these relations may be shown to follow from two very useful rules discovered by Baron Napier, the inventor of logarithms. For this purpose the right angle (not entering the formulas) is not taken into account, and we replace the hypotenuse and the two B B0 C a a A 90 A b b oblique angles by their respective complements; so that the five parts, called the circular parts, used in Napier's rules are a, b, An, c, Bc. The subscript c indicates that the complement is to be used. The first figure illustrates the ordinary method of representing a right spherical triangle. To emphasize the circular parts employed in Napier's rules, the same triangle might be represented as shown in the second figure. It is not necessary, however, to draw the triangle at all when using Napier's rules; in fact, it is found to be more convenient to simply write down the five parts in their proper order as on A, Bo the circumference of a circle, as shown in the third figure (hence the name circular parts). a b Any one of these parts may be called a middle part; then the two parts immediately adjacent to it are called acljacent parts, and the other two opposite parts. Thus, if a is taken as a middle part, A0 and b are the adjacent parts, while c, and Bo are the opposite parts. 200 SPHERICAL TRIGONOMETRY Napier's rules of circular parts. Rule I. The sine of any middle part is equal to the product of the tangents of the adjacent parts. Rule II. The sine of any middle part is equal to the product of the cosines of the opposite parts. These rules are easily remembered if we associate the first one with the expression " tan-adj." and the second one with " cos-opp." * Napier's rules may be easily verified by applying them in turn to each one of the five circular parts taken as a middle part, and comparing the results with (1) to (10). For example, let cc be taken as a middle part; then A, and B, are the adjacent parts, while a and b are the opposite parts. Then, by Rule I, sin cc = tan A0 tan B, Cc or, cos c = cot A cot B; Ac, Bg which agrees with (10), p. 197. By Rule II, sin cc = cos a cos b, a b or, cos c = cos a cos b; which agrees with (1), p. 196. The student should verify Napier's rules in this manner by taking each one of the other four circular parts as the middle part. Writers on Trigonometry differ as to the practical value of Napier's rules, but it is generally conceded that they are a great aid to the memory in applying formulas (1) to (10) to the solution of right spherical triangles, and we shall so employ them. 5. Solution of right spherical triangles. To solve a right spherical triangle, two elements (parts) must be given in addition to the right angle. For the sake of uniformity we shall continue to denote the right angle in a spherical triangle ABC by the letter C. General directions for solving right spherical triangles. Cc C, cc cc A B7? A? B Ac Bo AC Be b a b a b a a First step. Write down the five circular parts as in first figure. Second step. Underline the two given parts and the required unknown part. Thus, if A, and a are given to find b, we underline all three as is shown in the second figure. * Or by noting that a is the first vowel in the words " tangent" and "adjacent," while o is the first vowel in the words " cosine" and "opposite." RIGHT SPHERICAL TRIANGLES 201 Third step. Pick out the middle part (in this case b) and cross the line under it as indicated in the third figure. Fourth step. Use Rule I hf the other two parts are adjacent to the middle part (as in case illustrated), or Rule II if they are opposite, and solve for the unknown part. Check: Check with that rule which involves the three required parts. Careful attention must be paid to the algebraic signs of the functions when solving spherical triangles; the cosines, tangents, and cotangents of angles or arcs greater than 90~ being negative. When computing with logarithms we shall write (n) after the logarithms when the functions are negative. If the number of negative factors is even, the result will be positive; if it is odd, the result will be negative and (n) should be written after the resulting logarithm. In order to be able to show our computations in compact form, we shall write down all the logarithms of the trigonometric functions just as they are given in our table; that is, when a logarithm has a negative characteristic we will not write down - 10 after it.t Ex. 1. Solve the right spherical triangle, having given B = 33~ 50', a = 108~. Solution. Follow the above general directions. To find A cc A B, b a Using Rule II sin Ac = cos Bc cos a cosA = sin B cos a log sin B = 9.7457 log cos a = 9.4900 (n) log cosA = 9.2357 (n).. 1800-At = 800 6' and A = 99~ 54'. To find b Cc A, B, b a Using Rule I sin a = tan B, tan b tan b = sin a tan B log sin a = 9.9782 log tan B = 9.8263 log tan b = 9.8045.-. b = 32 31'. To find c cc Ac Be b a Using Rule I sin B, = tan cc tan a cot c = cos B cot a log cosB = 9.9194 log cot a = 9.5118 (n) log cot c = 9.4312 (n)... 180 - c = 740 54' and c = 1050 6'. The value of log cos A found is the same as that found in our first computation. The student should observe that in checking our work in this example * Thus, in above case, Ac and a are given; therefore we underline the three required parts and cross b as the middle part. Applying Rule II, cc and Be being opposite parts, we get sin b = cos cc cos Be, or, sin b = sin c sin B. t For example, as in the table, we will write log sin 240 = 9.6093. To be exact, this should be written log sin 240 = 9.6093 - 10, or, log sin 240 = 1.6093. $ Since cos A is negative, we get the supplement of A from the table. 202 SPHERICAL TRIGONOMETRY it was not necessary to look up any new logarithms. Hence the check in this case is only on the correctness of the logarithmic work.* Check: Using Rule I tc A *f Be sin Ac = tan b tan cc cosA = tan b cot c log tan b = 9.8045 log cot c = 9.4312 (n) log cos A = 9.2357 (n) b a In logarithmic computations the student should always write down an outline or skeleton of the computation before using his logarithmic table at all. In the last example this outline would be as follows: log sin B = log sin a = log cos B = log cos a = (n) log tan B = log cot a = (n) log cos A = (n) log tan b = log cot c = (n).. 180~- A =.. b =.. 180~-c = and A = and c = It saves time to look up all the logarithms at once, and besides it reduces the liability of error to thus separate the theoretical part of the work from that which is purely mechanical. Students should be drilled in writing down forms like that given above before attempting to solve examples. Ex. 2. Solve the right spherical triangle, having given c = 70~ 30', A = 100~. Solution. Follow the general directions. To find a cc Ac B. b a Using Rule II sin a = co cc cos Ac sin a = sin c sin A log sin c = 9.9743 log sin A = 9.9934 log sin a = 9.9677.-. 180~-at = 680 10' and a = 111~ 50'. To find b cO b a Using Rule I sin Ac = tan b tan cc tan b = cos A tan c log cos A = 9.2397 (n) log tan c = 0.4509 log tan b = 9.6906 (n).. 180 - b = 26~ 8' and b = 153~ 52'. To find B c b a Using Rule I sin cc = tan Ae tan Be cot B = cos c tan A log cos c = 9.5235 log tan A = 0.7537 (n) log cot B = 0.2772 (n).'. 1800 —B= 27~ 51' and B = 152~ 9'. The work of verifying the results is left to the student. * In order to be sure that the angles and sides have been correctly taken from the tables, in such an example as this, we should use them together with some of the given data in relations not already employed. t Since a is determined from its sine, it is evident that it may have the value 68~ 10' found from the table, or the supplementary value 111~ 50'. Since A > 900, however, we know from Th. II, p. 199, that a > 90~; hence a= 111 50' is the only solution. RIGHT SPHERICAL TRIANGLES 203 6. The ambiguous case. Two solutions. When the given parts of a right spherical triangle are an oblique angle and its opposite side, there are two triangles which c B satisfy the given conditions. For, in the triangle ABC, let C = 90~, and let A and CB A >A' (= a) be the given parts. If ---o we extend AB and AC to A', b it is evident that the triangle A'BC also satisfies the given conditions, since BCA'= 90~, A' = A, and BC = a. The remaining parts in A'BC are supplementary to the respective remaining parts in ABC. Thus A'B = 180~- c, A'C = 180- b, A'BC = 1800 -ABC. This ambiguity also appears in the solution illustrated in the following example: Ex. 3. Solve the right spherical triangle, having 128~ 33'. Solution. We proceed as in the previous examples. of the triangle, as is given A = 105~ 59', a = To find b cc Ac Be a sin b = tan a tan Ac sin b = tan a cotA log tan a = 0.0986 (n) log cotA = 9.4570 (n) log sin B = 9.5556.-. b = 21~ 4', or, 180~ - b = 158 56' = b'.* To find B 7 C b a sin Ac = cos a cos Bc cos A sin B = cos a log cos A = 9.4399 (n) log cosa = 9.7946 (n) log sin B = 9.6453.-. B = 26 14', or, 1800 - B = 1530 46' = B'. t To find c cc b a sin a = cosAc cos c, sin a sin c = sin A log sin a = 9.8932 log sin A = 9.9828 log sin c = 9.9104.-. c' = 54~ 27', or, 180 - c' = 1250 33'= c.t Hence the two solutions are: 1. b = 21~ 4', 2. b' = 158~ 56', c = 125~ 33', c' = 54~ 27', B = 26 14' (triangle ABC); B'= 153~ 46' (triangle A'BC). It is not necessary to check both solutions. We leave this to the student. * Since sin B is positive and B is not known, we cannot remove the ambiguity. Hence both the acute angle taken from the table and its supplement must be retained. t The two values of B must be retained, since b has two values which are supplementary. $ Since a > 90~ and b has two values, one > and the other < 90~, it follows from Th. I, p. 198, that c will have two values, the first one < 90~ and the second > 90~. 204 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following right spherical triangles: No. GIVEN PARTS REQUIRED PARTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 a = 132~ 6' a = 159~ A = 50~ 20' a = 160~ B = 80~ B = 112~ a = 61~ a = 61~ 40' A = 99~ 50' b= 15~ A = 62~ 59' A = 73~ 7' B = 144~ 54' B = 68~ 18' A = 161~ 52' a =-113~ 25' b = 77~ 51' c = 137~ 20' B = 122~ 40' b = 38~ 30' b = 67~ 40' c = 81~ 50' B = 123~ 40' b = 144~ 10' a = 112~ c = 152~ 20' B = 37~ 4' c = 114~ 32' b = 146~ 32' c,= 470 34' b = 131~ 8' b = 110~ 47' A = 131~ 27' B = 80~ 55' c = 98~ 7' A = 1480 5' a = 40~ 42' A = 149~ 41' A = 27~ 12' A'= 1520 48' A = 109~ 23' A = 66~ 12' A = 72~ 29' B = 27~ 7' B'= 152~ 53' A = 120~ 44' a = 41~ 6' a = 60~ 31' A = 78~ 47' A'= 1010 13' A = 30~ 32' a = 166~ 9' A = 112~ 3' A = 135~ 3' a = 146~ 33' A = 70~ B = 65~ 23' b = 134~ 31' B = 66~ 44' a = 25~ 25' a'= 154~ 35' a = 110~ 58' b = 127 17' B = 140~ 38' b = 25~ 24' b'= 154~ 36' a = 156~ 30' b = 26~ 25' B = 143~ 50' a = 70~ 10' a'= 109~ 50' a = 22~ 1' B = 101~ 49' B = 1090 12' b = 680 17' b = 109~ 48' B = 75~ 6' b = 38~ 1' c = 122~ 7' c = 137~ 20' c = 69~ 54'; or, c = 110~ 6' b = 1130 22' c = 107~ 5' c = 112~ 38' c = 109~ 46'; or, c'= 70~ 14' B = 33~ 53' c = 47~ 32' b = 147~ 32' c = 106~ 28'; or, c'= 73~ 32' b = 430 18' c = 50~ 18' c = 81~ 54' c = 105~ 44' c = 73~ 35' b = 74~ 7' a = 137~ 9' B = 74~ 51' A = 144~ 54' B = 1010 14' a = 69~ 18' c = 84~ 27' I 20. For more examples take any two parts in the above triangles and solve for the other three. 7. Solution of isosceles and quadrantal triangles. Plane isosceles triangles were solved by dividing each one into two equal right triangles and then solving one of the right triangles. Similarly, we may solve an isosceles spherical triangle by dividing it into two symmetrical (equal) right spherical triangles by an arc drawn from the vertex perpendicular to the base, and then solving one of the right spherical triangles. A quadrantal triangle is a spherical triangle one side of which is a quadrant (= 90~). By (f), p. 195, the polar triangle of a quadrantal triangle is a right triangle. Therefore, to solve a quadrantal triangle we have only to solve its polar triangle and take the supplements of the parts obtained by the calculation. Ex. 1. Solve the triangle, having given c = 90~, a = 67~ 38', b = 48~ 50'. Solution. This is a quadrantal triangle since one side c = 90~. We then find the corresponding elements of its polar triangle by (f), p. 195. They are C'= 900, A' = 112~ 22', B' = 131~ 10'. We solve this right triangle in the usual way. RIGHT SPHERICAL TRIANGLES 205 Construct the polar (right) triangle. Given A' = 112~ 22', B' = 131 10': To find a' c'c A' B' *b a' sinA' = cos a' cos Bc cosA' cos a' = — sin B' log cosA' = 9.5804 (n) log sin B' = 9.8767 log cos a' = 9.7037 (n) 180~ - a' = 59~ 38'. a' = 120~ 22'. Similarly, we get b'= 135~ 23', c' = 68~ 55'. Hence in the given quadrantal triangle we have A= 180 - a' = 59 38', B = 180 - b' = 44~ 37', C = 180~ - c' = 111~ 5'. EXAMPLES Solve the following quadrantal triangles: No. GIVEN PARTS REQUIRED PARTS 1 A =139~ b=143~ c=90 a = 1171' B = 153~ 42' C=132~ 34' 2 A=45~30' B=139~ 20' c=90~ a = 57~ 22' b = 129~ 42' C =57 53' 3 a =30~ 20' C =42~ 40' c = 90 A=2001' B =141~ 30' b =113~ 17' 4 B = 70~ 12' C =106~ 25' c= 90~ A =33~ 28' a =350 4' b= 78~ 47' 5 A = 105~ 53' a = 104~ 54' c = 90~ B = 69~16' b = 70~ C = 84~ 30'; or B = 110~ 44' b = 110~ C =95~ 30' Solve the following isosceles spherical triangles: No. GIVEN PARTS REQUIRED PARTS 6 a = 54~20' c =72054' A= B b = 54~ 20' A = B = 57~ 59' C = 93~ 59' 7 a = 54~ 30' C = 71 A = B b = 54~ 30' A = B —90~ c = 180~ 8 a=66~29' A B =500~17' b=66~29' c=11130' C= 128~ 42' 9 c =156040' C= 187 46' A =B a = b = 3~ 58' or 176~ 2' A = B = 89~ 12' or 90~ 48' Prove the following relations between the elements of a right spherical triangle (C = 90~): 10. cos2 A sin2 c = sin (c + a) sin (c - a). 13. sin (b + c) = 2 cos2 - A cos b sin c. 11. tan a cos c = sin b cot B. 14. sin (c - b) = 2 sin2 A A cos b sin c. 12. sin2 A = cos2 B + sin2 a sin2 B. CHAPTER II OBLIQUE SPHERICAL TRIANGLES 8. Fundamental formulas. In this chapter some relations between the sides and angles of any spherical triangle (whether right angled or oblique) will be derived. 9. Law of sines. In a spherical triangle the sines of the sides are proportional to the sines of the opposite angles. Proof. Let ABC be any spherical triangle, and draw the arc CD perpendicular to AB. There will be two cases according as CD falls C C b a b A,3n A, D i.F-, — D _~. -—..-JL upon AB (first figure) or upon AB produced (second figure). For the sake of brevity let CD = p, AD = n, BD = m, angle A CD = x, angle BCD = y. In the right triangle ADC (either figure) (A) sinp = sin b sin A. Rule II, p. 200 In the right triangle BCD (first figure) (B) sinp = sin a sin B. Rule II, p. 200 This also holds true in the second figure, for sin DBC = sin (180~ -B) = sin B. Equating the values of sinp from (A) and (B), si a sinB = sin b sin A, or, dividing through by sinA sinB, sin a_ sin b ~(C ) sinA sin B 206 OBLIQUE SPHERICAL TRIANGLES 207 In like manner, by drawing perpendiculars from A and B, we get sinb sin c (D) -.- =., and ~(D)> sinB sin C a (E) sinC - sin ' respectively. Writing (C), (D), (E) as a single statement, we get the law of sines, sin a sin b sin c sinA sin B sinC 10. Law of cosines. In a spherical triangle the cosine of any side is equal to the product of the cosines of the other two sides plus the product of the sines of these two sides and the cosine of their included angle. Proof. Using the same figures as in the last section, we have in the right triangle BDC, cos a = cos p cos m Rule II, p. 200 = cosp cos(c - n) cosp I cos c cos n + sin c sin n I (A) cos cos c cosn + cos sin c sin n. In the right triangle ADC, (B) cosp cos n = cos b. cos b Whence cosp = cos cos n and, multiplying both sides by sin n, cos b (C) cosp sin n = sin n = cos b tan n. cos ioCs i n n ()th~ Xr, But cos A = tan n cot b, or, Rule I, (D) tan = tanb cos A. Substituting value of tan n from (D) in (C), we have (E) cosp sin n = cos b tan b cos A = sin b cos A. Substituting the value of cosp cos n from (B) and the va cosp sin n from (E) in (A), we get the law of cosines, (F) cos a = cos b cos c + sin b sin c cos A. * Compare with the law of sines in Granville's Plane Trigonometry, p. 102. p. 200 lue of 208 SPHERICAL TRIGONOMETRY Similarly, for the sides b and c we may obtain (G) cos b = cos c cos a + sin c sin a cos B, (H) cos c cos a cos b + sin a sin b cos C. 11. Principle of Duality. Given any relation involving one or more of the sides a, b, c, and the angles A, B, C of any general spherical triangle. Now the polar triangle (whose sides are denoted by a', b', c', and angles by A', B', C') is also in this case a general spherical triangle, and the given relation must hold true for it also; that is, the given relation applies to the polar triangle if accents are placed upon the letters representing the sides and angles. Thus (F), (G), (H) of the last section give us the following law of cosines for the polar triangle: (A) cos at = cos bcoso c' + sin b' sin c' cos A '. (B) cos b' = cos c' cos a' + sin c' sin a' cos B'. (C) cos c' = cos a' cos b' + sin a' sin b' cos C1. But by (/), p. 195, a = 180-A, b'= 180 -B, c' = 180- C, A'= 180~- a, B' = 180~-b, C' = 180 - c. Making these substitutions in (A), (B), (C), which refer to the polar triangle, we get (D) cos (180~- A) = cos (180~ -B) cos (180~- C) + sin (180~ - B) sin (180~- C) cos (180~- a), (E) cos (180~ -B) = cos (180 - C) cos (180~ - A) + sin (180 - C) sin (180 - A) cos (180~ - ), (F) cos (180 - C) = cos (180 - A) cos (180 - B) + sin (180~- A) sin (180~ - B) cos (180~- c), which involve the sides and angles of the original triangle. The result of the preceding discussion may then be stated in the following form: Theorem. In any relation between the parts of a general spherical triangle, each part may be replaced by the supplement of the opposite part, and the relation thus obtained will hold true. OBLIQUE SPHERICAL TRIANGLES 209 The Principle of Duality follows when the above theorem is applied to a relation involving one or more of the sides and the supplements of the angles (instead of the angles themselves). Let the supplements of the angles of the triangle be denoted by a, p, *; that is, a=180~0-A, ==180~-B, y=180~- C, or, A=180~-a, B=1800~ —, C=180~ —y. c When we apply the above theorem to a rela- a tion between the sides and supplements of the angles of a triangle, we, in fact, replace a by a (= 180~-A), replace b by 3 (= 1800~-B), replace c by y (= 180- C), replace a (= 180~-A) by 180~- (180~- a)= a, replace / (= 180~- B) by 180~- (180~ - b) = b, replace y (= 180 ~-C) by 180~- (180~- c) = c, or, what amounts to the same thing, interchange the Greek and Roman letters. For instance, substitute A = 180- a, B = 180 -,, C = 180~-y in (F), (G), (H) of the last section. This gives the law of cosines for the sides in the new form (12) cos a = cos b cos c - sin b sin c cos a, (13) cos b = cos c cos a - sin c sin a cos f, (14) cos c = cos a cos b - sin a sin b cos y. [Since cos A = cos (180 - a)= - cos a, etc.] If we now apply the above theorem to these formulas, we get the law of cosines for the angles, namely, (15) cos a = cos p cos y -sin f sin y cos a, (16) cos p = cos y cos a - sin y sin a cos b, (17) cos y = cos a cos f -sin a sin f cos c, * a, 3, y are then the exterior angles of the triangle, as shown in the figure. 210 SPHERICAL TRIGONOMETRY that is, we have derived three new relations between the sides and supplements of the angles of the triangle.* We may now state the Principle of Duality. If the sides of a general spherical triangle are denoted by the Roman letters a, b, c, and the supplements of the corresponding opposite angles by the Greek letters a, f, y, then, from any given formula involving any of these six parts, we may write down a dual formula by simply interchanging the corresponding Greek and Roman letters. The immediate consequence of this principle is that formulas in Spherical Trigonometry occur in pairs, either one of a pair being the dual of the other. Thus (12) and (15) are dual formulas; also (13) and (16), or (14) and (17). If we substitute A = 180~-a, B = 180~-, C= 180~-y in the law of sines (p. 207), we get sin a sin b sin c sin a sin 3 sin y [Since sin A = sin (180~ - a)= sin a, etc.] Applying the Principle of Duality to this relation, we get sin a sin /3 sin y sin a sin b sin c which is essentially the same as the previous form. The forms of the law of cosines that we have derived involve algebraic sums. As these are not convenient for logarithmic calculations, we will reduce them to the form of products. 12. Trigonometric functions of half the supplements of the angles of a spherical triangle in terms of its sides. Denote half the sum of the sides of a triangle (i.e. half the perimeter) by s. Then (A) 2s = a + b + c, or, s = -(a + b + c). * If we had employed the interior angles of the triangle in our formulas (as has been the almost universal practice of writers on Spherical Trigonometry), the two sets of cosine formulas would not have been of the same form. That the method used here has many advantages will become more and more apparent as the reading of the text is continued. Not only are the resulting formulas much easier to memorize, but much labor is saved in that, when we have derived one set of formulas for the angles (or sides), the corresponding set of formulas for the sides (or angles) may be written down at once by mere inspection by applying this Principle of Duality. The great advantage of using this Principle of Duality was first pointed out by Mobius (1790-1868). OBLIQUE SPHERICAL TRIANGLES 211 Subtracting 2 c from both sides of (A), 2 s - 2 c =a + b + c - 2 c, or, (B) s - c = - (a + b - c). Similarly, (C) (D) s - b = -(a - b + c), and s-a = (- a+ + b + ) = -(1 + c - a). From Plane Trigonometry, (E) 2 sin2 a = 1 - cos a, (F) 2 cos2 a = 1 + cos a. But from (12), p. 209, solving for cos a, cos b cos c - cos a cos a = sin b sin c Ton1ero / 7.\ loPorwnm.I.\ 2 C cos b cos c -cos a 2 sin2 a - 1 — 2si2-a ~ sin b sin c sin b sin c - cos b cos c + cos a sin b sin c cos a - (cos b cos c - sin b sin c) sin b sin c cos a - cos (b + c) sin b sin c - 2 sin I-( +b c)sin (a - b c), =,or sin b sin c 2 sin 1 (a + b + c) sin - (b + c - a) (G) 2 sin2 a n in sin b sin c [Since sin (a- b - c)= - sin (-a + b + c)= -sin (b + c - a).] Substituting from (A) and (D) in (G), we get sin s sin (s - a) sin2 " a = - -.. — or, - sin b sin c (18) sin. a= Isin s sin (s- a) (18) sin I a = sinb sin sin b sin c *Let A=a A=a B=b+c B=b+c A+B=a+b+c A-B=a-b-c (A+B)= 2(a+b+c). (A- B)= (a- b-c). Hence, substituting in (65), p. 74, Granville's Plane Trigonometry, namely, cos A - cos B= - 2 sin (A + B) sin (A - B), get cos a - cos(b + c)= -2 sin (a b + c)sin(a- b - c). vwe 212 SPHERICAL TRIGONOMETRY Similarly, (F) becomes s a = + cos b cos c - cos a 2 cos21 a = 1 + sin b sin c sin b sin c + cos b cos c - cos a sin b sin c cos (b - c)- cos a sin b sin c - 2 sin I (a + b - c) sin - (b - c - a) o sin b sin c (H) 2 cos~21a _ 2 sinl (a + b -c) sin -(a-b + c) sin b sin c [Since sin (b - c - a)= - sin (- + c + a)= -sin (a - b + c).] Substituting from (B) and (C) in (H), we get Cos21 sin (s - c) sin (s - b) cos2 l a= -.-, or, 2 sin b sin c (19) cosa sin (s - b) sin (s - c) (19) cossia=n\ sin ---} sin b sin c sin Ia Since tan - a = - i2 we get from this, by substitution from (18) and (19), 2 (20) tan sin s sin (s - a) t 2(20) ^ - sin (s - b) sin(s - c) Let A= b - c A=b-c B=a B=a A4+B= a+b-c A-B= b-c-a j(A +B)= (a+b-c). (A-B)= (b-c-a). Hence, substituting in formula (65), found on p. 74, Granville's Plane Trigonometry, namely, cos A - cos B = - 2 sin (A + B) sin (A - B), we get cos (b - c)- cos a = - 2 sis (a + b- c) sin (b - c - a). t In memorizing these formulas it will be found an aid to the memory to note the fact that under each radical (a) only the sine function occurs. (b) The denominators of the sine and cosine formulas involve those two sides of the triangle which are'not opposite to the angle sought. (c) When reading the numerator and denominator of the fraction in the tangent formula, s comes first and then the differences s-a, s-b, s-c, in cyclical order; s and the first difference occurring also in the numerator of the corresponding sine formula, while the last two differences occur in the numerator of the corresponding cosine formula. OBLIQUE SPHERICAL TRIANGLES 213 In like manner, we may get ) n sin s sin (s - b) (21) inl = sincsin a o(22) Cof | sin (s - c) sin(s - a) ~(22)> 2 ~COSf 1 N,=sin c sin a / sin s sin (s - b) (23) tan 2 = N sin (s-c) sin (s- a) Also ~(24) s.in!i sin s sin (s - c) (24) sin - N sinasin b s(25) Cos'y = sin (s - a) sin (s - b) (25) cosjy -= I8 1 ine 1inb 2 N sin a sin b 6 t y sin (s - a) sin (s - b) In solving triangles it is sometimes more convenient to use other forms of (20), (23), (26). Thus, in the right-hand member of (20), multiply both the numerator and denominator of the fraction under the radical by sin(s - a). This gives tn = sin s sin2 (s - a) tan a 2 sin (s - a) sin (s - b) sin(s - c)... | sin s = sin (s- a) T ge 2 sin s in(ssin (s- - a) sin s-a) then tan 1 r == - Let t an -d d= (27) tan 2 d = Isin (s- a) sin (s- b)sin (s- c) v / 2 M sins (28) tana sin (s - a) (28) tn t= a - tan d sin (s - b) (29) tan sin s - b) tan d sin (s - c) (30) tan 2 y = I tan 2d n It may be shown that d = diameter of the circle inscribed in the spherical triangle. 214 SPHERICAL TRIGONOMETRY 13. Trigonometric functions of the half sides of a spherical triangle in terms of the supplements of the angles. By making use of the Principle of Duality on p. 208, we get at once from formulas (18) to (30), by replacing the supplement of an angle by the opposite side and each side by the supplement of the opposite angle, the following formulas: (sin = I sin i sin (o - a) v(31) sn2a A sin f sin y Isin (cr - f) sin (o- - y) (32) cosa - = sin n a sin? sin y (33) tan sin (r sin (cr - a) (33) tan ~ a = 8 sin (~ - P) sin ( -- y)' (34) sin 1 b = sin y sin a (35) cos~ 1b= sisin (o- - y) sin (o- - a) (35) cos I b = --- - f s y —i/a, 2v / ^ sin y sin a sin o' sin (or - B) (36) tan b = sin ( -- y) sin ( - a) v(37 /c = ~2sin Ii(o- - y)r sin (ir - y ) (37) sin = sin a sin(l — y) (37) sincasinL (39) tan,1 c sin = in (ai - y) 2 sin - a) sin (- -' (40) tan 6 = sin sin -r - y) sin ( - a) tan 1- ' (41) tan a = tan sin (ir - y) (43) tan 2 C = tan, 6 where = (c + + X) =- (180- A + 180~ -B + 180 - C) = 270~- (A +B + C). What we have done amounts to interchanging the corresponding Greek and Roman letters. * It may be shown that 8 is the supplement of the diameter of the circumscribed circle. OBLIQUE SPHERICAL TRIANGLES 215 14. Napier's analogies. Dividing (20) by (23), we get tan - a sins sin(s - a) _ sin s sin(s —b) tan/3 - -bsin(-)sin (s-c) ' sin (s-c) sin (s -a) sin _ a sin s sin (s - a) cos a _ sin (s - b) sin(s - c) or, cos sin (s - c) sin(s - a) sin a cos sn sin (s - a) sin 2 - C si 2 sin (s- e) Hence. cos - a sin s / sin (s - b) By composition and division, in proportion, sill a cos i + Cos I a sin - o _ sin (s - a) sin (s -b) sin - a cos -os - a sin sin (s - a)- sin (s - b) From (40), (41), p. 63, and (66), p. 74, Granville's Plane Trigonometry, the left-hand member equals sin ( a + - f)/ sin ( a _- ) and the right-hand member sin (s - a)+ sin(s - b) _tan [s -a + (s - b)] _ tan c sin (s - a) - sin (s - b) tan [s - a -(s - b)] tan I (b - a) Equating these results,we get, noting that tan -(b-a) =-tan -(a-b), sin 1 (a + f) _ tan - c - 2or sin ( -- ) tan (a - b)' or (44) tan (a - b) 2 - tan - c. sin i (a - t2c. In the same manner we may get the two similar formulas for tan - (b - c) and tan I (c - a). Multiplying (20) and (23), we get tan'I ata tan1, I Isin s sin (s - a) sin s sin (s - b) tan 2 atan,= -sn.... a 2 - '4sin(s - b) sin (s - c) sin (s - c)sin(s -a) sin I a sin I f sill s cos Ia cos f3 sin (s - c) By composition and division, in proportion, cos - a cos 1 - sin n a sin _ sin (s - ) - sin s cos a cos /3 + sin - c sin ~, sin (s - c) + sin s *For s-a+s-b=2s-a-b=a+b+c-a- b=c, and s-a-s+ b=b-a. 216 SPHERICAL TRIGONOMETRY From (42), (43), p. 63, and (66), p. 74, Granville's Plane Trigonometry, the left-hand member equals cos (a + I A). cos (a - 1/3) and the right-hand member sin (s - c)- sin s _ tan (s - c -s) tan (-c) * sin(s - c) + sin s tan I (s - c + s) tan I (a + b) Equating these results, we get, noting that tan f (-c) = -tan a c, cos (a + ~ ) _ tan I c cos - (a -) tan (a + b)' o (45) tan (a + b) = ( i) tan c. In the same manner we may get the two similar formulas for tan a (b + c) and tan (c + a). By making use of the Principle of Duality on p. 208,we get at once from formulas (44) and (45), sin' (a - b) (46) tan (a - = sin ( + b) tan y, ( sitn (a + #) = cos (a-b) 1 By changing the letters in cyclic order we may at once write down the corresponding formulas for tan (/3 —y), tan I (y-a-), tan (/3+y), and tan - (y + a). The relations derived in this section are known as XNapier's analogies. Since cos ~ (a - b) and tan I y = tan - (180~- C) = tan (90~- C) =cot - C are always positive, it follows from (47) that cos - (a + b) and tan 2 (a + /3) always have opposite signs; or, since tan (a + /3) = tan 1 (180~-A + 180~- B) = tan [360~-(A +B)] = tan [180~2 (A + B)] = - tan - (A + B), we may say that cos -(a + b) and tan I (A + B) always have the same sign. Hence we have the Theorem. In a spherical triangle e the s of any two sides is less than, greater than, or equal to 180~, according as the sumn of the corresponding opposite angles is less than, greater than, or equal to 180~. 15. Solution of oblique spherical triangles. We shall now take up the numerical solution of oblique spherical triangles. There are three cases to consider with two subdivisions under each case. * For and S -C-S= -C, s - c + s= 2s-c=a+b+c-c= a+b. OBLIQUE SPHERICAL TRIANGLES 217 CASE I. (a) Given the three sides. (b) Given the three angles. CASE II. (a) Given two sides and their included angle. (b) Given two angles and their included side. CASE III. (a) Given two sides and the angle opposite one of them. (b) Given two angles and the side opposite one of them. 16. Case I. (a) Given the three sides. Use formulas from p. 213, namely, (27) tan -d = sin (s - a) sin (s - b)sin(s - c) v / 2 N sin s sin (s - a) (28) tan- = tan d ' (29) tan! =r- tan d sin (s - c) (30) tan yI Si tan( d - to find a, fl,, and therefore A, B, C, and check by the law of sines, sin a sin b sin c sinA sin B sin C Ex. 1. Given a = 60~, b = 137~ 20', c = 116; find A, B, C. Solution. a = 60~ To find log tan I d use (27) b = 137~ 20' log sin (s - a) = 9.9971 c = 116~ log sin (s- b) = 9.5199 2 s = 313~ 20' log sin (s - c) = 9.8140 s = 156~ 40'. 29.3310 s - a = 96~ 40'. log sin s = 9.5978 s- b =19~ 20'. 21|19.7332 s - c = 40~ 40'. log tan l d = 9.8666 To find A use (28) To find B use (29) To find C use (30) logsin(s - a) = 9.9971 logsin(s - b) = 9.5199 logsin(s - c) = 9.8140 log tan I d = 9.8666 log tan I d = 9.8666 log tan I d = 9.8666 log tan 2 a = 0.1305 log tan I P = 9.6533 log tan - = 9.9474 a = 53~ 29'. - p = 24~ 14'. = 410 32'. a = 106~ 58'. p = 48~ 28. y = 83~ 4'... A =1800-106058'=7302'... B= 1800-48028'=131032'... C=-.1800-8304'=96056'. Check: log sin a = 9.9375 log sinb = 9.8311 log sin c = 9.9537 log sin A = 9.9807 log sin B = 9.8743 log sin C = 9.9969 9.9568 9.9568 9.9568 This checks up closer than is to be expected in general. There may be a variation of at most two units in the last figure when the work is accurate. 218 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles: No. GIVEN PARTS REQUIRED PARTS 1 a =38~ b =51~ c=42~ A= 51~ 58' B=83 54' C=58~ 53' 2 a = 01 b = 49 c = 60 A= 142~ 32' B =27~ 52' C = 32~ 28' 3 a=61~ b =39 c = 92~ A= 35~ 32' B=24 42' C= 138 24' 4 a =62~ 20' b =54 10' c = 97 50' A =47~ 22' B=42 20' C=124~ 38' 5 a =58 b = 800 c = 96~ A= 55~ 58' B =74~ 14' C = 10336' 6 a =46~ 30' b=62040' c = 83~ 20' A =43~ 58' B = 580 14' C = 108 6' 7 a =71~ 15' b=39~ 10' c=40035' A=130 36' B=30~26' C=31026' 8 a=47~ 30' b=55~40' c=60010' A =56~ 32' B=6907' C=78~ 58' 9 a =43~ 30' b =720 24' c =870 50' A =41~ 27' B=66 26' C=10603' 10 a = 110~ 40' b= 45~ 10' c = 730 30' A =144~ 27' B = 26 9' C = 36~ 35' 17. Case I. (b) Given the three angles. Use formulas from p. 214, namely,* (40) (41) (42) (43) tan - 8 = sin (a- - a) sin (o- - ) sin(o - -7) tan 1 a =sin (o — -a) 2 tan -- 8 tan - b sin (a- - P) tan - 8 - sin (o - y) tan i c = v, tan ~ 8 to find a, b, c; and check by the law of sines, sin a sin b sin c sillA sin B sin C Ex. 1. Given A = 70~, B = 131~10', C = 94~ 50'; find a, b, c. Solution. Here we use the supplements of the angles. cr = 180~- A = 110~ To find log tan ~ 8 use (40) p=1800~- B= 48~ 50' 'y=180O0-(= 85~010/ log sin ( - a)= 9.3179 2 - =2440 log sin (r - ) = 9.9810 log sin (a — 7)= 9.7778 = 1220. r= 1202. 29.0767 3- a = 12~. log sin = 9.9284 - = 73~10'. 219.1483 -P == 36~50~'. logtan = 9.5742 * These formulas may be written down at once from those used in Case I, (a), p. 217, by simply interchanging the corresponding Greek and Roman letters. OBLIQUE SPHERICAL TRIANGLES 219 To find a use (41) log sin (a - a) = 9.3179 log tan ~ a = 9.5742 log tan 1 a = 9.7437 ~a = 29~..'. a = 58~. To find b use (42) To find c use (43) log sin ( - 3) = 9.9810 log sin (o - y) = 9.7778 log tan 1 8 = 9.5742 log tan 1 8 = 9.5742 log tan 1 b = 0.4068 log tan - c = 0.2036 b = 680 36'. - c = 570 58'. b = 137~ 12'. c = 115~ 56'. Check: log sin a = 9.9284 log sin A = 9.9730 9.9554 log sin b = 9.8321 log sin B = 9.8767 9.9554 log sin c = 9.9539 log sin C = 9.9985 9.9554 EXAMPLES Solve the following oblique spherical triangles: No. GIVEN PARTS REQUIRED PARTS 1 A=75~ B=82~ C=61~ a = 67 52' b=71044' c=57~ 2 A = 120 B = 130~ C = 80~ a = 144~ 10' b = 148049' c = 4144' 3 A = 91 10' B=85~ 40' C=72~ 30' a =89~ 51' b = 85~ 49' c = 72 32' 4 A= 138~ 16' B = 31 11' C = 35~ 53' a = 100~ 5' b = 49~ 59' c = 60~ 6' 5 A = 78~ 40' B=63~ 50' C=46~ 20' a =39~ 30' b = 35~ 36' c = 27~ 59' 6 A=121 B = 102~ C= 68 a =130~ 50' b=120~18' c = 54~ 56' 7 A = 130 B = 110 C = 80~ a=139~ 21' b = 1260 58' c = 560 52' 8 A=28~ B=92~ C= 85~ 26' a=27~ 56' b=85~ 40' c=8402' 9 A = 59018' B = 108~ C =76~ 22' a = 61~44' b = 1034' c = 84032' 10 A=100 B= 100~ C 50 a=1120 14' b=112~ 14' c=4604' 18. Case II. (a) Given two sides and their included angle, as a, b, C. Use formulas on p. 216, namely, sin'(a - b) (46) tan ( - ) =-. ( + b) tan, (47) t an -coso (a + b)t, to find cc and p and therefore A and B; and from p. 215 use (44) solved for tan ~ c, namely, (44) tan c = - 2 sin - (a + i3) tan - (a - b) sin ( - /3) rnes. to find c. Check by the law of sio Ex. 1. Given a = 64~ 24', b = 42~ 30', C = 58~ 40'; find A, B, c. Solution. y = 180~ - C = 1210 20'. -.. = 60~ 40'. a= 64~ 24' b = 42~ 30' a + b = 106~ 54'.. 1 (a + b) = 53027'. a = 64~ 24' b = 420 30' a - b = 21~ 54'.'. (a - b) = 10~ 57'. 220 SPHERICAL TRIGONOMETRY To find i (a - 3) use (46) log sin 2 (a - b) = 9.2786 log tan ~ y = 0.2503 9.5289 log sin i (a + b) = 9.9049 log tan j (a - 3) = 9.6240 (n)... ( (a- p)=-220 49'. To find A and B i (ca +)= 108~ 49' ~ (a —P) =- 22049' Adding, a = 86~ Subtracting, P = 1310 38'..'. A = 1800 - a = 940. B = 180~ - P = 48~ 22'. To find (a + - ) use (47) log cos I (a - b) = 9.9920 log tan - 7 = 0.2503 10.2423 log cos-(a + b) = 9.7749 log tan ~ (a + 3) = 0.4674 (n) 1800- - (a + ) = 71 11'.t... ~ (a + /) = 108~ 49'. To find c use (44) logsin (a + 3) = 9.9761 log tan (a - b) = 9.2867 19.2628 log sin (a -3) = 9.5886 (n) logtan c= 9.6742t c = 250 17'... c = 50~ 34'. Check: log sin a = 9.9551 log sin A = 9.9989 9.9562 log sin b = 9.8297 log sin B = 9.8735 9.9562 log sin c = 9.8878 log sin C = 9.9315 9.9563 If c only is wanted, we may find it from the law of cosines, (14), p. 209, without previously determining A and B. But this formula is not well adapted to logarithmic calculations. Another method is illustrated below, which depends on the soluB^ ~tion of right spherical triangles, and hence requires only those formulas which follow C / \from applying Napier's rules of circular a/ p a parts, p. 200. Through B draw an arc of a great circle per L. L X C pendicular to A C, intersecting A C at D. Let BD = p, CD = m, AD = n. Applying Rule I, p. 200, to the right spherical triangle BCD, we have cos C = tan m cot a. or. (A) tan m = tan a cos C. Applying Rule II, p. 200, to BCD, cOS a = cOS m cos-p, or, (B) cosp = cos a sec m. * Since tan (a -IP) is negative, ~(a -1) may be an angle in the second or fourth quadrant. But a > b, therefore A > B and a < 3, since a and 3 are the supplements of A and B. Hence (ac - 3) must be a negative angle numerically less than 90o. t Here (a +,3) must be a positive angle less than 180~. Since tan (a + 3) is negative, (a + 3) must lie in the second quadrant, and we get its supplement from the table. t tan a c is positive, since sin (a-/3) is negative and there is a minus sign before the fraction. OBLIQUE SPHERICAL TRIANGLES Applying the same rule to ABD, cos c = cos n cosp, or, 221 (C) Cosp = cos c seen. Equating (B) and (C), cos c sec n = cos a sec m, or, cos c = cos a sec m, cos n. But n = b - m; therefore (D) cos c = cos a see m cos ( - m). Now c may be computed from (A) and (D), namely, (48) tan m = tan a cos C, cos a cos (b - m) cos c = COS m (49) Ex. 2. Given a= 98~, b = 80~, C= 110~; find c. Solution. Apply the method just explained. To find b - m use (48) log tan a = 0.8522 (n) log cos C = 9.5341 (n) log tan m = 0.3863 m = 67~ 40'..-. b- m = 12~ 20'. To find c use (49) log cos a = 9.1436 (n) log cos (b - m) = 9.9899 19.1335 log cos m = 9.5798 log cos = 9.5537 (n) 180~ - c = 69~ 2'. c = 110~ 58'. EXAMPLES Solve the following oblique spherical triangles No. GIVEN PARTS REQUIRED PARTS 1 a = 137~ 20' c = 116~ A = 70 B = 1310 17' C = 94~ 48' a = 57~ 57' 2 a=72 b = 47 C = 33 A =121~ 33' B= 40~ 57' c = 37~ 26' 3 a=980- c=60~ B=110 A =87 C=60051' b =111~17' 4 b = 120~ 20' c = 70~ 40' A = 50 B = 134 57' C = 50~ 41' a = 69~ 9' 5 a = 125~ 10' b = 153~ 50' C = 140~ 20' A =147~ 29' B = 163 9' c=7608' 6 a = 93~ 20' b = 56~ 30' C = 74~ 40' A = 101~ 24' B = 54~ 58' c =79~ 10' 7 b = 76 30' c = 47~ 20' A = 92~ 30' B = 78~ 21' C = 47~ 47' a = 82~ 42' 8 c = 40~ 20' a = 100~ 30' B = 46~ 40' A = 131~ 29' C = 29~ 33' b = 72~ 40' 9 b =76~ 36' c =110~ 26' A = 46~ 50' B =57~ 43' C = 125~ 28' a = 57~ 13' 10 a =84~ 23' b = 124~ 48' C = 62 A = 68~ 27' B = 129~ 51' c =70~ 52' 222 SPHERICAL TRIGONOMETRY 19. Case II. (b) Given two angles and their included side, as A, B, c. Use formulas * on pp. 215, 216, namely, (44) tan (a - b)=- si2 2 (~- ) tan I C 2.^,sin I (a 2 /) (45) tan I ( +)- oo ( )tan c to find a and b; and frormn p. 216, use (46) solved for tan ~ y, namely, sin I (a + b) tan 1 (a -, (46) tan y = 2 - 2 v / 2 ' sin 1 (a - b) to find y and therefore C. Check by the law of sines. Ex. 1. Given c = 116~, A = 70~, B= 131~ 20'; find a, b, C. Solution. a = 1800 - A = 110~, and P = 1800 - B = 480 40'. a = 110~ /p= 48 40' ct + 3 = 1580 40'... ~( + f) = 790 20'. To find (a- b) use (44) log sin - (ac- /) = 9.7076 log taln - c = 0.2042 9.9118 log sin 1 (a + /) = 9.9924 log tan - (a - b) = 9.9194 (n) (a - b) = - 390 43'.t To find a and b - (a + b) = 970 39' - (a - b) = - 390 43' Adding, a = 570 56' Subtracting, b = 137~ 22'. Check: log sin a = 9.9281 cl log sinA = 9.9730 lo, 9.9551 a= 1100 / = 48~40' a-/3= 61~20'.'. (a-/3) = 300 40'. c = 116~.. c = c 580. To find I (a + b) use (45) log cos (a - ) = 9.9346 log tan I c = 0.2042 10.1388 log cos - (a + /) = 9.2674 log tan ~ (a + b) = 0.8714 (n) 1800 - -i(a + b) = 82 21'..-. (a + b)=97039'. To find C use (46) log sin I (a + b) = 9.9961 log tan (a - 3) = 9.7730 19.7691 log sin (a - b) = 9.8055 (n) log tan - = 9.9636 -7 = 42~ 36'. y = 85 12'..C. G = 180~- y = 94~ 48'. )g sin b = 9.8308 g sin B = 9.8756 9.9552 log sin c = 9.9537 log sin C = 9.9985 9.9552 * Same as those used in Case II, (a), p. 219, with Greek and Roman letters interchanged. t Since A < B it follows that a < b, and (a- b) is negative. OBLIQUE SPHERICAL TRIANGLES 223 If C only is wanted, we can calculate it without previously determining a and b, by dividing the given triangle into two right spherical triangles, as was illustrated on p. 220. Through B draw an arc of a great x circle perpendicular to AC, intersecting y\ AC at D. Let BD = p, angle ABD = x, angle CBD= y. Applying Rule I of Napier's rules, p. 200, to the right spher- A 0 ical triangle ABD, we have b cos c = cot x cot A, or, cot x = tan A cos c. (A) Applying Rule II, p. 200, to ABD, we have cos A = cosp sin x, or, (B) cosp = cos A csc x. Applying the same rule to CBD, cos C = cosp sin y, or, cosp = cos C csc y. (C) Equating (B) and (C), cos C csc y = cos A csc x, or, cos C = cos A csc x sin y. But y == B- x; therefore (D) cos C = cos A csc x sin (B - x). Now C may be computed from (A) and (D), namely, (50) cot x = tan A cos c. (51) cos A sin (B - x) cos C = sin x Ex. 2. Given A = 35~ 46', B = 115~ 9', c = 51~ 2'; find C. Solution. Apply the method just explained. Tofind B - x use (50) To find C use (51) log tanA = 9.8575 log cosA= 9.9093 log cos c = 9.7986 log sin(B - x) = 9.8811 log cotx = 9.6561 19.7904 x = 650 38'. log sin x = 9.9595.. B - x = 49~ 31'. logcos C= 9.8309 C= 47~ 21'. 224 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles: No. GIVEN PARTS REQUIRED PARTS 1 A=67~30' B=45~50' c=-74~20' a=63~15' b=43~54' C=95~1' 2 B=98~ 30' C = 67~ 20' a= 60~ 40' b = 86~ 40' c = 68~ 40' A =59~ 44' 3 C =110 A=94~ b = 44~ a =114~ 10' c =120~ 46' B=130 34' 4 =70~ 20' B =43~ 50' a =50~ 46' b=32~ 59' c=47045' A=80~ 14' 5 A=78~ B=41~ c=108~ a=95~ 38' b=41~52' C=110~ 49' 6 B =135 C = 50~ a =70~ 20' b = 120~ 17' c = 69~ 20' A =50~ 26' 7 A=31~ 40' C=122~ 20' b=40~40' a =34~ 3' c =64~19' B=37~ 40' 8 A1080 12' B = 145~ 46' c= 126~ 32' a =69 4' b= 146~ 26' C = 125~ 12' 9 A = 130 36' B = 30~ 26' c = 4035' a=71~15' b =39~ 10' C=31~26' 10 A=51~ 58' B =83~ 54' c = 42~ a =38~ b =51 C = 58~ 53' 20. Case III. (a) Given two sides and the angle opposite one of them, as a, b, B (ambiguous case *). From the law of sines, p. 207, we get sin a sin B (11) sinA = ---— sin sin b which gives At. To find C we ztse, from p. 216, formula (46), solved for tan i y, namely, sin (a + b) tan 1L(a- l) (46) tan y 2 n 2 -sin 1 (a - b) To find c, solve (44), p. 215, for tan 2 c, namely, sin 1 (a+ ) tan 1 (a -b) (44) tan I c s in ~ 2~~~~~~~~~__,_ Check by the law of sines. Ex. 1. Given a = 58~, b = 137~ 20', Solution. To find A use (11) log sin a = 9.9284 log sin B= 9.8756 19.8040 log sin b = 9.8311 logsinA = 9.9729.. A = 69~ 58', or, A2 = 180~ - A1 = 110~ 2'. sin = a - p) B = 131~ 20'; find A, C, c. a=58~ a=58~ b=137~ 20' b=137~ 20' a + b =195~ 20' a-b= -79~ 20' -1(a- b)=97~40'. (a-b)-=-39~40'. p = 180~ - B = 48~ 40'. Since a<b and both A1 and A2 are < B, it follows that we have two solutions. * Just as in the corresponding case in the solution of plane oblique triangles (Granville's Plane Trigonometry, pp. 105, 161), there may be two solutions, one solution, or no solution, depending on the given data. t Since the angle A is here determined from its sine, it is necessary to consider both of the values found. If a > b then A > B; and if a < b then A < B. Hence [next page] OBLIQUE SPHERICAL TRIANGLES 225 First solution. al = 180~ - A1 = 110~ 2'. cr = 110~ 2' /,= 48 40' a1 + p = 1580 42' (al + ) = 790 21'. Tofind C1 use (46) log siln - (a + b) = 9.9961 log sil log tan - (ac -1P) = 9.7733 log t 19.7694 log sin - (a - b) = 9.8050 (n) log si] log tan l71 = 9.9644 ^ 1 = 420 39'. 71 = 85~ 18'.. C = 180~ -y 7 = 94~ 42'. Check: log sin a = 9.9284 log sin b = 9.8311 log sinA1 = 9.9729. log sinB = 9.8756 9.9555 9.9555 a1 = 110~ 2' / = 48~ 40' al- = 61~ 22' (a - ) = 30~ 41'. To find cl use (44) n-(a1 +/p)= 9.9924 an (a - b) = 9.9187 (n) 19.9111 an1 (i - p)= 9.7078 log tan 1 Cl = 10.2033 cl = 570 57'... cl = 115~ 54'. log sin cl = 9.9541 log sin C1 = 9.9985 9.9556 Second solution. a2 = 180~ - A2 = 69~ 58'. a2 = 69~ 58' / = 48040' a2 + /p = 118~ 38' (a2 + p) = 59~ 19'. To find C2 use (46) log sin 1 (a + b) = 9.9961 log si log tan - (a2 - /) = 9.2743 log t 19.2704 log sin - (a - b) = 9.8050 (n) log si: log tan y2 = 9.4654 2 72 = 16~ 17'. 72 = 32~ 34'... C2 = 180 - 72 = 147~ 26'. Check: log sin a = 9.9284 log sin b = 9.8311 log sinA2 = 9.9729 log sinB = 9.8756 9.9555 9.9555 a2 = 69~ 58' p = 48~ 40' a2 - / = 21~ 18' (a2 - ) = 10~ 39'. Tofind cl use (44) I(a2 +/) = 9.9345 an -(a - b)= 9.9187 (n) 19.8532 n- (c2 - /) = 9.2667 log tan - c2 = 10.5865 I c2 = 75~ 28'..C. c2 = 150~ 56'. log sin C2 = 9.6865 log sin C2 = 9.7310 9.9555 If the side c or the angle C is wanted without first calculating the value of A, we may resolve the given triangle into two right triangles and then apply Napier's rules, as was illustrated under Cases II, (a), and II, (b), pp. 220, 223. Theorem. Only those values of A should be retained which are greater or less than B according as a is greater or less than b. If log sin A = a positive number, there will be no solution. 226 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles: No. GIVEN PARTS REQUIRED PARTS 1 a=43 20' b=48030' A=58040' B1= 68 47' C1=70040' cl=49018' B2=111 13' C2 = 14 29' c2 =11036' 2 a=56040' b=30050' A=103040' B=36036' C= 520 c=42039' 3 a=30 20' b=46~30' A=36040' B1= 59 4' C = 97 39' cl 56 57' B2 = 120~ 57' C2 =28~ 5' c = 230 28' 4 b=990 40' c=64020' B=95~ 40' C=65~ 30' A=97~ 20' a=100 45' 5 a=40~ b=1180 20' A=29 40' B1=42 40' C1]=159054' c1=153~30' B2=137~ 20' C2=50~ 21' c2-90~ 10' 6 a=1150 20' c=1460 20' C= 141~ 10' Impossible 7 a= 109 20' c=82~ A=107040' C=90~ B=113037' b=114052' 8 b=108 30' c=40050' C=39~ 50' B1=68018' A1=132034' a1=131016' B2 =111~ 42' A2=77~ 5' a2=95~ 50' 9 a=162020' b=15040' B=125~ Impossible 10 a =55 c = 138~ 10' A =42~ 30' C=146~38' B=551' b = 96034' 21. Case III. (b) Given two angles and the side opposite one of them, as A, B, b (ambiguous case *). From the law of sines, p. 207, we get sin A sin b (11) sin a = sin B which gives a.t To find c we use, from p. 215, the formula (44), solved for tan - c, namely, sin 2 (a + 8) tan 1 (a - b) (44) tan c -- — 2 2sin (a — f To find C, solve (46), p. 216, for tan y, namely, sin4 (a + b)tan' (a- ) (46) tan y = -- -2 2 - - 2 / 2' sin - (a - b) Check by the law of sines. * Just as in Case II, (b), we may have two solutions, one solution, or no solution, depending on the given data. f Since the side is here determined from its sine, it is necessary to examine both of the values found. If A > B then a > b; and if A < B then a < b. Hence we have the Theorem. Only those values of a should be retained which are greater or less than b according as A is greater or less than B. If log sine a = a positive number, there will be no solution. $ Same as those used in Case III, (a), p. 224, when the Greek and Roman letters are interchanged. OBLIQUE SPHERICAL TRIANGLES 227 Ex. 1. Given A = 110~, B = 131~ 20', b = 137~ 20'; find a, b, C. Solution. a = 180~ -A = 70~, and P = 180~-B = 48~ 40'. To find a use (11) log sinA = 9.9730 logsinb = 9.8311 19.8041 log sinB= 9.8756 log sin a = 9.9285.. al = 58~ 1', or, a2 = 180~ - ai = 121~ 59'. First solution. al= 58~ 1' b = 137~ 20' ai + b = 1950 21' ~ (al + b) = 97~ 41'. To find cl use (44) log sin - (a + 38) = 9.9346 log tan ~ (al - b) = 9.9187 (n) 19.8533 log sin (a - P) = 9.2674 log tan cl = 10. 5859 cl - 750 27'... cl = 150~ 54'. a= 70~ r = 70~ = 48~ 40' p = 48~ 40' a = 118~ 40' a - = 21~ 20' 1 (C + ) = 59~ 20'. ~ (a —/) = 10~ 40'. Since A <B and both al and a2 are < b, it follows that we have two solutions. a = 58~ 1' b = 137~ 20' al - b =- 790 19' a (a - b) - 390 40'. To find C1 use (46) log sin - (ai + b) = 9.9961 log tan 2 (a - /) = 9.2750 19.2711 log sin 1 (al - b) = 9.8050 (n) log tan y71 = 9.4661 - - = 16~ 18'. 71 = 32~ 36'... C1 = 180~- y = 147~ 24'. Check: log sin al = 9.9285 log sin b = 9.8311 log sin cl = 9.6869 log sin A = 9.9730 log sinB = 9.8756 log sin C = 9.7314 9.9555 9.9555 9.9555 Second solution. This gives c2 = 64~ 8', and C2 = 85~ 18'. Remembering that a2 = 121~ 59', we may now check the second solution. Check: log sin a2 = 9.9285 log sin b = 9.8311 log sin c2 =9.9542 log sin A = 9.9730 log sin B = 9.8756 log sin C2 = 9.9985 9.9555 9.9555 9.9557 Hence the two solutions are a1 = 58~ 1' c1 = 150~ 54' C = 147~ 23', and a2= 121 59' c2 = 64 8' C2 = 85~ 18'. If the angle C or the side c is wanted without first computing a, we may resolve the given triangle into two right triangles and then apply Napier's rules, as was illustrated under Cases II, (a), and II, (b), pp. 220, 223. 228 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles: No. GIVEN PARTS REQUIRED PARTS 1 A =108~ 40' B-=134~ 20' a = 145~ 36' b = 154~ 45' c = 34 9' C = 70 18 2 B=116~ C=80~ c=84~ b=114 50' A=79 20' a = 8256' 3 A=132~ B=140~ b=127~ a1=67024' C1=16406' c1=160 6' a2=112~ 36' C= 128 21' c2= 1030 2' 4 A=62 C= 102~ a=640 30' c=90~ B=63043' b=66026' 5 A=1330 50' B=66~ 30' a=81010' Impossible 6 B=22020' C=146040' c=138020' b=27022' A=47~21' a-11709' 7 A=61040' C=140020' c=150020' a1=4303' B1=89024' b1=12908' a2= 136~ 57' B2=26~ 59' b2 = 20~ 36' 8 B=73~ C=81~ 20' b= 122~ 40' Impossible 9 B=36020' C=46~30' b=42012' A1=164044' al=162038' cl=124041' A2=119~19' a2=81~19' c2=55019' 10 A=110~ 10' B=133~ 18' a=147~ 6' b= 1555' c=330 2' C=70021' 22. Length of an arc of a circle in linear units. From Geometry we know that the length of an arc of a circle is to the circumference of the circle as the number of degrees in the arc is to 360. That is L: 27rR::N: 360, or, L 7TRN (52) L=, 180 where L = length of arc, N = number of degrees in are, R = length of radius. In case the length of the arc is given to find the number of degrees in it, we instead solve for N, giving (53) 180 L N=7 7TR Considering the earth as a sphere, an arc of one minute on a great circle is called a geographical mile or a nautical mile.* Hence there are 60 nautical miles in an arc of 1 degree, and 360 x 60 = 21,600 nautical miles in the circumference of a great circle of the earth. If we assume the radius of the earth to be 3960 statute miles, the length * In connection with a ship's rate of sailing a nautical mile is also called a knot. OBLIQUE SPHERICAL TRIANGLES 229 of a nautical mile (= 1 min. = 6 of a degree) in statute miles will be, from (52), 3.1416 X 3960 X.5mi. L = 1.15 mi. 180 Ex. 1. Find the length of an arc of 22~ 30' in a circle of radius 4 in. Solution. Here N = 22~ 30' = 22.5~, and R = 4 in. 3.1416 x 4 x 22.5 Substituting in (52), L = =x 4 x 1.67 in. Ans. 180 Ex. 2. A ship has sailed on a great circle for 51 hr. at the rate of 12 statute miles an hour. How many degrees are there in the arc passed over? Solution. Here L = 56 x 12 = 66 mi., and R = 3960 mi. Substituting in (53), N = 180 =.955~= 57.3. Ans. 3.1416 x 3960 23. Area of a spherical triangle. From Spherical Geometry we know that the area of a spherical triangle is to the area of the surface of the sphere as the number of degrees in its spherical excess * is to 720. That is, Area of triangle: 4 7rR2:: E:720, or, 7rR2E (54) Area of spherical triangle = 0 180 In case the three angles of the triangle are not given, we should first find them by solving the triangle. Or, if the three sides of the triangle are given, we may find E directly by Lhuilier's formula,t namely, (55) tan E = Vtan 2 s tan 2 (s - a) tan 2 (s - b) tan 2 (s - c), where a, b, e denote the sides and s = ~ (a + b + c). The area of a spherical polygon will evidently be the sum of the areas of the spherical triangles formed by drawing arcs of great circles as diagonals of the polygon. Ex. 1. The angles of a spherical triangle on a sphere of 25-in. radius are A = 74~ 40', B = 67~ 30', C = 49~ 50'. Find the area of the triangle. Solution. Here E = (A + B + C) - 180~ = 12~. Substituting in (54), Area = 3.1416 x (25)2 x 12 = 130.9 sq. in. Ans. 180 The spherical excess (usually denoted by E) of a spherical triangle is the excess of the sum of the angles of the triangle over 180~. Thus, if A, B, and C are the angles of a spherical triangle, E=A+B+ C-1800. t Derived in more advanced treatises. 230 SPHERICAL TRIGONOMETRY EXAMPLES 1. Find the length of an arc of 5~ 12' in a circle whose radius is 2 ft. 6 in. Ans. 2.72 in. 2. Find the length of an arc of 75~ 30' in a circle whose radius is 10 yd. Ans. 13.17 yd. 3. How many degrees are there in a circular arc 15 in. long, if the radius is 6 in.? Ans. 143~ 18'. 4. A ship sailed over an arc of 4 degrees on a great circle of the earth each day. At what rate was the ship sailing? Ans. 11.515 mi. per hour. 5. Find the perimeter in inches of a spherical triangle of sides 48~, 126~, 80~, on a sphere of radius 25 in. Ans. 110.78 in. 6. The course of the boats in a yacht race was in the form of a triangle having sides of length 24 mi., 20 mi., 18 mi. If we assume that they sailed on arcs of great circles, how many minutes of arc did they describe? Ans. 53.85 min. 7. The angles of a spherical triangle are A = 63~, B = 84~ 21', C = 79~; the radius of the sphere is 10 in. What is the area of the triangle? Ans. 80.88 sq. in. 8. The sides of a spherical triangle are a = 6.47 in., b = 8.39 in., c 9.43 in.; the radius of the sphere is 25 in. What is the area of the triangle? Ans. 26.9 sq. in. Hint. Find E by using formula (55). 9. In a spherical triangle A = 75~ 16', B = 39~ 20', c = 26 ft.; the radius of the sphere is 14 ft. Find the area of the triangle. Ans. 158.45 sq. ft. 10. Two ships leave Boston at the same time. One sails east 441 mi. and the other 287 mi. E. 38~ 21' N. the first day. If we assume that each ship sailed on an arc of a great circle, what is the area of the spherical triangle whose vertices are at Boston and at the positions of the ships at the end of the day? Ans. 41,040 sq. mi. 11. A steamboat traveling at the rate of 15 knots per hour skirts the entire shore line of an island having the approximate shape of an equilateral triangle in 18 hr. What is the approximate area of the island? Ans. 34,960 sq. mi. 12. Find the areas of the following spherical triangles, having given (a) a = 470 30', b = 55~ 40', c =60~ 10'; R = 10 ft. Ans. 42.96 sq. ft. (b) a = 430 30', b = 72~ 24', c = 87~ 50'; R = 10 in. 59.19 sq. in. (c) A = 74~ 40', B = 67~ 30', C = 490 50'; R = 100 yd. 2094 sq. yd. (d) A = 112~0 30', B = 83~ 40', C = 70~ 10'; R = 25 cm. 941.2 sq. cm. (e) a = 640 20', b = 42~ 30', C = 50~ 40'; R = 12 ft. 46.73 sq. ft. (f) C =110, A = 94, b = 44; R = 40 rd. 709.2 sq. rd. (g) a = 430 20', b = 48~ 30', A = 58~ 40'; R = 100 rd. 24.88 acres. (h) A = 108~ 40', B = 134~ 20', a = 145~ 36'; R = 3960 mi. 36,460,000 sq. mi. CHAPTER III APPLICATIONS OF SPHERICAL TRIGONOMETRY TO THE CELESTIAL AND TERRESTRIAL SPHERES 24. Geographical terms. In what follows we shall assume the earth to be a sphere of radius 3960 statute miles. The meridian of a place on the earth is that great circle of the earth which passes through the place and the north and south poles. (_Norith pole) N Thus, in the fig rg (Greeathc) te m n Greenwich BS is the meridian of Boston and CS is the med(West)\ X i / I(East)) \ -Equator 1 S, \ X5 T / \\\> \ / ~(Cape Town) S (Southi pole) Thus, in the figure representing the earth, NGS is. the meridian of Greenwich, NTBS is the meridian of Boston, and NCS is the Ineridian of Cape Town. The latitude of a place is the are of the meridian of the place extending from the equator to the place. Latitude is measured north or south of the equator from 0~ to 90~. Thus, in the figure, the arc LB measures the north latitude of Boston, and the arc TC measures the south latitude of Cape Town. The longitude of a place is the arc of the equator extending from the zero meridian * to the meridian of the place. Longitude is * As in this case, the zero meridian, or reference meridian, is usually the meridian passing through Greenwich, near London. The meridians of Washington and Paris are also used as reference meridians. 231 232 SPHERICAL TRIGONOMETRY measured east or west from the~Greenwich meridian from 0~ to 180~. Thus, in the figure, the arc MT measures the east longitude of Cape Town, while the arc ML measures the west longitude of Boston. Since the arcs MT and ML are the measures of the angles MNT and MNL respectively, it is evident that we can also define the longitude of a place as the angle between the reference meridian and the meridian of the place. Thus, in the figure, the angle MATT is the east longitude of Cape Town, while the angle MNL is the west longitude of Boston. The bearing of one place from a second place is the angle between the arc of a great circle drawn from the second place to the first place, and the meridian of the second place. Thus, in the figure, the bearing of Cape Town from Boston is measured by the angle CBN or the angle CBL, while the bearing of Boston from Cape Town is measured by the angle NCB or the angle SCB.* 25. Distances between points on the surface of the earth. Since we know from Geometry that the shortest distance on the surface of a (North pole) N (W es- 1 l( (East) E^1 W ar-fo7~~T^ '^ "T^' / (Cap Town) S/ (South pole) sphere between any two points on that surface is the arc, not greater than a semicircumference, of the great circle that joins them, it is evident that the shortest distance between two places on the earth is measured in the same way. Thus, in the figure, the shortest * The bearing or course of a ship at any point is the angle the path of the ship makes with the meridian at that point. APPLICATIONS OF SPHERICAL TRIGONOMETRY 233 distance between Boston and Cape Town is measured on the arc BC of a great circle. We observe that this arc BC is one side of a spherical triangle of which the two other sides are the arcs BN and CN. Since arc BN = 9~- - arc LB = 90~ - north latitude of Boston, arc CN = 90~ + arc TC = 90~ + south latitude of Cape Town, and angle BNC = angle MNL + angle MNT = west longitude of Boston + east longitude of Cape Town = difference in longitude of Boston and Cape Town, it is evident that if we know the latitudes and longitudes of Boston and Cape Town, we have all the data necessary for determining two sides and the included angle of the triangle BNC. The third side BC, which is the shortest distance between Boston and Cape Town, may then be found as in Case II, (a), p. 219. In what follows, north latitude will be given the sign + and south latitude the sign -. Rule for finding the shortest distance between two points on the earth and the bearing of each from the other, the latitude and longitude of each point being given. First step. Subtract the latitude of each place from 90~.* The results will be the two sides of a spherical triangle. Second step. Find the difference of longitude of the two places by subtracting the lesser longitude from the greater if both are E. or both are TF., but add the two if one is E. and the other is Ti. This gives the included angle of the triangle.t Third step. Solving the triangle by Case II, (a), p. 219, the third side gives the shortest distance between the two points in degrees of arc,? and the angles give the bearings. * Note that this is algebraic subtraction. Thus, if the two latitudes were 250 N. and 42~ S., we would get as the two sides of the triangle, 90 - 25~ = 650 and 90~ -(- 42)= 90~ + 42~= 132~. t If the difference of longitude found is greater than 180~, we should subtract it from 360~ and use the remainder as the included angle. $ The number of minutes in this arc will be the distance between the two places in geographical (nautical) miles. The distance between the two places in statute miles is given by the formula 3.1416 x 3960 x N L= 180 where -N= the number of degrees in the arc. 234 SPHERICAL TRIGONOMETRY Ex. 1. Find the shortest distance along the earth's surface between Boston (lat. 42~ 21' N., long. 71~ 4' W.) and Cape Town (lat. 33~ 56' S., long. 180 26' E.), NV and the bearing of each city from the other. Solution. Draw a spherical triangle in agreement with o 859o0 the figure on p. 232. c)/ \\\ i First step. B.) COO c = 900 - 42~ 21'= 470 39', b = 90~- (- 33~ 56') = 123~ 56'. Second step. C N = 71~ 4' + 18~ 26' = 89~ 30' = difference in long. Third step. Solving the triangle by Case II, (a), p. 219, we get n = 68~ 14' = 68.23~ = 4094 nautical miles, C = 52~ 43' = bearing of Boston from Cape Town, and B = 116~ 43' = bearing of Cape Town from Boston. Hence a ship sailing from Boston to Cape Town on the arc of a great circle sets out from Boston on a course S. 63~ 17' E. and approaches Cape Town on a course S. 52~ 43' E.* EXAMPLES 1. Find the shortest distance between Baltimore (lat. 39~ 17' N., long. 76~ 37' W.) and Cape Town (lat. 33~ 56' S., long. 18026' E.), and the bearing of each from the other. Ans. Distance = 65~ 48' = 3947 nautical miles, S. 64~ 58' E. = bearing of Cape Town from Boston, N. 57~ 42'W. = bearing of Boston from Cape Town. 2. What is the distance from New York (lat. 400 43' N., long. 74~ W.) to Liverpool (lat. 530 24' N., 3~ 4' W.)? Find the bearing of each place from the other. In what latitude will a steamer sailing on a great circle from New York to Liverpool cross the meridian of 50~ WV., and what will be her course at that point? Ans. Distance = 47~ 50' = 2870 nautical miles, N. 75~ 7' W. = bearing of New York from Liverpool, N. 490 29' E. = bearing of Liverpool from New York. Lat. 51~ 13' N., with course N. 66~ 54' E. 3. Find the shortest distance between the following places in geographical miles: (a) New York (lat. 40~ 43' N., long. 74~ W.) and San Francisco (lat. 37~ 48'N., long. 1220 28' W.). Ans. 2230. (b) Sandy Hook (lat. 40~ 28' N., long. 74~ 1' W.) and Madeira (lat. 32~ 28' N., long. 160 55' W.). Ans. 2749. (c) San Francisco (lat. 37~ 48' N., long. 122~ 28' W.) and Batavia (lat. 6~ 9' S., long. 106~ 53' E.). Ans. 7516. (d) San Francisco (lat. 37~ 48' N., long. 122~ 28' W.) and Valparaiso (lat. 33~ 2' S., long. 71~41' W.) Ans. 5109. * A ship that sails on a great circle (except on the equator or a meridian) must be continually changing her course. If the ship in the above example keeps constantly on the course S. 630 17' E., she will never reach Cape Town. APPLICATIONS OF SPHERICAL TRIGONOMETRY 235 4. Find the shortest distance in statute miles (taking diameter of earth as 7912 mi.) between Boston (lat. 42~ 21' N., long. 71~ 4' W.) and Greenwich (lat. 51~ 29' N.), and the bearing of each place from the other. Ans. Distance = 3276 mi., N. 53~ 7' E. = bearing of Greenwich from Boston, N. 710 39' W. = bearing of Boston from Greenwich. 5. As in last example, find the shortest distance between and bearings for Calcutta (lat. 22~ 33' N., long. 88~ 19' E.) and Valparaiso (lat. 33~ 2' S., long. 71~ 42' W.). Ans. Distance = 10,860 mi., S. 640 20.5' E. = bearing of Calcutta from Valparaiso, S. 54~ 54.5' W. = bearing of Valparaiso from Calcutta. 6. Find the shortest distance in statute miles from Oberlin (long. 82~ 14' W.) to New Haven (long. 72~ 55' W.), the latitude of each place being 41~ 17' N. Ans. 483.2 mi. 7. From a point whose latitude is 17~ N. and longitude 130~ W. a ship sailed an arc of a great circle over a distance of 4150 statute miles, starting S. 54~ 20' W. Find its latitude and longitude, if the length of 1~ is 691 statute miles. Ans. Lat. 19~ 42' S., long. 178~ 21' W. 26. Astronomical problems. One of the most important.applications of Spherical Trigonometry is to Astronomy. In:fact, Trigonometry was first developed by astronomers, and for centuries was studied only in connection with Astronomy. We shall take up the study of a few simple problems in Astronomy. 27. The celestial sphere. When there are no clouds to obstruct the view, the sky appears like a great hemispherical vault, with the observer at the center. The stars seem to glide upon the inner surface of this sphere from east to west,* their paths being parallel circles whose planes are perpendicular to the polar axis of the earth, and having their centers in that axis produced. Each star t makes a complete revolution, called its diurnal (daily) motion, in 23 hr. 56 min., ordinary clock time. We cannot estimate the distance of the surface of this sphere from us, further than to perceive that it must be very far away indeed, because it lies beyond even the remotest terrestrial objects. To an observer the stars all seem to be at the same enormous distance from him, since his eyes can judge their directions only and not their distances. It is therefore natural, and it is extremely convenient from a mathematical point of view, to regard this imaginary sphere on which all the heavenly bodies seem to be projected, as having a radius of unlimited length. This * This apparent turning of the sky from east to west is in reality due to the rotation of the earth in the opposite direction, just as to a person on a swiftly moving train the objects outside seem to be speeding by, while the train appears to be at rest. The sky is really motionless, while the earth is rotating from west to east. t By stars we shall mean fixed stars and nebulse whose relative positions vary so slightly that it takes centuries to make the change perceptible. 236 SPHERICAL TRIGONOMETRY sphere, called the celestial sphere, is conceived of as having such enormous proportions that the whole solar system (sun, earth, and planets) lies at its center, like a few particles of dust at the center of a great spherical balloon. The stars seem to retain the same relative positions with respect to each other, being in this respect like places on the earth's surface. As viewed from the earth, the sun, moon, planets, and comets are also projected on the celestial sphere, but they are changing their apparent positions with respect to the stars and with respect to each other. Thus, the sun appears to move eastward with respect to the stars about one degree each day, while the moon moves about thirteen times as far. The following figure represents the celestial sphere, with the earth at the center showing as a mere dot. (Zenith) z ( orth '..stA ( celestial/ v / t elest 'pole) /A. East) \ (Th North aof an observr is rthe point on te (South d irectly overhead. A plumb line held by the observer and extended Of /I of horizon) horizo h zn) \ I / /A~" / P'(South and from the nadir. A plane celestial pole) ZI (Nadir) The zenith of an observer is the point on the celestial sphere directly overhead. A plumb line held by the observer and extended upwards will pierce the celestial sphere at his zenith (Z in figure). The nadir is the point on the celestial sphere which is diametrically opposite to the zenith (Z' in the figure). The horizon of an observer is the great circle on the celestial sphere having the observer's zenith for a pole; hence every point on the horizon (SWNE in the figure) will be 90~ from the zenith and from the nadir. A plane tangent* to a surface of still water On account of the great distance, a plane passed tangent to the earth at the place of the observer will cut the celestial sphere in a great circle which (as far as we are concerned) coincides with the observer's horizon. APPLICATIONS OF SPHERICAL TRIGONOMETRY 237 at the place of the observer will cut the celestial sphere in his horizon. All points on the earth's surface have different zeniths and horizons. Every great circle passing through the zenith will be perpendicular to the horizon; such circles are called vertical circles (as ZMHZ' and ZQSP'Z' in figure). The celestial equator or equinoctial is the great circle in which the plane of the earth's equator cuts the celestial sphere (EQWQ' in the figure). The poles of the celestial equator are the points (P and P' in the figure) where the earth's axis, if produced, would pierce the celestial sphere. The poles may also be defined as those two points on the sky where a star would have no diurnal (daily) motion. The Pole Star is near the north celestial pole, being about 14~ from it. Every point on the celestial equator is 90~ from each of the celestial poles. All points on the earth's surface have the same celestial equator and poles. The geographical meridian of a place on the earth was defined as that great circle of the earth which passes through the place and the north and south poles. The celestial meridian of a point on the earth's surface is the great circle in which the plane of the point's geographical meridian cuts the celestial sphere (ZQSP'Z'tQtP in the figure). It is evidently that vertical circle of an observer which passes through the north and south points of his horizon. All points on the surface of the earth which do not lie on the same north-and-south line have different celestial meridians. The hour circle of a heavenly body is that great circle of the celestial sphere which passes through the body * and through the north and south celestial poles. In the figure PMDP' is the hour circle of the star M. The hour circles of all the heavenly bodies are continually changing with respect to any observer. The spherical triangle PZM, having the north pole, the zenith, and a heavenly body at its three vertices, is a very important triangle in Astronomy. It is called the astronomical triangle. 28. Spherical coordinates. When learning how to draw (or plot) the graph of a function, the student has been taught how to locate a point in a plane by measuring its distances from two fixed and mutually perpendicular lines called the axes of coordinates, the two distances being called the rectangular coordinates of the point. * By this is meant that the hour circle passes through that point on the celestial sphere where we see the heavenly body projected. 238 SPHERICAL TRIGONOMETRY If we now consider the surface to be spherical instead of plane, a similar system of locating points on it may be employed, two fixed and mutually perpendicular great circles being chosen as reference circles, and the angular distances of a point from these reference circles being used as the spherical coordinates of the point. Since the reference circles are perpendicular to each other, each one of them passes through the poles of the other. In his study of Geography the student has already employed such a system for locating points on the earth's surface, for the latitude and longitude of a point on the earth are really the spherical coordinates of the point, the two reference circles being the equator and the zero meridian (usually the meridian of Greenwich). Thus, in the figure on p. 231, we may consider the spherical coordinates of Boston to be the arcs ML (west longitude) and LB (north latitude); and of Cape Town the spherical coordinates would be the arcs MT (east longitude) and TC (south latitude). Similarly, we have systems of spherical coordinates for determining the position of a point on the celestial sphere, and (Zenith) z we shall now take up the study of the more N \ (Noon) important of these. (North p / \ \ 29. The horizon and ~pole)o '^ |! / \ / \ meridian system. In KNOB go E! ~ M /p y \ this case the two fixed M o --- — (t S —u7 \ and mutually perpent' '_ _ - ^ -Earth A - S dicular great circles of.Y / ^ __ \ /_] - reference are the horiHor'izon.W F H z on of the observer (Sunset) (Suns) (SHWNE) and his meridian (SM12ZPN), and the spherical coordinates of a heavenly body are its altitude and azimuth. The altitude of a heavenly body is its angular distance above the horizon measured on a vertical circle from 0~ to 90~.* Thus the altitude of the sun Ml is the arc HM. The distance of a heavenly body from the zenith is called its zenith distance (ZM in the figure), and it is evidently the complement of its altitude. The altitude of the zenith is 90~. The altitude of the sun at sunrise or sunset is zero. The azimuth of a heavenly body is the angle between its vertical circle and the meridian of the observer. This angle is usually * At sea the altitude is usually measured by the sextant, while on land a surveyor's transit is used. APPLICATIONS OF SPHERICAL TRIGONOMETRY 239 measured along the horizon from the south point westward to the foot of the body's vertical circle.* Thus the azimuth of the sun M is the angle SZH, which is measured by the arc SI1. The azimuth of the sun at noon is zero and at midnight 180~. The azimuth of a star directly west of an observer is 90~, of one north 180~, and of one east 270~. Knowing the azimuth and altitude (spherical coordinates) of a heavenly body, we can locate it on the celestial sphere as follows. From the south point of the horizon, as S (which may be considered the origin of coordinates, since it is an intersection of the reference circles), lay off the azimuth, as SH. Then on the vertical circle passing through H lay off the altitude, as HM. The body is then located at M. Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Azimuth Altitude Azimuth Altitude (a) 45~ 450 (j) 0~ 0~ (b) 60~ 300 (k) 180~ 0~ (c) 90~ 600 (1) 0~ 90 (d) 120~ 750 (m) 900 0~ (e) 180~ 55~ (n) 2700 00 (f) 225~ 00 (o) 360~ 00 (g) 300~ 60~ (p) 330~ 45~ (h) 315~ 15~ (q) 75~ 75~ (i) 178~ 82~ (r) 900 900 Since any two places on the earth have, in general, different meridians and different horizons, it is evident that this system of spherical coordinates is purely local. The sun rises at M1 on the eastern horizon (altitude zero), mounts higher and higher in the sky, on a circle (M1M2M3) parallel to the celestial equator, until it reaches the observer's meridian Mi (at noon, when its altitude is a maximum), then sinks downward to M3 and sets on the western horizon. Similarly, for any other heavenly body, so that all are continually changing their altitudes and azimuths. To an observer having the zenith shown in the figure, a star in the northern sky near the north pole will not set at all, and to the same observer a star near the south pole will not rise at all. If its path for one day were traced on the celestial sphere, it would be a circle (as ABC) with its center in the polar axis and lying in a plane parallel to the plane of the equator. * That is, azimuth is measured from 0~ to 360~ clockwise. 240 SPHERICAL TRIGONOMETRY 30. The equator and meridian system. In this case the two fixed and mutually perpendicular great circles of reference are the celestial equator (EQDWQ') and the meridian of the observer (NPZQSP'Z'Q'); and the spherical coordinates of a heavenly body are its declination and hour angle. The declination of a heavenly body is its angular distance north or south of the celestial equator measured on the hour circle of the body from 0~ to 90~.* z Thus, in the figure, the /~o~, <T\arc DM is a measure of (Worth r /'/ / '\ nthe north declination SVole) "'/ / of the star M. North /[ \ 7 X A, St) N. declination is always./ "N / -k XoD \ considered positive and - 'a -/ -.-^/ south declination negaN. - - ' /'-'A i Earth _\ I f '\ tive. Hence the decli> 5\ / / ^ ^ nation of the north pole is + 90~, while \^/)// / ^^ ''Y / that of the south pole is - 90~. M\, I ^ - The declinations of Q0^.\~ 1 >^the sun, moon, and Q,V~~~] — j~/ planets are continually changing, but the declination of a fixed star changes by an exceedingly small amount in the course of a year. The angular distance of a heavenly body from the north celestial pole, measured on the hour circle of the body, is called its north polar distance (P3M in figure). The north polar distance of a star is evidently the complement of its declination. The hour angle of a heavenly body is the angle between the meridian of the observer and the hour circle of the star measured westward from the meridian from 0~ to 360~. Thus, in the figure, the hour angle of the star Mi is the angle QPD (measured by the arc QD). This angle is commonly used as a measure of time, hence the name hour angle. Thus the star M makes a complete circuit in 24 hours; that is, the hour angle QPD continually increases at the uniform rate of 360~ in 24 hours, or 15~ an hour. For this reason the hour angle of a heavenly body is usually reckoned in hours from * The declinations of the sun, moon, planets, and some of the fixed stars, for any time of the year, are given in the NTautical Almanac or American Ephemeris, published by the United States government. APPLICATIONS OF SPHERICAL TRIGONOMETRY 241 0 to 24, one hour being equal to 15~.* When the star is at M1 (on the observer's meridian) its hour angle is zero. Then the hour angle increases until it becomes the angle M1PM (when the star is at AM). When the star sets on the western horizon its hour angle becomes M1PM2. Twelve hours after the star is at Mi it will be at 3M, when its hour angle will be 180~ (= 12 hours). Continuing on its circuit, the star rises at M4 and finally reaches M1, when its hour angle has become 360~ (= 24 hours), or 0~ again. Knowing the hour angle and declination (spherical coordinates) of a heavenly body, we can locate it on the celestial sphere as follows. From the point, as Q, where the reference circles intersect, lay off the hour angle (or arc), as QD. Then on the hour circle passing through D lay off the declination, as DM. The body is then located at M. Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Hour angle Declination Hour angle Declination (a) 450 N. 30~ (j) 60~ S. 450 (b) 60~ N. 60~ (k) 00 0~ (c) 900 S. 450 (1) 180~ 0~ (d) 120~ S. 30~ (m) 90~ N. 900 (e) 180~ N. 500 (n) 270~ 0~ (f) 5 hr. N. 75~ (o) 12 hr. S. 10~ (g) 15 hr. - 25~ (p) 3 hr. + 80~ (h) 6 hr. + 79~ (q) 9 hr. - 45~ (i) 0 hr. - 90~ (r) 20 hr. + 60~ 31. Practical applications. Among the practical applications of Astronomy the most important are: (a) To determine the position of an observer on the surface of the earth (i.e. his latitude and longitude). (b) To determine the meridian of a place on the surface of the earth. (c) To ascertain the exact time of day at the place of the observer. (d) To determine the position of a heavenly body. The first of these, when applied to the determination of the place of a ship at sea, is the problem to which Astronomy mainly owes its economic importance. National astronomical observatories have been * On account of the yearly revolution of the earth about the sun, it takes the sun about 4 minutes longer to make the circuit than is required by any particular fixed star. Hence the solar day is about 4 minutes longer than the sidereal (star) day, but each is divided into 24 hours; the first giving hours of ordinary clock time, while the second gives sidereal hours, which are used extensively in astronomical work. When speaking of the sun's hour angle it shall be understood that it is measured in hours of ordinary clock time, while the hour angle of a fixed star is measured in sidereal hours. In either case 1 hour = 15~. 242 SPHERICAL TRIGONOMETRY established, and yearly nautical almanacs are being published by the principal nations controlling the commerce of the world, in order to supply the mariner with the data necessary to determine his position accurately and promptly. 32. Relation between the observer's latitude and the altitude of the celestial pole. To an observer on the earth's equator (latitude zero) the pole star is on the horizon; that is, the altitude of the star is zero. If the observer is traveling northward, the pole star will gradually rise; that is, the latitude of the observer and the altitude of the star are both increasing. Finally, when the observer reaches the north pole of the earth his latitude and the altitude of the star have both increased to 90~. The place of the pole in the sky then r_ '~%e (To Zenith) \a \ -%; Earth depends in some way on the observer's latitude, and we shall now prove that the altitude of a celestial pole is equal to the latitude of the observer. Let 0 be the place of observation, say some place in the northern hemisphere; then the angle QCO (or arc QO) measures its north latitude. Produce the earth's axis CP until it pierces the celestial sphere at the celestial north pole. A line drawn from 0 in the direction (as OP2) of the celestial north pole will be parallel to CP1, since the celestial north pole is at an unlimited distance from the earth (see ~ 27, p. 235). The angle NOP2 measures the altitude of the north pole. But CO is perpendicular to ON and CQ is perpendicular to OP2 (since it is perpendicular to the parallel line CP,); hence the angles NOP2 and QCO are equal, and we find that the altitude of the pole as observed at 0 is equal to the latitude of 0. 33. To determine the latitude of a place on the surface of the earth. If we project that part of the celestial sphere which lies above the APPLICATIONS OF SPHERICAL TRIGONOMETRY 243 horizon on the plane of the observer's celestial meridian, the horizon will be projected into a line (as NS), and the upper half of the celestial equator will also be projected into a line (as OQ). From the last section we know that the latitude of the observer equals the altitude of the elevated celestial pole (arc NP in figure), or, what amounts to the same thing, equals the angular distance between the zenith and the celestial equator (arc ZQ in figure). If then the elevated pole could be seen as a definitely marked point in the sky, the observer's latitude would be A Q found by simply measuring / \ the angular distance of that / pole above the horizon. But B there are no fixed stars visible at the exact points where the N Horizon 0 S ( Observer) polar axis pierces the celestial sphere, the so-called polar star being about 14~ from the celestial north pole. Following are some methods for determining the latitude of a place on the surface of the earth. First method. To determine latitude by observations on circumpolar stars. The most obvious method is to observe with a suitable instrument the altitude of some star near the pole (so near the pole that it never sets; as, for instance, the star whose path in the sky is shown as the circle ABC in figure, p. 238) at the moment when it crosses the meridian above the pole, and again 12 hours later, when it is once more on the meridian but below the pole. In the first case its elevation will be the greatest possible; in the second, the least possible. The mean of the two observed altitudes is evidently the latitude of the observer. Thus, in the figure on this page, if NA is the maximum altitude and NB the minimum altitude of the star, then NA + NB N 2 +N = NP = altitude of pole latitude of place of observation. Ex. 1. The maximum altitude of a star near the pole star was observed to be 54~ 16', and 12 hours later its minimum altitude was observed to be 40~ 24'. What is the latitude of the place of observation? Solution. 54~ 16' + 40~ 24' = 94~ 40'. 940 40' Therefore = 470 20' = altitude of north pole f2~ = north latitude of place of observation. 244 SPHERICAL TRIGONOMETRY Second method. To determine latitude from the meridian altitude of a celestial body whose declination is known. The altitude of a star M is measured when it is on,X ^ M(Star) the observer's meridian. If we J'(star) subtract this meridian altitude (arc SM in figure) from 90~, / we get the star's zenith distance (ZM). In the Nautical Almanac we now look up the N s Horizon o0 S star's declination at the same (Observer) instant; this gives us the arc QM. Adding the declination of the star to its zenith distance, we get QM + MZ = QZ = NP = altitude of pole = latitude of place. Therefore, when the observer is on the northern hemisphere and the star is on the meridian south of zenith, North latitude = zenith distance + declination.* If the star is on the meridian between the zenith and the pole (as at M" t), we will have North latitude = NP = ZQ = QM" — ZM"' = declination - zenith distance. If the observer is on the southern hemisphere and the z_ Star) star M is on his meridian between the zenith and south / \p So u <(Souwth pole, we would have / le) South latitude = SP = SM- MP' \ = SM - (90~- QM) ( Observer) = altitude - co-declination, if we consider only the numerical value of the declination. In working out examples the student should depend on the figure rather than try to memorize formulas to cover all possible cases. Ex. 2. An observer in the northern hemisphere measured the altitude of a star at the instant it crossed his celestial meridian south of zenith, and found it to be 63~ 40'. The declination of the star for the same instant was given by the Nautical Almanac as 21~ 15' N. What was the latitude of the observer? * If the star is south of the celestial equator (as at M'), the same rule will hold, for then the declination is negative (south), and the algebraic sum of the zenith distance and declination will still give the arc QZ. t Maximum altitude, if a circumpolar star. APPLICATIONS OF SPHERICAL TRIGONOMETRY 245 Solution. Draw the semicircle NZSO. Lay off the arc SM = altitude = 63~ 40', which locates the star at M. Since the declination of the star is north, the celestial (ar equator may be located by laying off the Q arc MQ = declination = 21~ 15' towards the south. The line QO will then be the projection of the celestial equator, and OP, drawn perpendicular to QO, will locate the N\ north pole P. 0 Zenith distance = ZM = 90~ - SM (alt.) = 90~ - 63~ 40' = 26~ 20'..-. North latitude of observer = NP = ZQ = ZM (zen. dist.) + MQ (dec.) = 26~ 20' + 21~ 15' = 47~ 35'. Third method. To determine latitude when the altitude, declination, and hour angle of a celestial body are known. Referring to the astronomical (spherical) triangle PZM, we see that Z side MZ = 90~ - Ri (alt.) = co-altitude, (\ u!*tar),,/\y^ / \^ the altitude of the star '/ '; i/ j' / \ being found by measure/ Z K / / \ ment. Also,,-/ 'KI _ _ Earthside PM = 90~ - DM (dec.) ilorizo-n - H = co-declination, the declination of the star being found from the Nautical Almanac. Angle ZPM = hour angle, which is given. This hour angle will be the local time when the observation is made on the sun. We then have two sides and the angle opposite one of them given in the spherical triangle PZM. Solving this for the side PZ, by Case III, (a), p. 224, we get Latitude of observer = NP = 90 ~- PZ. Ex. 3. The declination of a star is 69~ 42' N. and its hour angle 60~ 44'. What is the north latitude of the place if the altitude of the star is observed to be 49~ 40'? Solution. Referring to the above figure, we have, in this example, side MZ = co-alt. = 90~ - 49~ 40' = 40~ 20', side PM = co-dec. = 90~ - 69~ 42' = 20~ 18', angle ZPM = hour angle = 60~ 44'. Solving for the side PZ by Case III, (a), p. 224, we get side PZ=47~ 9'=co-lat..'. 90~ - 47~ 9' = 42~ 51' = north latitude of place. The angle MZP is found to be 27~ 53'; hence the azimuth of the star (angle SZH) is 180~ - 27~ 53' = 152~ 7'. 246 SPHERICAL TRIGONOMETRY EXAMPLES 1. The following observations for altitude have been made on some north circumpolar star. What is the latitude of each place? Maximum altitude Minimum altitude (a) New York 500 46' 30~ 40' (b) Boston 440 22' 400 20' (c) New Haven 580 24' 24~ 10' (d) Greenwich 64~ 36' 38~ 22' (e) San Francisco 550 6' 20~ 30' (f) Calcutta 24~ 18' 20~ 48' North latitude Ans. 40~ 43' 42~ 21' 41~ 17' 51~ 29' 37~ 48' 22~ 33' 2. In the following examples the altitude of some heavenly body has been measured at the instant when it crossed the observer's celestial meridian. What is the latitude of the observer in each case, the declination being found from the Nautical Almanac? Hemisphere Meridian altitude Declination Body is (a) Northern (b) Northern (c) Northern (d) Northern (e) Northern (f) Northern (g) Southern (h) Southern (i) Southern (j) Southern 60~ 75~ 40' 43~ 27' 380 6' 500 28~ 46' 67~ 45~ 26' 72~ 22~ 18' N. 20~ N. 32~ 13' S. 10~ 52' S. 44~ 26' N. 62~ N. 73~ 16' S. 59~ S. 81~ 48' S. 8~ N. 46~ 25' S. of zenith S. of zenith S. of zenith S. of zenith N. of zenith N. of zenith S. of zenith S. of zenith N. of zenith N. of zenith Latitude Ans. 50~ N. 46~ 33' N. 35~ 41' N. 70 28' N. 22~ N. 12~ 2' N. 36~ S. 37~ 14' S. 26~ S. 21~ 17' S. 3. In the following examples the altitude of some heavenly body not on the observer's celestial meridian has been measured. The hour angle and declination is known for the same instant. Find the latitude of the observer in each case. Hemisphere (a) Northern (b) Northern (c) Northern (d) Northern (e) (f) (g) (h) (i) Northern Northern Northern Northern Southern A Ititude 400 15~ 52~ 64~ 42' 00 25~ 00 90 26' 380 19~ 46~ 18' 00 57~ 36' Declination N. 10~ S. 8~ N. 19~ N. 24~ 20' S. 50 00 N. 110 14' 00 S. 12~ N. 70 S. 15~23' N. 14~ 00 Hour angle 500 65~ 2 hr. 345~ 5 hr. 21 hr. 68~ 54' 72~ 22/ 52~ 3 hr. 326~ 38~ 2 hr. Latitude Ans. 2702' N. 35~ 38' N. 48~ 16' N. 3~ 34' N. or 46~ 36' N. 770 37' N. 53~ 18' N. No solution 570 14'N. 33~ 56/ S. or 4~ 8' S. 52~ 56' S. 49~ 14' S. 72~ 26' S. 12~ 50' S. (j) Southern (k) Southern (1) Southern (m) Southern APPLICATIONS OF SPHERICAL TRIGONOMETRY 247 34. To determine the time of day. A very simple relation exists between the hour angle of the sun and the time of day at any place. The sun appears to move from east to west at the uniform rate of 15~ per hour, and when the sun is on the meridian of a place it is apparent noon at that place. Comparing, Hour angle of sun 00 15~ 300 450 900 180~ 1950 210~ 270~ 300~ 360~ Time of day Noon 1 P.M. 2 P.M. 3 P.M. 6 P.M. Midnight 1 A.M. 2 A.M. 6 A.M. 8 A.M. Noon The hour angle of the sun M is the angle at P in the astronomical (spherical) triangle PZM. We may find this hour angle (time of (Zenith) Z (Niorth pole) P31 / (North).....ar// - -...(South) / / // (}Vest) r\. \ \ / /'^~ \ ^\ / x /^~ i /P' day) by solving the astronomical triangle for the angle at P, provided we know three other elements of the triangle. 248 SPHERICAL TRIGONOMETRY DM = declination of sun, and is found from the Nautical Almanac..'. Side PM = 90~ - DM = co-declination of sun. HM = altitude of sun, and is found by measuring the angular distance of the sun above the horizon with a sextant or transit.. side MZ = 90~ - HM = co-altitude of sun. NP = altitude of the celestial pole = latitude of the observer (p. 243).. Side PZ = 90~- NP = co-latitude of observer. Hence we have Rule for determining the time of day at a place whose latitude is known, when the declination and altitude of the sun at that time and place are known. First step. Take for the three sides of a spherical triangle the co-altitude of the sun, the co-declination of the sun, the co-latitude of the place. Second step. Solve this spherical triangle for the angle opposite the first-mentioned side. This will give the hour angle in degrees of the sun, if the observation is made in the afternoon. If the observation is made in the forenoon, the hour angle will be 360~ - the angle found. Third step. When the observation is made in the afternoon the time of day will be hour angle P.M. 15-* P.M. 15 When the observation is made in the forenoon the time of day will be /hour angle 12)A.M. 15 Ex. 1. In New York (lat. 400 43' N.) the sun's altitude is observed to be 30~ 40'. Having given that the sun's declination is 10~ N. and that the observaZenith tion is made in the afternoon, what is the time,^^ sB of day? c(,/ Solution. First step. Draw the triangle. / b Side a = co-alt. = 90~ - 30~ 40' = 59~ 20'. A Co-dec. Side b = co-dec. = 90~ - 10~ = 80~. PoSe Sun Pole Side c = co-lat. = 90~ - 40~ 43' = 49~ 17'. Second step. As we have three sides given, the solution of this triangle comes under Case I, (a), p. 217. But as we only want the angle A (hour angle), some APPLICATIONS OF SPHERICAL TRIGONOMETRY 249 labor may be saved by using one of the formulas (18), (19), (20), pp. 211, 212. Let us use (18), a = 59~ 20' b = 800. sin s sin (s - a) c - 49~ 17' sm ^ \ — Si —I a = c= 490 17' s sin b sin c 2s = 188~ 37' 2 37log sin I a = I [log sin s + log sin (s - a) - {log sin b + log sin c]. s = 94 19'. 2 2 s - a = 34~ 59'. log sin s = 9.9988 log sin b = 9.9934 log sin (s - a) = 9.7584 log sin c = 9.8797 log numerator = 19.7572 log denominator = 19.8731 log denominator = 19.8731 9.8841 21 19.8841 logsin a = 9.9421 = 610 4'. a = 122~ 8'... A = 180~ - a = 57~ 52'= hour angle of sun. hour angle Third step. Time of day = P.m. = 3 hr. 51 min. P.M. Ans. 15 EXAMPLES 1. In Milan (lat. 45030' N.) the sun's altitude at an afternoon observation is 26~ 30'. The sun's declination being 8~ S., what is the time of day? Ans. 2 hr. 33 min. P.M. 2. In New York (lat. 40~ 43' N.) a forenoon observation on the sun gives 30~ 40' as the altitude. What is the time of day, the sun's declination being 10~ S.? Ans. 9 hr. 46 min. A.M. 3. A mariner observes the altitude of the sun to be 60~, its declination at the time of observation being 6~ N. If the latitude of the vessel is 12~ S., and the observation is made in the morning, find the time of day. Ans. 10 hr. 24 min. A.M. 4. A navigator observes the altitude of the sun to be 35~ 23', its declination being 10~ 48' S. If the latitude of the ship is 26~ 13' N., and the observation is made in the afternoon, find the time of day. Ans. 2 hr. 45 min. P.M. 5. At a certain place in latitude 40~ N. the altitude of the sun was found to be 41~. If its declination at the time of observation was 20~ N., and the observation was made in the morning, how, long did it take the sun to reach the meridian? Ans. 3 hr. 31 min. 6. In London (lat. 51~ 31' N.) at an afternoon observation the sun's altitude is 15~ 40'. Find the time of day, given that the sun's declination is 12~ S. Ans. 2 hr. 59 min. P.M. 7. A government surveyor observes the sun's altitude to be 21~. If the latitude of his station is 27~ N. and the declination of the sun 16~ N., what is the time of day if the observation was made in the afternoon? Ans. 4 hr. 57 min. P.M. 8. The captain of a steamship observes that the altitude of the sun is 26~ 30'. If he is in latitude 45~ 30' N. and the declination of the sun is 18~ N., what is the time of day if the observation was made in the afternoon? Ans. 4 hr. 41 min. P.M. 250 SPHERICAL TRIGONOMETRY 35. To find the time of sunrise or sunset. If the latitude of the place and the declination of the sun is known, we have a special case of the preceding problem; for at sunrise or sunset the sun is on the horizon and its altitude is zero. Hence the co-altitude, which is one side of the astronomical triangle, will be 90~, and the triangle will be a quadrantal triangle (p. 204). The triangle may then be solved by the method of the last section or as a quadrantal triangle. EXAMPLES 1. At what hour will the sun set in Montreal (lat. 45~ 30' N.), if its declination at sunset is 18~ N.? Ans. 7 hr. 17 min. P.M. 2. At what hour will the sun rise in Panama (lat. 8~ 57' N.), if its declination at sunrise is 23~ 2' S.? Ans. 6 hr. 15 min. A.M. 3. About the first of April of each year the declination of the sun is 4~ 30' N. Find the time of sunrise on that date at the following places: (a) New York (lat. 40~ 43' N.). Ans. 5 hr. 45 min. A.M. (b) London (lat. 51~ 31' N.). 5 hr. 37 min. A.M. (c) St. Petersburg (lat. 60~ N.). 5 hr. 29 min. A.M. (d) New Orleans (lat. 29~ 58' N.). 5 hr. 50 min. A.M. (e) Sydney (lat. 33~ 52' S.). 5 hr. 48 min. A.M. 36. To determine the longitude of a place on the earth. From the definition of terrestrial longitude given on p. 231 it is evident that the meridians on the Zy — 9 & earth are projected into O ~bserver,! sQ hour circles on the " " /]\ ~celestial sphere. Hence /,Sun)g/ \ the same angle (or arc) / \. / D \ which measures the / '\ E, —,/ R \ \ angle between the celes7-'t' - --- tial meridians (hour N _- go / " \ — " s circles) of the place -ree o/ n>h '\ of observation and of \ / / /y \ \ / Greenwich may be \ X / / \ As taken as a measure of \. | / /!p' the longitude of the place. Thus, in the figure, if PQP' is the meridian (hour circle) of Greenwich and PDP' the meridian (hour circle) of the place of observation, then the angle QPD (or arc QD) measures the west longitude of the place. If PMP' is the hour circle of the sun, it is evident that APPLICATIONS OF SPHERICAL TRIGONOMETRY 251 angle QPM =hour angle of sun for Greenwich = local time at Greenwich; angle DPM = hour angle of sun for observer = local time at place of observation. Also, angle QPM - angle DPM = angle QPD = longitude of place. Hence the longitude of the place of observation equals the difference * of local times between the standard meridian and the place in question. Or, in general, we have the following Rule for finding longitude: The observer's longitude is the amount by which noon at Greenwich is earlier or later than noon at the place of observation. If Greenwich has the earlier time, the longitude of the observer is east; if it has the later time, then the longitude is west. We have already shown (p. 248) how the observer may find his own local time. It then remains to determine the Greenwich time without going there. The two methods which follow are those in general use. First method. Find Greenwich time by telegraph (wire or wireless). By far the best method, whenever it is available, is to make a direct telegraphic comparison between the clock of the observer and that of some station the longitude of which is known. The difference between the two clocks will be the difference in longitude of the two places. Ex. 1. The navigator on a battleship has determined his local time to be 2 hr. 25 min. P.M. By wireless he finds the mean solar time at Greenwich to be 4 hr. 30 min. P.M. What is the longitude of the ship? Solution. Greenwich having the later time, 4 hr. 30 min. 2 hr. 25 min. Subtracting, 2 hr. 5 min. = west longitude of the ship. Reducing this to degrees and minutes of arc, 2 hr. 5 min. 15 Multiplying, 31~ 15' = west longitude of ship. Second method. Find Greenwich time from a Greenwich chronometer. The chronometer is merely a very accurate watch. It has been set to Greenwich time at some place whose longitude is known, and thereafter keeps that time wherever carried. * This difference in time is not taken greater than 12 hours. If a difference in time between the two places is calculated to be more than 12 hours, we subtract it from 24 hours and use the remainder instead as the difference. 252 SPHERICAL TRIGONOMETRY Ex. 2 An exploring party have calculated their local time to be 10 hr. A.M. The Greenwich chronometer which they carry gives the time as 8 hr. 30 min. A.M. What is their longitude? Solution. Greenwich has here the earlier time. 10 hr. 8 hr. 30 min. 1 hr. 30 min. = 22~ 30'= east longitude. Subtracting, EXAMPLES 1. In the following examples we have given the local time of the observer and the Greenwich time at the same instant. Find the longitude of the observer in each case. Observer's local time (a) Noon. (b) Noon. (c) Midnight. (d) 4 hr. 10 min. P.M. (e) 8 hr. 25 min. A.M. (f) 9 hr. 40 min. P.M. (g) 2 hr. 15 min. P.M. (h) 10 hr. 26 min. A.M. (i) 1 hr. 30 min. P.M. (j) Noon. (k) 6 hr. P.M. (1) 5 hr. 45 min. A.IM. (m) 10 hr. 55 min. P.M. Corresponding Greenwich time 3 hr. 30 min. P.M. 7 hr. 20 min. A.M. 10 hr. 15 min. P.M. Noon. Noon. Midnight. 11 hr. 20 min. A.M. 5 hr. 16 min. A.M. 7 hr. 45 min. P.M. Midnight. 6 hr. A.A. 7 hr. 30 min. P.-M. 8 hr. 35 min. A.M. Longitude of observer Ans. 52~ 30' W. 70~ E. 26~ 15' E. 620 30' E. 53~ 45'W. 350 W. 43~ 45' E. 770 30' E. 930 45' W. 180~ W. or E. 180~ E. or W. 153~ 45' E. 145~ W. 2. If the Greenwich time is 9 hr. 20 min. P.M., January 24, at the same instant that the time is 3 hr. 40 min. A.M., January 25, at the place of observation, what is the observer's longitude? Ans. 95~ E. 3. The local time is 4 hr. 40 min. A.M., March 4, and the corresponding Greenwich time is 8 hr. P.M., March 3. What is the longitude of the place? Ans. 130~ E. 4. In the following examples we have given the local time of the observer and the local time at the same instant of some other place whose longitude is known. Find the longitude of the observer in each case. Observer's local time (a) 2 hr. P.M. (b) 10 hr. A.M. (c) 5 hr. 20 min. P.M. (d) 8hr. 25min. A.M. (e) 9 hr. 45 min. P.M. (f) 7 hr. 40 min. P.M. (g) 4 hr. 50 min. P.M. Corresponding time and longitude of the other place Longitude of observer 5 hr. P.M. at Havana (long. 82~ 23'W.) Ans. 1270 23' W. 3 hr. P.M. at Yokohama (long. 139~ 41' E.) 64~ 41' E. 11 hr. 30 min. P.M. at Glasgow (long. 4~ 16' W.) 96~ 46' W. 6 hr. 35 min. A.M. at Vera Cruz (long. 96~ 9'W.) 68~ 39' W. Midnight at Batavia (long. 106~ 52' E.) 73~ 7 E. Noon at Gibraltar (long. 5 21' W.) 109~ 39' E. Noon at Auckland (long. 174~ 50' E.) 112~ 40' W. APPLICATIONS OF SPHERICAL TRIGONOMETRY 253 5. What is the longitude of each place mentioned in the examples on p. 249, the Greenwich time for the same instant being given below? Example, p. 249 Greenwich time Longitude of place (a) Ex. 3 2 hr. 12 min. P.M. Ans. 57~ W. long. (vessel) (b) Ex. 4 4 hr. 52 min. P.M. 31~45'W. long. (vessel) (c) Ex. 5 5 hr. 9 min. A.M. 50~ E. long. (observer) (d) Ex. 7 10 hr. 33 min. P.M. 84~ W. long. (surveyor) (e) Ex. 8 6 hr. 25 min. P.M. 26~ W. long. (ship) 37. The ecliptic and the equinoxes. The earth makes a complete circuit around the sun in one year. To us, however, it appears as if the sun moved and the earth stood still, the (apparent) yearly path of the sun among the stars being a great circle of the celestial sphere which we call the ecliptic. Evidently the plane of the earth's orbit K Earth cuts the celestial sphere in the ecliptic. The plane of the equator and the plane of the ecliptic are inclined to each other at an angle of about 23~0 (= e), called the obliquity of the ecliptic (angle LVQ in figure). The points where the ecliptic intersects the celestial equator are called the equinoxes. The point where the sun crosses the celestial equator when moving northward (in the spring, about March 21) is called the vernal equinox, and the point where it crosses the celestial equator when moving southward (in the fall, about September 21) is called the autumnal equinox. If we project the points V and A in our figure on the celestial sphere, the point V will be projected in the vernal equinox and the point A in the autumnal equinox. 38. The equator and hour circle of vernal equinox system." The two fixed and mutually perpendicular great circles of reference are in * Sometimes called the equator system. 254 SPHERICAL TRIGONOMETRY this case the celestial equator (QVQ') and the hour circle of the vernal equinox (P VP'), also called the equinoctial colure; and the spherical coordinates of a (Vorth celestial pole) p heavenly body are its KK ^ -: V ' ^ declination and right ascension. I~/ \ ~ d \M \The declination of a a/ \(sar, - s")L ^ heavenly body has al/ / \ ready been defined on p. 240 as its angular,-, ADZ';\ st'j / A m l distance north or south QP — >^~ / Y Q} ~of the celestial equator \ XatorX7 —> D ) t measured on the hour \ x (Vei i al / circle of the body from I^-^c ^00 to 90~, positive if north and negative if \|8 / y\/ south. In the figure I, /r K DM is the north decliPfP'~ ~nation of the star M. The right ascension of a heavenly body is the angle between the hour circle of the body and the hour circle of the vernal equinox measured eastward from the latter circle from 0~ to 360~, or in hours from 0 to 24. In the figure, the angle VPD (or the arc VD) is the right ascension of the star M. The right ascensions of the sun, moon, and planets are continually changing.* The angle LVQ (= e) is the obliquity of the ecliptic (= 23-~). Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Right ascension Declination Right ascension Declinzation (a) 0~ 00 (j) 90~ 00 (b) 180~ 0~ (k) 270~ 0~ (c) 900 N. 90~ (1) 90~ S. 900 (d) 45~ N. 45~ (m) 45~ S. 45~ (e) 60~ N. 60~ (n) 900 S. 30~ (f) 120~ + 30~ (o) 240~ + 60~ (g) 300~ - 600 (p) 330~ - 45~ (h) 12 hr. + 450 (q) 6 hr. + 15~ (i) 20 hr. 0~ (r) 9 hr. - 750 * The right ascensions of the sun, moon, and planets may be found in the Nautical Almanac for any time of. the year. APPLICATIONS OF SPHERICAL TRIGONOMETRY 255 Ex. 2. The right ascension of a planet is 10 hr. 40 min. and its declination S. 6~. Find the angular distance from this planet to a fixed star whose right ascension is 3 hr. 20 min. and declination N. 48~. Solution. Locate the planet and the star on the celestial sphere. Draw the spherical triangle whose vertices are at the north pole, the planet, and the fixed star. Then Angle A = difference of right ascensions = 10 hr. 40 min. - 3 hr. 20 min. = 7 hr. 20 min. = 110~. Side b = co-declination of star = 90~ - 480 = 420. Side c = co-declination of planet = 900 - (- 6~) = 96~. To find side a. Pole 00 -0 b Star a Plan.,,t As we have two sides and the included angle given, the solution of this triangle comes under Case II, (a), p. 219. Since a only is required, the shortest method is that illustrated on p. 220, the solution depending on the solution of right spherical triangles. On solving, we get a= 107~ 48'. Ans. 39. The system having for reference circles the ecliptic and the great circle KVKf passing through the pole of the ecliptic and the vernal (North celestial pole) ecliptic) s '\ ~, I McStar) (Verna\,.\./5' equinox \ \ P' equinox.* The spherical coordinates of a heavenly body in this case are its latitude and longitude. f The latitude of a heavenly body is its angular distance north or south of the ecliptic, measured on the great circle passing through * Sometimes called the ecliptic system. t Sometimes called celestial latitude and longitude in contradistinction to the latitude and longitude of places on the earth's surface (terrestrial latitude and longitude), which were defined on p. 231, and which have different meanings. 256 SPHERICAL TRIGONOMETRY the body and the pole of the ecliptic. Thus, in the figure, the arc TM measures the north latitude of the star M. The longitude of a heavenly body is the angle between the great circle passing through the body and the pole of the ecliptic, and the great circle passing through the vernal equinox and the pole of the ecliptic, measured eastward from the latter circle from 0~ to 360~. In the figure, the angle VKT (or the arc VT) is the longitude of the star M. The latitudes and longitudes of the sun, moon, and planets are continually changing. The angle LVQ (= e) is the obliquity of the ecliptic (= 23-~ = arc KP). Since the ecliptic is the apparent yearly path of the sun, the celestial latitude of the sun is always zero. The declination of the sun, however, varies from N. 23-~ (= arc QL) on the longest day of the year in the northern hemisphere (June 21), the sun being then the highest in the sky (at L), to S. 23-~ (arc Q'L') on the shortest day of the year (December 22), the sun being then the lowest in the sky (at L'). The declination of the sun is zero at the equinoxes (March 21 and September 21). Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Celestial longitude Celestial latitude (a) 0~ o0 (b) 900 N. 900 (c) 180~ N. 450 (d) 270~ 00 (e) 450 S. 300 (f) 135~ + 15~ (g) 315~ + 60~ (h) 6 hr. - 450 (i) 15 hr. + 450 Celestial longitude (j) 90~ (k) 180~ (1) 00 (m) 60~ (n) 1200 (o) 270~ (p) 300 (q) 9 hr. (r) 18 hr. Celestial latitude 00 00 S. 60~ N. 300 N. 450 -75~ - 60~ 00 + 300 Ex. 2. Given the right ascension of a star 2 hr. 40 min. and its declination 24~ 20' N., find its celestial latitude and longitude. Solution. Locate the star on the celestial sphere. Consider the spherical triangle KPM on the next page. Angle KPM = Z Q'PV + Z VPD = 90~ + right ascension = 90~ + 2 hr. 40 min. = 90~ + 400 = 1300. Side PM = co-declension = 90- 24~ 20' = 65~ 40'. APPLICATIONS OF SPHERICAL TRIGONOMETRY 257 Side KP = LQ = e = 23~ 30'. To find side KM = co-latitude of the star, and angle PKM = co-longitude of the star. (N\orth pole) P Y'\ _ \ E Star) /' (V er. Eq.) L' As we have two sides and the included angle given, the solution of this triangle comes under Case II, (a), p. 219. Solving, we get Side KM = 81~ 52' and Z PKM = 44~ 52'..-. 90 - KM = 90~ - 81~ 52' = 8~ 8' = TM = latitude of star, and 90~ - PM = 90~ - 44 52' = 45~ 8' = VT = longitude of star. EXAMPLES 1. Find the distance in degrees between the sun and the moon when their right ascensions are respectively 12 hr. 39 min., 6 hr. 56 min., and their declinations are 9~ 23' S., 22~ 50' N. Ans. 90~. 2. Find the distance between Regulus and Antares, the right ascensions being 10 hr. and 16 hr. 20 min., and the polar distances 77~ 19' and 116~ 6'. Ans. 99~ 56'. 3. Find the distance in degrees between the sun and the moon when their right ascensions are respectively 15 hr. 12 min., 4 hr. 45 min., and their declinations are 21~ 30' S., 5~ 30' N. Ans. 154~ 19'. 4. The right ascension of Sirius is 6 hr. 39 min., and his declination is 16 31' S.; the right ascension of Aldebaran is 4 hr. 27 min., and his declination is 16~ 12' N. Find the angular distance between the stars. Ans. 46 2'. 5. Given the right ascension of a star 10 hr. 50 min., and its declination 12~ 30' N., find its latitude and longitude. Take e = 23~ 30'. Ans. Latitude = 18~ 24' N., longitude = 281~ 7'. 6. If the moon's right ascension is 4 hr. 15 min. and its declination 6~ 20' N., what is its latitude and longitude? Ans. Latitude = 14~ 43' N., longitude = 62~ 58'. 258 SPHERICAL TRIGONOMETRY 7. The sun's longitude was 59~ 40'. What was its right ascension and declination? Take e = 23~ 27'. Ans. Right ascension = 3 hr. 50 min., declination = 20~ 38' N. Hint. The latitude of the sun is always zero, since it moves in the ecliptic. Hence in the triangle KPM (figure, p. 257), KM= 900, and it is a quadrantal triangle. This triangle may then be solved by the method explained on p. 204. 8. Given the sun's declination 16~ 1' N., find the sun's right ascension and longitude. Take e = 23~ 27'. Ans. Right ascension = 9 hr. 14 min., longitude = 136~ 6'. 9. The sun's right ascension is 14 hr. 8 min.; find its longitude and declination. Take e = 23~ 27'. Ans. Longitude = 214~ 16', declination = 12~ 56' S. 10. Find the length of the longest day of the year in latitude 42~ 17' N. Ans. 15 hr. 6 min. Hint. This will be the time from sunrise to sunset when the sun is the highest in the sky, that is, when its declination is 23~ 27' N. 11. Find the length of the shortest day in lat. 42~ 17' N. Ans. 8 hr. 54 min. Hint. The sun will then be the lowest in the sky, that is, its declination will be 23~ 27' S. 12. Find the length of the longest day in New Haven (lat. 41~ 19' N.). Take e = 23~ 27'. Ans. 15 hr. 13. Find the length of the shortest day in New Haven. Ans. 9 hr. 14. Find the length of the longest day in Stockholm (lat. 59~ 21' N.). Take e = 23~ 27'. Ans. 18 hr. 16 min. 15. Find the length of the shortest day in Stockholm. Ans. 5 hr. 48 min. 40. The astronomical triangle. We have seen that many of our most important astronomical problems depend on the solution of (Zenith) Z (Nor h Tole) (North)N -- - S' — outh) y' the astronomical triangle PZM. In any such problem the first thing to do is to ascertain which parts of the astronomical triangle APPLICATIONS OF SPHERICAL TRIGONOMETRY 259 are given or can be obtained directly from the given data, and which are required. The different magnitudes which may enter into such problems a HM= altitude of the heavenly body, DM = declination of the heavenly body, angle ZPMl = hour angle of the heavenly body, angle SZM = azimuth of the heavenly body, NP = altitude of the celestial pole = latitude of the observer. As parts of the astronomical triangle PZM we then have side MZ = 900~- HM = co-altitude, side PM = 90 - DM = co-declination, side PZ = 90~ - NP = co-latitude, angle ZPM = hour angle, angle PZM = 180~0- azimuth (angle SZM).* The student should be given practice in picking out the known and unknown parts in examples involving the astronomical triangle, and in indicating the case under which the solution of the triangle comes. For instance, let us take Ex. 15, p. 261. f Latitude = 51~ 32' N..'. side PZ = 90~- 51~ 32'= 38~ 28'. Altitude = 35~ 15'. Given parts parts side MZ = 90~- 35~ 15'= 540 45t. Declination = 21~ 27' N.. side MP= 90~- 21~ 27'= 68~ 33'. Required: Local time = hour angle = angle ZPM. Since we have three sides given to find an angle, the solution of the triangle comes under Case I, (a), p. 217. This gives angle ZPM = 59~ 45'= 3 hr. 59 min. P.M. 41. Errors arising in the measurement of physical quantities. Errors of some sort will enter into all data obtained by measurement. For instance, if the length of a line is measured by a steel tape, account must be taken of the expansion due to heat as well as the sagging of the tape under various tensions. Or, suppose the navigator of a ship * When the heavenly body is situated as in the figure. If the body is east of the observer's meridian, we would have angle PZil[= azimuth - 180~. t In this connection the student is advised to read ~ 93 in Granville's Plane Trigonometry. 260 SPHERICAL TRIGONOMETRY at-sea is measuring the altitude of the sun by means of a sextant. The observed altitude should be corrected for errors due to the following causes: 1. Dip. Owing to the observer's elevation above the sea level (on the deck or bridge of the ship), the observed altitude will be too great on account of the dip (or lowering) of the horizon. 2. Index error of sextant. As no instrument is perfect in construction, each one is subject to a certain constant error which is determined by experiment. 3. Refraction of light. Celestial bodies appear higher than they really are because of the refraction of light by the earth's atmosphere. This refraction will depend on the height of the celestial body above the horizon, and also on the state of the barometer and thermometer, since changes in the pressure and temperature of the air affect its density. 4. Semidiameter of the sun. As the observer cannot be sure where the center of the sun is, the altitude of (say) the lower edge of the sun is observed and to that is added the known semidiameter of the sun for that day found from the Nautical Almanac. 5. Parallax. The parallax of a celestial body is the angle subtended by the radius of the earth passing through the observer, as seen from the body. As viewed from the earth's surface, a celestial body appears lower than it would be if viewed from the center, and this may be shown to depend on the parallax of the body. We shall not enter into the detail connected with these corrections, as that had better be left to works on Field Astronomy; our purpose here is merely to call the attention of the student to the necessity of eliminating as far as possible the errors that arise when measuring physical quantities. For the sake of simplicity we have assumed that the necessary corrections have been applied to the data given in the examples found in this book. MISCELLANEOUS EXAMPLES 1. The continent of Asia has nearly the shape of an equilateral triangle. Assuming each side to be 4800 geographical miles and the radius of the earth to be 3440 geographical miles, find the area of Asia. Ans. About 13,333,000 sq. mi. 2. The distance between Paris (lat. 48~ 50' N.) and Berlin (lat. 52~ 30' N.) is 472 geographical miles, measured on the arc of a great circle. What time is it at Berlin when it is noon at Paris? Ans. 44 min. past noon. 3. The altitude of the north pole is 45~, and the azimuth of a star on the horizon is 135~. Find the polar distance of the star. Ans. 60~. APPLICATIONS OF SPHERICAL TRIGONOMETRY 261 4. What will be the altitude of the sun at 9 A.M. in Mexico City (lat. 19025' N.), if its declination at that time is 8~ 23' N.? Ans. 37~ 41'. 5. Find the altitude of the sun at 6 hr. A.M. at Munich (lat. 48~ 9' N.) on the longest day of the year. Ans. Altitude = 17~ 15'. 6. Find the time of day when the sun bears due east and due west on the longest day of the year at St. Petersburg (lat. 59~ 56' N.). Ans. 6 hr. 58 min. A.M., 5 hr. 2 min. P.M. 7. What is the direction of a wall in lat. 52~ 30' N. which casts no shadow at 6 A.M. on the longest day of the year? Ans. 75~ 11', reckoned from the north point of the horizon. 8. Find the latitude of the place at which the sun rises exactly in the northeast on the longest day of the year. Ans. 55~ 45' N. 9. Find the latitude of the place at which the sun sets at 10 hr. P.M. on the longest day. Ans. 63~ 23' N. or S. 10. Given the latitude of the place of observation 52~ 30' N., the declination of a star 38~, its hour angle 28~ 17'. Find the altitude of the star. Ans. Altitude = 65~ 33'. 11. Given the latitude of the place of observation 51~ 19' N., the polar distance of a star 67~ 59', its hour angle 15~ 8'. Find the altitude and azimuth of the star. Ans. Altitude = 580 22', azimuth = 27~ 48'. 12. Given the declination of a star 7~ 54' N., its altitude 22~ 45', its azimuth 50~ 14'. Find the hour angle of the star and the latitude of the observer. Ans. Hour angle = 45~ 41', latitude = 67~ 59' N. 13. The latitude of a star is 51~ N., and its longitude 3150. Find its declination. Take e = 23~ 27'. Ans. Declination = 32~ 23' N. 14. Given the latitude of the observer 440 50' N., the azimuth of a star 41~ 2', its hour angle 20~. Find its declination. Ans. Declination = 20~ 49' N. 15. Given the latitude of the place of observation 51~ 32' N., the altitude of the sun west of the meridian 35~ 15', its declination 21~ 27' N. Find the local time. Ans. 3 hr. 59 min. P.M. CHAPTER IV RECAPITULATION OF FORMULAS SPHERICAL TRIGONOMETRY 42. Right spherical triangles, pp. I96-I97. (1) cos c = (2) sin a = B (3) sin b = (4) cos A = a (5) cos B = (6) cosA = (7 cos B = (8) sin b= (9) Sin a = (10) cos c = COS a cos b, sin c sin A, sin c sin B, cos a sin B, cos b sin A,: tan b cot c,: tan a cot c,: tan a cot A,: tan b cot B, cot A cot B. General directions for solving right spherical triangles by Napier's rules of circular parts are given on p. 200. Spherical isosceles and quadrantal triangles are discussed on p. 204. 43. Relations between the sides and angles of oblique spherical triangles, pp. 206-216. a = 180~ -A, 3 = 1800 - B, y = 180~- C. = - (a + b + c), o- = (+ +/ d = diameter of inscribed circle. 8 = 180~- diameter of circumscribed circle. Law of sines, p. 207. sin a sin b sin c (11) s a = sl sin A sin B sin C sin a sin b sin c or = = - sin ca sin / sin y Law of cosines for the sides, p. 209. (12) cos a = cos b cos c - sin b sin c -7). cos a. 262 RECAPITULATION OF FORMULAS 263 Law of cosines for the angles, p. 209. (15) cos a = cos / cos y - sin f sin y cos a. Functions of a, 3lf, ~ y in terms of the sides, pp. 211-213. (18) sina= sin s sin (s - a) (18) sin - a = i sin c' (19) Cos I= sin(s - b)sin (s - c) (19) CO a2 sin b sin c.(20) tan 17 a= sin s sin (s - a) (20) tan 1 =^ sins - ) sin (s - ) (27) tan d = sin(s-a) sin (s-b)sin(s-c) v/2 s sin s sin (s - a) (28) tan = s tan( - 2 sin(s - h) (29) tan 3 ta =) 2 la' tan Id sin(s - c) (30) tan y= tan ' d Functions of the half sides in terms of a, f, y, p. 214. (31) sin a= sin sin(o — a) (331) ta - -sin sin (32) cos a sin (-a - &)sin(cr -y) (-40)2 sin /3 sinly sin a) (sin e- sin (o - a) (33) ta- m ^ osin(o- - a)sin (o- -$sinlr - y) (40) tan I8 = sin o- sin- ) (41) tan a= sin ( — ) v/2 tan 8 sin (a - /) (42) tan tan = - (v4L3)~ t tan I 8 sin (o- - y) (43) tan a= tanl 2* / 3 tan i l 264 SPHERICAL TRIGONOMETRY Napier's Analogies, p. 215. sin (a +/3) (44) tan (a -b) =- costan 1 c. sin I2(a -- b) (46) tan (a + ) =- + ) tan -. cos (a - b) (46) tan - (a ) = 2 (a b) tany pp. 2 16-227. CASE I. (a) Given the three sides, p. 217. (b) Given the three angles, p. 218. CASE II. (a) Given two sides and their included angle, p. 219. (b) Given two angles and their included side, p. 222. CASE III. (a) Given two sides and the angle opposite one of them, p.224. (b) Given two angles and the side opposite one of them, p. 226. 45. Length of an arc of a circle in linear units, p. 228. (52) L = - 180 N = number of degrees in angle. 46. Area of a spherical triangle, p. 229. (54) Area = 180 E= A + B+C -180~. (55) tan 1 E = tan s tan (s - a) tan (s - b) tan (s-c) FOUR-PLACE TABLES OF LOGARITHMS COMPILED BY WILLIAM ANTHONY GRANVILLE, PH.D. SHEFFIELD SCIENTIFIC SCHOOL, YALE UNIVERSITY GINN AND COMPANY BOSTON * NEW YORK * CHICAGO * LONDON ENTERED AT STATIONERS' HALL COPYRIGHT, 1908, BY WILLIAM ANTHONY GRANVILLE ALL RIGHTS RESERVED 79.3 GtIe gtbeneum Wregs GINN AND COMPANY PROPRIETORS ~ BOSTON U.S.A. CONTENTS PAGES TABLE I. LOGARITHMS OF NUMBERS... 1-5 RULES FOR FINDING THE LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS OF ANGLES NEAR 0~ AND 90~.. 6 TABLE II. LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS, THE ANGLE BEING EXPRESSED IN DEGREES AND MINUTES...... 7-16 CONVERSION TABLES FOR ANGLES..... 17 TABLE III. LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS, THE ANGLE BEING EXPRESSED IN DEGREES AND THE DECIMAL PART OF A DEGREE.... 19-37 TABLE OF NATURAL VALUES OF THE TRIGONOMETRIC FUNCTIONS FOR EVERY DEGREE... 38 111 TABLE I FOUR-PLACE LOGARITHMS OF NUMBERS This table gives the mantissas of the common logarithms (base 10) of the natural numbers (integers) from 1 to 2000, calculated to four places of decimals. A logarithm found from this table by interpolation may be in error by one unit in the last decimal place. 1 2 TABLE I. LOGARITHMS OF NUMBERS 0 No. 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 '131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 No. 0, 0000 1 0004 2 0009 3 0013 4 0017 5 0022 6 0026 7 0030 8 0035 9 9039 0043 0048 0052 0056 0086 0090 0095 0099 0128 0133 0137 0141 0170 0212 0253 0294 0334 0374 0175 0216 0257 0298 0338 0378 0179 0220 0261 0302 0342 0382 0183 0224 0265 0306 0346 0386 0060 0103 0145 0187 0228 0269 0310 0350 0390 0065 0107 0149 0191 0233 0273 0314 0354 0394 0069 0111 0154 0195 0237 0278 0318 0358 0398 0073 0116 0158 0199 0241 0282 0322 0362 0402 0077 0120 0162 0204 0245 0286 0326 0366 0406 0082 0124 0166 0208 0249 0290 0330 0370 0410 0414 0418 0422 0426 0430 0434 0438 0441 0445 0449 0453 0457 0461 0465 0469 0473 0477 0481 0484 0488 0492 0496 0500 0504 0508 0512 0515 0519 0523 0527 0531 0535 0538 0542 0546 0550 0554 0558 0561 0565 0569 0573 0577 0580 0584 0588 0592 0596 0599 0603 0607 0611 0615 0618 0622 0626 0630 0633 0637 0641 0645 0648 0652 0656 0660 0663 0667 0671 0674 0678 0682 0686 0689 0693 0697 0700 0704 0708 0711 0715 0719 0722 0726 0730 0734 0737 0741 0745 0748 0752 0755 0759 0763 0766 0770 0774 0777 0781 0785 0788 0792 0795 0799 0803 0806 0810 0813 0817 0821 0824 0828 0831 0835 0839 0842 0846 0849 0853 0856 0860 0864 0867 0871 0874 0878 0881 0885 0888 0892 0896 0899 0903 0906 0910 0913 0917 0920 0924 0927 0931 0934 0938 0941 0945 0948 0952 0955 0959 0962 0966 0969 0973 0976 0980 0983 0986 0990 0993 0997 1000 1004 1007 1011 1014 1017 1021 1024 1028 1031 1035 1038 1041 1045 1048 1052 1055 1059 1062 1065 1069 1072 1075 1079 1082 1086 1089 1093 1096 1099 1103 1106 1109 1113 1116 1119 1123 1126 1129 1133 1136 1139 1143 1146 1149 1153 1156 1159 1163 1166 1169 1173 1176 1179 1183 1186 1189 1193 1196 1199 1202 1206 1209 1212 1216 1219 1222 1225 1229 1232 1235 1239 1242 1245 1248 1252 1255 1258 1261 1265 1268 1271 1274 1278 1281 1284 1287 1290 1294 1297 1300 1303 1307 1310 1313 1316 1319 1323 1326 1329 1332 1335 1339 1342 1345 1348 1351 1355 1358 1361 1364 1367 1370 1374 1377 1380 1383 1386 1389 1392 1396 1399 1402 1405 1408 1411 1414 1418 1421 1424 1427 1430 1433 1436 1440 1443 1446 1449 1452 1455 1458 1461 1464 1467 1471 1474 1477 1480 1483 1486 1489 1492 1495 1498 1501 1504 1508 1511 1514 1517 1520 1523 1526 1529 1532 1535 1538 1541 1544 1547 1550 1553 1556 1559 1562 1565. 1569 1572 1575 1578 1581 1584 1587 1590 1593 1596 1599 1602 1605 1608 1.611 1614 1617 1620 1623 1626 1629 1632 1635 1638 1641 1644 1647 1649 1652 1655 1658 1661 1664 1667 1670 1673 1676 1679 1682 1685 1688 1691 1694 1697 1700 1703 1706 1708 1711 1714 1717 1720 1723 1726 1729 1732 1735 1738 1741 1744 1746 1749 1752 1755 1758 Prop. Parts -4 - 5 1 0.5 2 1.0 3 1.5 4 2.0 5 2.5 6 3.0 7 3.5 8 4.0 9 4.5 4 1 0.4 2 0.8 3 1.2 4 1.6 5 2.0 6 2.4 7 2.8 8 3.2 9 3.6 3 1 0.3 2 0.6 3 0.9 4 1.2 5 1.5 6 1.8 7 2.1 8 2.4 9 2.7 2 1 0.2 2 0.4 3 0.6 4 0.8 5 1.0 6 1.2 7 1.4 8 1.6 9 1.8 1761 1764 1767 1770 1772 1775 1778 1781 1784 1787 0 1 2 3 4 5 6 7 8 9 I TABLE I. LOGARITHMS OF NUMBERS 3 No. 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 No. 0 1 2 3 4 5 6 7 8 9 Prop. Parts -. —.- ~ — 1761 176411767 1770 1772 1775 17781 17811 17841 1787 1790 1793 1796 1798 1801 1804 1807 1818 1821 1824 1827 1830 1833 1836 1847 1850 1853 1855 1858 1861 1864 1875 1903 1931 1959 1987 2014 1878 1906 1934 1962 1989 2017 1881 1909 1937 1965 1992 2019 1884 1912 1940 1967 1995 2022 1886 1915 1942 1970 1998 2025 1889 1917 1945 1973 2000 2028 1892 192C 194S 1976 2003 203C 1810 1838 1867 1895 1923 1951 1978 2006 2033 1813 1816 1841 1844 1870 1872 1898 1926 1953 1981 2009 2036 1901 1928 1956 1984 2011 2038 *a Ns 1 2 3 4 5 6 7 8 9 o QD 0.9 1.2 1.5 1.8 2.1 2.4 2.7 2041 2044 2047 2049 2052 2055 2057 2060 2063 2066 2068 2071 2074 2076 2079 2082 2084 2087 2090 2092 2095 2098 2101 2103 2106 2109 2111 2114 2117 2119 2122 2125 2127 2130 2133 2135 2138 2140 2143 2146 2148 2151 2154 2156 2159 2162 2164 2167 2170 2172 2175 2177 2180 2183 2185 2188 2191 2193 2196 2198 2201 2204 2206 2209 2212 2214 2217 2219 2222 2225 2227 2230 2232 2235 2238 2240 2243 2245 2248 2251 2253 2256 2258 2261 2263 2266 2269 2271 2274 2276 2279 2281 2284 2287 2289 2292 2294 2297 2299 2302 2304 2307 2310 2312 2315 2317 2320 2322 2325 2327 2330 2333 2335 2338 2340 2343 2345 2348 2350 2353 2355 2358 2360 2363 2365 2368 2370 2373 2375 2378 2380 2383 2385 2388 2390 2393 2395 2398 2400 2403 2405 2408 2410 2413 2415 2418 2420 2423 2425 2428 2430 2433 2435 2438 2440 2443 2445 2448 2450 2453 2455 2458 2460 2463 2465 2467 2470 2472 2475 2477 2480 2482 2485 2487 2490 2492 2494 2497 2499 2502 2504 2507 2509 2512 2514 2516 2519 2521 2524 2526 2529 2531 2533 2536 2538 2541 2543 2545 2548 2550 2553 2555 2558 2560 2562 2565 2567 2570 2572 2574 2577 2579 2582 2584 2586 2589 2591 2594 2596 2598 2601 2603 2605 2608 2610 2613 2615 2617 2620 2622 2625 2627 2629 2632 2634 2636 2639 2641 2643 2646 2648 2651 2653 2655 2658 2660 2662 2665 2667 2669 2672 2674 2676 2679 2681 2683 2686 2688 2690 2693 2695 2697 2700 2702 2704 2707 2709 2711 2714 2716 2718 2721 2723 2725 2728 2730 2732 2735 2737 2739 2742 2744 2746 2749 2751 2753 2755 2758 2760 2762 2765 2767 2769 2772 2774 2776 2778 2781 2783 2785 2788 2790 2792 2794 2797 2799 2801 2804 2806 2808 2810 28 2815 2817 2819 2822 2824 2826 2828 2831 2833 2835 2838 2840 2842 2844 2847 2849 2851 2853 2856 2858 2860 2862 2865 2867 2869 2871 2874 2876 2878 2880 2883 2885 2887 2889 2891 2894 2896 2898 2900 2903 2905 2907 2909 2911 2914 2916 2918 2920 2923 2925 2927 2929 2931 2934 2936 2938 2940 2942 2945 2947 2949 2951 2953 2956 2958 2960 2962 2964 2967 2969 2971 2973 2975 2978 2980 2982 2984 2986 2989 2991 2993 2995 2997 2999 3002 3004 3006 3008 2 1 0.2 2 0.4 3 0.6 4 0.8 5 1.0 6 1.2 7 1.4 8 1.6 9 1.8 3010 3012 3015 3017 3019 3021 3023 3025 3028 3030 0 1 2 3 4 5 6 7 8 9 4 TABLE I. LOGARITHMS OF NUMBERS No. 0-1 2 3 4 5 6 7 8 9 Prop. Parts No.i0'1 2 3 4I5 6 7 8 9 Prop. Partsj 20 21 22 23 24 25.26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 3010 3032 3054 30751 3096 13118 31391 3160 3181 3201 3222 3243 3263 3284 3304 3424 3444 3464 3483 3502 3617 3636 3655 3674 3692. - A r- - I 3802 3979 4150 4314 4472 4624 3820 3997 4166 4330 4487 4639 3838 4014 4183 4346 4502 4654 3856 3874 4031 4048 4200 4216 33z24 3522 3711 3892 4065 4232 4393 4548 4698 3345 3365 3385 3541 3560 3579 3729 3747 3766 3909 4082 4249 4409 4564 4713 3927 4099 4265 4425 4579 4728 3945 4116 4281 4440 4594 4742 3404 3598 3784 3962 4133 4298 4456 4609 4757.,4 o3.,4 d - F1 1 2 3 4 5 -6 7 8 9 Difference 4362 4518 4669 4378 4533 4683 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 4786 4928 5065 5198 5328 5453 5575 5694 5809 5922 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 4814 4955 5092 5224 5353 5478 5599 5717 5832 5944 4829 4969 5105 5237 5366 5490 5611 5729 5843 5955 4843 4983 5119 5250 5378 5502 5623 5740 5855 5966 4857 4997 5132 5263 5391 5514 5635 5752 5866 5977 4871 5011 5145 5276 5403 5527 5647 5763 5877 5988 4886 5024 5159 5289 5416 5539 5658 5775 5888 5999 4900 5038 5172 5302 5428 5551 5670 5786 5900 6010 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 20 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 19 1.9 3.8 5.7 7.6 9.5 11.4 13.3 15.2 17.1 6021 6128 6232 6335 6435 6532 6628 6721 6812 6902 6031 6138 6243 6345 6444 6542 6637 6730 6821 6911 6042 6149 6253 6355 6454 6551 6646 6739 6830 6920 6053 6160 6263 6365 6464 6561 6656 6749 6839 6928 6064 6170 6274 6375 6474 6571 6665 6758 6848 6937 6075 6180 6284 6385 6484 6580 6675 6767 6857 6946 6085 6191 6294 6395 6493 6590 6684 6776 6866 6955 6096 6201 6304 6405 6503 6599 6693 6785 6875 6964 6107 6212 6314 6415 6513 6609 6702 6794 6884 6972 6117 6222 6325 6425 6522 6618 6712 6803 6893 6981 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 8261 8267 8274 8280 8287 829823 8299 8306 8312 8319 8325 8331 8338 8344 8351 '8357 8363 8370 8376 8382 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 -7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 14 1.4 2.8 4.2 5.6 7.0 8.4 9.8 11.2 12.6 12 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 &Sf1 1 -457t 5 /63 0 4 /U 6/O 83'3Zl 84+~ &V94y 853.UU 05UO No. 0 1 2 3 4 5 6 7 8 9 imiil TABLE I. LOGARITHMS OF NUMBERS 5 iNo. 0 1 7Part No. 0 i 1 2 3 4 5 6 7 8 9 Prop. Parts 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 84511 84571 84631 8470 8476 8482 84 8488494 8500 8506 85.13 8573 8633 8692 8751 8808 8865 8921 8976 8519 8579 8639 8698 8756 8814 8871 8927 8982 8525 8585 8645 8704 8762 8820 8876 8932 8987 8531 8591 8651 8710 8768 8825 8882 8938 8993 8537 8597 8657 8716 8774 8831 8887 8943 8998 8543 8603 8663 8722 8779 8837 8893 8949 9004 8549 8609 8669 8727 8785 8842 8899 8954 9009 8555 8615 8675 8733 8791 8848 8904 8960 9015 8561 8621 8681 8739 8797 8854 8910 8965 9020 8567 8627 8686 8745 8802 8859 8915 8971 9025 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 Ad, AesA n FRA AL~ Ex. dig. 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 Difference 10 9 1.0 0.9 2.0 1.8 3.0 2.7 4.0 3.6 5.0 4;5 6.0 5.4 7.0 6.3 8.0 7.2 9.0 8.1 8 7 0.8 0.7 1.6 1.4 2.4 2.1 3.2 2.8 4.0 3.5 4.8 4.2 5.6 4.9 6.4 5.6 7.2 6.3 6 5 0.6 0.5 1.2 1.0 1.8 1.5 2.4 2.0 3.0 2.5 3.6 3.0 4.2 3.5 4.8 4.0 5.4 4.5 4 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 1 2 3 4 5 6 7 8 9 00001 0004 00091 00131 0017 0022 0026 00301 00351 0039 No.0 1 2 3 4 5 6 7 8 9.,,., RULES FOR FINDING THE LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS OF ANGLES NEAR 0~ AND 90~ The derivation of the following rules will be found on page 182, Granville's Plane Trigonometry. If the angle is given in degrees, minutes, and seconds, it should first be reduced to degrees and the decimal part of a degree. For this purpose use the conversion table on page 17. Rule I. To find the Logarithms of the Functions of an Angle near 0~.* log sin x~ = 2.2419 + log x. log tan x~ = 2.2419 + log x. log cot x~ = 1.7581 - log x. log cos x~ is found from the tables in the usual way. Rule II. To find the Logarithms of the Functions of an Angle near 90~.-t log cos x = 2.2419 + log (90 - x). log cot x~ = 2.2419 + log (90- x). log tanx~ = 1.7581 - log (90- x). log sin x~ is found from the tables in the usual way. These rules will give results accurate to four decimal places for all angles between 0~ and 1.1~ and between 88.9~ and 90~. * Example 1, page 182, Granville's Plane Trigonometry, illustrates the application of this rule. t Example 2, page 183, Granville's Plane Trigonometry, illustrates the application of this rule. 6 TABLE II FOUR-PLACE LOGARITHMS OF TRIGONOMETRIC FUNCTIONS, THE ANGLE BEING EXPRESSED IN DEGREES AND MINUTES This table gives the common logarithms (base 10) of the sines, cosines, tangents, and cotangents of all angles from 0~ to 5~ and from 85~ to 90~ for each minute; and from 5~ to 85~ at intervals of 10 minutes, all calculated to four places of decimals. In order to avoid the printing of negative characteristics, the number 10 has been added to every logarithm in the first, second, and fourth columns (those having log sin, log tan, and log cos at the top). Hence in writing down any logarithm taken from these three columns - 10 should be written after it. Logarithms taken from the third column (having log cot at the top) should be used as printed. A logarithm found from this table by interpolation may be in error by one unit in the last decimal place, except for angles between 0~ and 18' or between 89~ 42' and 90~, when the error may be larger. In the latter cases the table refers the student to the formulas on page 6 for more accurate results. 7 8 TABLE II. LOGARITHMIC SINES '~__ 0~ Angle log sin diff. l' log tan | dOi log cot I log cos m 0~ 0' 00 1' 00 2' 00 3' 0~ 4' 0~ 5' 00 6' 00 7' 00 8' 0~ 9' 0~ 10' 0~ 11' 0~ 12' 0~ 13' 00 14' 00 15' 0~ 16' 00 17' 0~ 18' 0~ 19' 0~ 20' 0~ 21' 0~ 22' 00 23' 0~ 24' 0~ 25' 0~ 26' 0~ 27' 0~ 28' 00 29' 00 30' 0~ 31' 00 32' 00 33' 00 34' 00 35' 0~ 36' 00 37' 00 38' 00 39' 0~ 40' 00 41' 00 42' 00 43' 00 44' 00 45' 00 46' 00 47' 00 48' 00 49' 0~ 50' 00 51' 00 52' 0~ 53' 00 54' 00 55' 00 56' 00 57' 00 58' 00 59' 00 60' 6.4637 6.7648 6.9408 7.0658 7.1627 7.2419 7.3088 7.3668 7.4180 7.4637 7.5051 7.5429 7.5777 7.6099 7.6398 7.6678 7.6942 7.7190 7.7425 7.7648 7.7859 7.8061 7.8255 7.8439 7.8617 7.8787 7.8951 7.9109 7.9261 7.9408 7.9551 7.9689 7.9822 7.9952 8.0078 8.0200 8.0319 8.0435 8.0548 8.0658 8.0765 8.0870 8.0972 8.1072 8.1169 8.1265 8.1358 8.1450 8.1539 8.1627 8.1713 8.1797 8.1880 8.1961 8.2041 8.2119 8.2196 8.2271 8.2346 8.2419.35 2 co 223 211 202 194 184 178 170 164 158 152 147 143 138 133 130 126 122 119 116 113 110 107 105 102 100 97 96 93 92 89 88 86 84 83 81 80 78 77 75 75 73.~ O c3 l 1 c s A- Q *H U 6.4637 6.7648 6.9408 7.0658 7.1627 7.2419 7.3088 7.3668 7.4180 7.4637 7.5051 7.5429 7.5777 7.6099 7.6398 7.6678 7.6942 7.7190 7.7425 7.7648 7.7860 7.8062 7.8255 7.8439 7.8617 7.8787 7.8951 7.9109 7.9261 7.9409 7.9551 7.9689 7.9823 7.9952 8.0078 8.0200 8.0319 8.0435 8.0548 8.0658 8.0765 8.0870 8.0972 8.1072 8.1170 8.1265 8.1359 8.1450 8.1540 8.1627 8.1713 8.1798 8.1880 8.1962 8.2041 8.2120 8.2196 8.2272 8.2346 8.2419...20 - L a *.,ct oF;H ol 235 223 212 202 193 184 178 170 164 158 152 148 142 138 134 129 126 122 119 116 113 110 107 105 102 100 98 95 94 91 90 87 86 85 82 82 79 79 76 76 74 73 3.5363 3.2352 3.0592 2.9342 2.8373 2.7581 2.6912 2.6332 2.5820 2.5363 2.4949 2.4571 2.4223 2.3901 2.3602 2.3322 2.3058 2.2810 2.2575 2.2352 2.2140 2.1938 2.1745 2.1561 2.1383 2.1213 2.1049 2.0891 2.0739 2.0591 2.0449 2.0311 2.0177 2.0048 1.9922 1.9800 1.9681 1.9565 1.9452 1.9342 1.9235 1.9130 1.9028 1.8928 1.8830 1.8735 1.8641 1.8550 1.8460 1.8373 1.8287 1.8202 1.8120 1.8038 1.7959 1.7880 1.7804 1.7728 1.7654 1.7581 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 90~ 00' 89~ 59' 89~ 58' 89~ 57' 89~ 56' 89~ 55' 89~ 54' 89~ 53' 89~ 52' 89~ 51' 89~ 50' 89~ 49' 89~ 48' 89~ 47' 89~ 46' 89~ 45' 89~ 44' 89~ 43' 89~ 42' 89~ 41' 890 40' 89~ 39' 89~ 38' 89~ 37' 89~ 36' 89~ 35' 89~ 34' 89~ 33' 89~ 32' 89~ 31' 89~ 30' 89~ 29' 89~ 28' 89~ 27' 89~ 26' 89~ 25' 89~ 24' 89~ 23' 89~ 22' 89~ 21' 89~ 20' 89~ 19' 89~ 18' 89~ 17' 89~ 16' 89~ 15' 89~ 14' 89~ 13' 89~ 12' 89~ 11' 890 10' 89~ 9' 89~ 8' 89~ 7 89~ 6' 89~ 5' 89~ 4' 89~ 3' 89~ 2' 89~ 1' 89~ 0' log cos diff. 'log cot sdjifi, log tan log sin Angle 89~ 0 COSINES, TANGENTS, AND COTANGENTS 9 ' 1~ Angle flog sin diff. log tan di log cot log cos I~ ~~~~o t a n!- diff, 1' 0 o 0 O 1~ 0' 1~ 1' 1~ 2' 1~ 3' 1~ 4' 1~ 5' 1~ 6' 1~ 7' 1~ 8' 1~ 9' 1~ 10' 1~ 11' 1~ 12' 1~ 13' 1~ 14' 1~ 15' 1~ 16' 1~ 17' 1~ 18' 1~ 19' 1~ 20' 1~ 21' 1~ 22' 1~ 23' 1~ 24' 1~ 25' 1~ 26' 1~ 27' 1~ 28' 1~ 29' 1~ 30' 10 31' 1~ 32' 1~ 33' 1~ 34' 1~ 35' 1~ 36' 1~ 37' 1~ 38' 1~ 39' 1~ 40' 1~ 41' 1~ 42' 1~ 43' 1~ 44' 1~ 45' 1~ 46' 1~ 47' 1~ 48' 1~ 49' 1~ 50' 10 51' 1~ 52' 1~ 53' 1~ 54' 1~ 55' 1~ 56' 1~ 57' 1~ 58' 1~ 59' 1~ 60' 8.2419 8.2490 8.2561 8.2630 8.2699 8.2766 8.2832 8.2898 8.2962 8.3025 8.3088 8.3150 8.3210 8.3270 8.3329 8.3388 8.3445 8.3502 8.3558 8.3613 8.3668 8.3722 8.3775 8.3828 8.3880 8.3931 8.3982 8.4032 8.4082 8.4131 8.4179 8.4227 8.4275 8.4322 8.4368 8.4414 8.4459 8.4504 8.4549 8.4593 8.4637 8.4680 8.4723 8.4765 8.4807 8.4848 8.4890 8.4930 8.4971 8.5011 8.5050 8.5090 8.5129 8.5167 8.5206 8.5243 8.5281 8.5318 8.5355 8.5392 8.5428 71 71 69 69 67 66 66 64 63 63 62 60 60 59 59 57 57 56 55 55 54 53 53 52 51 51 50 50 49 49 48 48 47 46 46 45 45 45 44 44 43 43 42 42 41 42 40 41 40 39 40 39 38 39 37 38 37 37 37 36 8.2419 8.2491 8.2562 8.2631 8.2700 8.2767 8.2833 8.2899 8.2963 8.3026 8.3089 8.3150 8.3211 8.3271 8.3330 8.3389 8.3446 8.3503 8.3559 8.3614 8.3669 8.3723 8.3776 8.3829 8.3881 8.3932 8.3983 8.4033 8.4083 8.4132 8.4181 8.4229 8.4276 8.4323 8.4370 8.4416 8.4461 8.4506 8.4551 8.4595 8.4638 8.4682 8.4725 8.4767 8.4809 8.4851 8.4892 8.4933 8.4973 8.5013 8.5053 8.5092 8.5131 8.5170 8.5208 8.5246 8.5283 8.5321 8.5358 8.5394 8.5431 72 71 69 69 67 66 66 64 63 63 61 61 60 59 59 57 56 56 55 55 54 53 53 52 51 51 50 50 49 49 48 47 47 47 46 45 45 45 44 43 44 43 42 42 42 41 41 40 40 40 39 39 39 38 38 37 38 37 36 37 1.7581 1.7509 1.7438 1.7369 1.7300 1.7233 1.7167 1.7101 1.7037 1.6974 1.6911 1.6850 1.6789 1.6729 1.6670 1.6611 1.6554 1.6497 1.6441 1.6386 1.6331 1.6277 1.6224 1.6171 1.6119 1.6068 1.6017 1.5967 1.5917 1.5868 1.5819 1.5771 1.5724 1.5677 1.5630 1.5584 1.5539 1.5494 1.5449 1.5405 1.5362 1.5318 1.5275 1.5233 1.5191 1.5149 1.5108 1.5067 1.5027 1.4987 1.4947 1.4908 1.4869 1.4830 1.4792 1.4754 1.4717 1.4679 1.4642 1.4606 1.4569 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9997 9.9997 9.9997 9.9997 88~ 60' 88~ 59' 88~ 58' 88~ 57' 88~ 56' 88~ 55' 88~ 54' 88~ 53' 88~ 52' 88~ 51' 880 50' 88~ 49' 88~ 48' 88~ 47' 88~ 46' 88~ 45' 88~ 44' 88~ 43' 88~ 42' 88~ 41' 88~ 40' 88~ 39' 88~ 38' 88~ 37' 88~ 36' 88~ 35' 88~ 34' 88~ 33' 88~ 32' 88~ 31' 88~ 30' 88~ 29' 88~ 28' 88~ 27' 88~ 26' 88~ 25' 88~ 24' 88~ 23' 88~ 22' 88~ 21' 88~ 20' 88~ 19' 88~ 18' 88~ 17' 88~ 16' 88~ 15' 88~ 14' 88~ 13' 88~ 12' 88~ 11' 88~ 10' 88~ 9' 88~ 8' 88~ 7' 88~ 6' 88~ 5' 88~ 4' 88~ 3' 88~ 2' 88~ 1' 88~ 0' I I I I I I log cos diff. 1' log cot di. l' log tan log sin Angle 88~ 10 TABLE II. LOGARITHMIC SINES Angle log sin diff. 'log tan diffm log cot log cos 2~ 0' 2~ 1' 2~ 2' 2~ 3' 2~ 4/ 2~ 5' 2~ 6' 2~ 7' 2~ 8' 2~ 9' 2~ 10' 2~ 11' 2~ 12' 2~ 13' 2~ 14' 2~ 15' 2~ 16" 2~ 17' 2~ 18/ 2~ 19' 2~ 20' 2~ 21' 2~ 22' 2~ 23' 2~ 24' 2~ 25' 2~ 26' 2~ 27' 2~ 28' 2~ 29' 20 30' 2~ 31' 2~ 32' 2~ 33' 2~ 34' 2~ 35' 2~ 36' 2~ 37' 2~ 38' 2~ 39' 20 40' 2~ 41' 2~ 42' 2~ 43' 2~ 44' 2~ 45' 2~ 46' 2~ 47' 2~ 48' 2~ 49' 2~ 50' 2~ 51' 2~ 52' 2~ 53' 2~ 54' 2~ 55' 2~ 56' 2~ 57' 2~ 58' 2~ 59' 20 60' 8.5428 8.5464 8.5500 8.5535 8.5571 8.5605 8.5640 8.5674 8.5708 8.5742 8.5776 8.5809 8.5842 8.5875 8.5907 8.5939 8.5972 8.6003 8.6035 8.6066 8.6097 8.6128 8.6159 8.6189 8.6220 8.6250 8.6279 8.6309 8.6339 8.6368 8.6397 8.6426 8.6454 8.6483 8.6511 8.6539 8.6567 8.6595 8.6622 8.6650 8.6677 8.6704 8.6731 8.6758 8.6784 8.6810 8.6837 8.6863 8.6889 8.6914 8.6940 8.6965 8.6991 8.7016 8.7041 8.7066 8.7090 8.7115 8.7140 8.7164 8.7188 36 36 35 36 34 35 34 34 34 34 33 33 33 32 32 33 31 32 31 31 31 31 30 31 30 29 30 30 29 29 29 28 29 28 28 28 28 27 28 27 27 27 27 26 26 27 26 26 25 26 25 26 25 25 25 24 25 25 24 24 8.5431 8.5467 8.5503 8.5538 8.5573 8.5608 8.5643 8.5677 8.5711 8.5745 8.5779 8.5812 8.5845 8.5878 8.5911 8.5943 8.5975 8.6007 8.6038 8.6070 8.6101 8.6132 8.6163 8.6193 8.6223 8.6254 8.6283 8.6313 8.6343 8.6372 8.6401 8.6430 8.6459 8.6487 8.6515 8.6544 8.6571 8.6599 8.6627 8.6654 8.6682 8.6709 8.6736 8.6762 8.6789 8.6815 8.6842 8.6868 8.6894 8.6920 8.6945 8.6971 8.6996 8.7021 8.7046 8.7071 8.7096 8.7121 8.7145 8.7170 8.7194 36 36 35 35 35 35 34 34 34 34 33 33 33 33 32 32 32 31 32 31 31 31 30 30 31 29 30 30 29 29 29 29 28 28 29 27 28 28 27 28 27 27 26 26 27 26 26 26 25 26 25 25 25 25 25 25 24 25 24 1.4569 1.4533 1.4497 1.4462 1.4427 1.4392 1.4357 1.4323 1.4289 1.4255 1.4221 1.4188 1.4155 1.4122 1.4089 1.4057 1.4025 1.3993 1.3962 1.3930 1.3899 1.3868 1.3837 1.3807 1.3777 1.3746 1.3717 1.3687 1.3657 1.3628 1.3599 1.3570 1.3541 1.3513 1.3485 1.3456 1.3429 1.3401 1.3373 1.3346 1.3318 1.3291 1.3264 1.3238 1.3211 1.3185 1.3158 1.3132 1.3106 1:3080 1.3055 1.3029 1.3004 1.2979 1.2954 1.2929 1.2904 1.2879 1.2855 1.2830 1.2806 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 870 60' 87~ 59' 87~ 58' 87~ 57' 87~ 56' 87~ 55' 87~ 54' 87~ 53' 87~ 52' 87~ 51' 87~ 50' 87~ 49' 87~ 48' 87~ 47' 87~ 46' 87~ 45' 87~ 44' 87~ 43' 87~ 42' 87~ 41' 87~ 40/ 87~ 39' 87~ 38' 87~ 37' 87~ 3'6 87~ 35' 87~ 34' 87~ 33' 87~ 32' 87~ 31' 87~ 30/ 87~ 29' 87~ 28' 87~ 27' 87~ 26' 87~ 25' 87~ 24' 87~ 23' 87~ 22' 87~ 21' 87020/ 87~ 19' 87~ 18' 87~ 17' 87~ 16' 87~ 15' 87~ 14' 87~ 13' 87~ 12' 87~ 11' 87~ 10' 87~ 9' 87~ 8' 87~ 7' 87~ 6' 87~ 5' 87~ 4' 87~ 3Y 87~ 2' 87~ 1' 87~ 0' log cos diff. I log cot sd j' log tan log sin Angle 87~ COSINES, TANGENTS, AND COTANGENTS 11 30 0 Angle 30 0' 30 1' 30 2' 30 3/ 30 4' 30 5/ 30 6' 30 7' 30 8' 30 9/ 30 10' 30 11' 30 12' 30 13' 30 14' 30 15' 30 16' 30 17' 30 18' 30 19' 30 20' 30 21' 30 22' 30 23' 3~ 24' 30 25' 30 26' 30 27' 30 28' 30 29' 30 30' 30 31' 30 32' 30 33' 30 34' 30 35' 30 36' 30 37' 30 38' 30 39' 3~ 40' 30 41' 30 42' 30 43' 30 44' 3~ 45' 30 46' 30 47' 30 48' 30 49' 30 50' 30 51' 30 52' 30 53' 30 54' 30 55' 30 56' 3~ 57' 30 58' 30 59' 30 60' log sin 8.7188 8.7212 8.7236 8.7260 8.7283 8.7307 8.7330 8.7354 8.7377 8.7400 8.7423 8.7445 8.7468 8.7491 8.7513 8.7535 8.7557 8.7580 8.7602 8.7623 8.7645 8.7667 8.7688 8.7710 8.7731 8.7752 8.7773 8.7794 8.7815 8.7836 8.7857 8.7877 8.7898 8.7918 8.7939 8.7959 8.7979 8.7999 8.8019 8.8039 8.8059 8.8078 8.8098 8.8117 8.8137 8.8156 8.8175 8.8194 8.8213 8.8232 8.8251 8.8270 8.8289 8.8307 8.8326 8.8345 8.8363 8.8381 8.8400 8.8418 8.8436 diff.l' 24 24 24 23 24 23 24 23 23 23 22 23 23 22 22 22 23 22 21 22 22 21 22 21 21 21 21 21 21 21 20 21 20 21 20 20 20 20 20 20 19 20 19 20 19 19 19 19 19 19 19 19 18 19 19 18 18 19 18 18 log tan 8.7194 8.7218 8.7242 8.7266 8.7290 8.7313 8.7337 8.7360 8.7383 8.7406 8.7429 8.7452 8.7475 8.7497 8.7520 8.7542 8.7565 8.7587 8.7609 8.7631 8.7652 8.7674 8.7696 8.7717 8.7739 8.7760 8.7781 8.7802 8.7823 8.7844 8.7865 8.7886 8.7906 8.7927 8.7947 8.7967 8.7988 8.8008 8.8028 8.8048 8.8067 8.8087 8.8107 8.8126 8.8146 8.8165 8.8185 8.8204 8.8223 8.8242 8.8261 8.8280 8.8299 8.8317 8.8336 8.8355 8.8373 8.8392 8.8410 8.8428 8.8446 I 24 24 24 24 23 24 23 23 23 23 23 23 22 23 22 23 22 22 22 21 22 22 21 22 21 21 21 21 21 21 21 20 21 20 20 21 20 20 20 20 20 20 19 20 19 20 19 19 19 19 19 19 18 19 19 18 19 18 18 18 do iflm log cot diff. 1' 1.2806 1.2782 1.2758 1.2734 1.2710 1.2687 1.2663 1.2640 1.2617 1.2594 1.2571 1.2548 1.2525 1.2503 1.2480 1.2458 1.2435 1.2413 1.2391 1.2369 1.2348 1.2326 1.2304 1.2283 1.2261 1.2240 1.2219 1.2198 1.2177 1.2156 1.2135 1.2114 1.2094 1.2073 1.2053 1.2033 1.2012 1.1992 1.1972 1.1952 1.1933 1.1913 1.1893 1.1874 1.1854 1.1835 1.1815 1.1796 1.1777 1.1758 1.1739 1.1720 1.1701 1.1683 1.1664 1.1645 1.1627 1.1608 1.1590 1.1572 1.1554 log cos 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9989 860 60' 86~ 59' 860 58' 86~ 57' 86~ 56' 860 55' 86~ 54' 86~ 53' 86~ 52' 860 51' 860 50' 86~ 49' 86~ 48' 86~ 47' 86~ 46' 86~ 45' 860 44' 86~ 43' 86~ 42' 860 41' 86~ 40' 860 39' 86~ 38' 86~ 37' 860 36' 86~ 35' 86~ 34' 860 33' 860 32' 86~ 31' 86~ 30' 86~ 29' 86~ 28' 86~ 27' 86~ 26' 860 25' 860 24' 860 23' 86~ 22' 86~ 21' 860 20' 86~ 19' 86~ 18' 86~ 17' 860 16' 86~ 15' 860 14' 86~ 13' 86~ 12' 86~ 11' 86~ 10' 86~ 9' 86~ 8' 86 7' 860 6' 860 5' 860 4' 860 3/ 860 2' 860 1' 860 0' log cos diff. ' log cot dicf.1, log tan log sin Angle 860 12 TABLE II. LOGARITHMIC SINES 40 0 Angle log sin diff. i| log tan cot log og cos |o l l l l l l~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 4~ 0' 40 1' 40 2' 4~ 3' 4~ 4' 4~ 5' 4~ 6' 4~ 7/ 40 8' 40 9' 4~ 10' 40 11' 4~ 12' 4~ 13' 4~ 14' 4~ 15' 4~ 16' 4~ 17' 4~ 18' 4~ 19' 40 20' 4~ 21' 4~ 22' 4~ 23' 4~ 24' 4~ 25' 4~ 26' 4~ 27'. 4~ 28' 4~ 29' 40 30' 4~ 31' 4~ 32' 4~ 33' 4~ 34' 4~ 35' 4~ 36' 4~ 37' 4~ 38' 4~ 39' 4~ 40' 4~ 41' 40 42' 4~ 43' 4~ 44' 4~ 45' 4~ 46' 4~ 47' 4~ 48' 4~ 49' 4~ 50' 4~ 51' 4~ 52' 4~ 53' 4~ 54' 4~ 55' 4~ 56' 4~ 57' 4~ 58' 4~ 59' 4~ 60' 8.8436 8.8454 8.8472 8.8490 8.8508 8.8525 8.8543 8.8560 8.8578 8.8595 8.8613 8.8630 8.8647 8.8665 8.8682 8.8699 8.8716 8.8733 8.8749 8.8766 8.8783 8.8799 8.8816 8.8833 8.8849 8.8865 8.8882 8.8898 8.8914 8.8930 8.8946 8.8962 8.8978 8.8994 8.9010 8.9026 8.9042 8.9057 8.9073 8.9089 8.9104 8.9119 8.9135 8.9150 8.9166 8.9181 8.9196 8.9211 8.9226 8.9241 8.9256 8.9271 -8.9286 8.9301 8.9315 8.9330 8.9345 8.9359 8.9374 8.9388 8.9403 18 18 18 18 17 18 17 18 17 18 17 17 18 17 17 17 17 16 17 17 16 17 17 16 16 17 16 16 16 16 16 16 16 16 16 16 15 16 16 15 15 16 15 16 15 15 15 15 15 15 15 15 15 14 15 15 14 15 15 15 8.8446 8.8465 8.8483 8.8501 8.8518 8.8536 8.8554 8.8572 8.8589 8.8607 8.8624 8.8642 8.8659 8.8676 8.8694 8.8711 8.8728 8.8745 8.8762 8.8778 8.8795 '8.8812 8.8829 8.8845 8.8862 8.8878 8.8895 8.8911 8.8927 8.8944 8.8960 8.8976 8.8992 8.9008 8.9024 8.9040 8.9056 8.9071 8.9087 8.9103 8.9118 8.9134 8.9150 8.9165 8.9180 8.9196 8.9211 8.9226 8.9241 8.9256 8.9272 8.9287 8.9302 8.9316 8.9331 8.9346 8.9361 '8.9376 8.9390 8.9405 8.9420 19 18 18 17 18 18 18 17 18 17 18 17 17 18 17 17 17 17 16 17 17 17 16 17 16 17 16 16 17 16 16 16 16 16 16 16 15 16 16 15 16 16 15 15 16 15 15 15 15 16 15 15 14 15 15 15 15 14 15 15 1.1554 1.1535 1.1517 1.1499 1.1482 1.1464 1.1446 1.1428 1.1411 1.1393 1.1376 1.1358 1.1341 1.1324 1.1306 1.1289 1.1272 1.1255 1.1238 1.1222 1.1205 1.1188 1.1171 1.1155 1.1138 1.1122 1.1105 1.1089 1.1073 1.1056 1.1040 1.1024 1.1008 1.0992 1.0976 1.0960 1.0944 1.0929 1.0913 1.0897 1.0882 1.0866 1.0850 1.0835 1.0820 1.0804 1.0789 1.0774 1.0759 1.0744 1.0728 1.0713 1.0698 1.0684 1.0669 1.0654 1.0639 1.0624 1.0610 1.0595 1.0580 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9983 85~ 60' 85~ 59' 85~ 58' 85~ 57' 85~ 56' 85~ 55' 85~ 54' 85~ 53' 85~ 52' 85~ 51' 85~ 50' 85~ 49' 85~ 48' 85~ 47' 85~ 46' 85~ 45' 85~ 44' 85~ 43' 85~ 42' 85~ 41' 85~ 40' 85~ 39' 85~ 38' 85~ 37' 85~ 36' 85~ 35' 85~ 34' 85~ 33' 85~ 32' 85~ 31' 85~ 30' 85~ 29' 85~ 28' 85~ 27' 85~ 26' 85~ 25' 85~ 24' 85~ 23' 85~ 22' 85~ 21' 85~ 20' 85~ 19' 85~ 18' 85~ 17' 85~ 16' 85~ 15' 85~ 14' 85~ 13' 85~ 12' 85~ 11' 85~ 10' 85~ 9' 85~ 8' 85~ 7' 85~ 6' 85~ 5' 85~ 4' 850 3' 85~ 2' 85~ 1' 85~ 0' _ _ log cos di-f. 1j log cot |om' log tlog sin Angle 85~ 0 COSINES, TANGENTS, AND COTANGENTS 13 5~- 150 Angle log sin diff. 1' log tan log co cot I log cos diff. m 50 0O 50 10' 50 20' 5~ 30' 5~ 40' 50 50' 6~ 0' 60 10' 6~ 20' 60 30' 60 40' 60 50' 70 0' 70 10' 7~ 20' 7~ 30' 7~ 40' 70 50' 80 0O 80 10' 80' 20' 8~ 30' 8~ 40' 8~ 50' 90 0' 90 10' 90 20' 90 30' 90 40' 90 50' 100 0' 10~ 10' 100 20' 100 30' 100 40' 10~ 50' 110 0O 110 10' 110 20' 110 30' 110 40' 110 50' 120 0' 12~ 10' 120 20' 120 30' 120 40' 120 50' 130 0' 130 10' 13~ 20' 130 30' 130 40' 13~ 50' 14~ 0' 14~ 10' 14~ 20' 140 30' 140 40' 140 50' 150 0' 8.9403 8.9545 8.9682 8.9816 8.9945 9.0070 9.0192 9.0311 9.0426 9.0539 9.0648 9.0755 9.0859 9.0961 9.1060 9.1157 9.1252 9.1345 9.1436 9.1525 9.1612 9.1697 9.1781 9.1863 9.1943 9.2022 9.2100 9.2176 9.2251 9.2324 9.2397 9.2468 9.2538 9.2606 9.2674 9.2740 9.2806 9.2870 9.2934 9.2997 9.3058 9.3119 9.3179 9.3238.9.3296 9.3353 9.3410 9.3466 9.3521 9.3575 9.3629 9.3682 9.3734 9.3786 9.3837 9.3887 9.3937 9.3986 9.4035 9.4083 9.4130 14.2 13.7 13.4 12.9 12.5 12.2 11.9 11.5 11.3 10.9 10.7 10.4 10.2 9.9 9.7 9.5 9.3 9.1 8.9 8.7 8.5 8.4 8.2 8.0 7.9 7.8 7.6 7.5 7.3 7.3 7.1 7.0 6.8 6.8 6.6 6.6 6.4 6.4 6.3 6.1 6.1 6.0 5.9 5.8 5.7 5.7 5.6 5.5 5.4 5.4 5.3 5.2 5.2 5.1 5.0 5.0 4.9 4.9 4.8 4.7 8.9420 8.9563 8.9701 8.9836 8.9966 9.0093 9.0216 9.0336 9.0453 9.0567 9.0678 9.0786 9.0891 9.0995 9.1096 9.1194 9.1291 9.1385 9.1478 9.1569 9.1658 9.1745 9.1831 9.1915 9.1997 9.2078 9.2158 9.2236 9.2313 9.2389 9.2463 9.2536 9.2609 9.2680 9.2750 9.2819 9.2887 9.2953 9.3020 9.3085 9.3149 9.3212 9.3275 9.3336 9.3397 9.3458 9.3517 9.3576 9.3634 9.3691 9.3748 9.3804 9.3859 9.3914 9.3968 9.4021 9.4074 9.4127 9.4178 9.4230 9.4281 14.3 13.8 -13.5 13.0 12.7 12.3 12.0 11.7 11.4 11.1 10.8 10.5 10.4 10.1 9.8 9.7 9.4 9.3 9.1 8.9 8.7 8.6 8.4 8.2 8.1 8.0 7.8 7.7 7.6 7.4 7.3 7.3 7.1 7.0 6.9 6.8 6.6 6.7 6.5 6.4 6.3 6.3 6.1 6.1 6.1 5.9 5.9 5.8 5.7 5.7 5.6 5.5 5.5 5.4 5.3 5.3 5.3 5.1 5.2 5.1 1.0580 1.0437 1.0299 1.0164 1.0034 0.9907 0.9784 0.9664 0.9547 0.9433 0.9322 0.9214 0.9109 0.9005 0.8904 0.8806 0.8709 0.8615 0.8522 0.8431 0.8342 0.8255 0.8169 0.8085 0.8003 0.7922 0.7842 0.7764 0.7687 0.7611 0.7537 0.7464 0.7391 0.7320 0.7250 0.7181 0.7113 0.7047 0.6980 0.6915 0.6851 0.6788 0.6725 0.6664 0.6603 0.6542 0.6483 0.6424 0.6366 0.6309 0.6252 0.6196 0.6141 0.6086 0.6032 0.5979 0.5926 0.5873 0.5822 0.5770 0.5719 9.9983 9.9982 9.9981 9.9980 9.9979 9.9977 9.9976 9.9975 9.9973 9.9972 9.9971 9.9969 9.9968 9.9966 9.9964 9.9963 9.9961 9.9959 9.9958 9.9956 9.9954 9.9952 9.9950 9.9948 9.9946 9.9944 9.9942 9.9940 9.9938 9.9936 9.9934 9.9931 9.9929 9.9927 9.9924 9.9922 9.9919 9.9917 9.9914 9.9912 9.9909 9.9907 9.9904 9.9901 9.9899 9.9896 9.9893 9.9890 9.9887 9.9884 9.9881 9.9878 9.9875 9.9872 9.9869 9.9866 9.9863 9.9859 9.9856 9.9853 9.9849.1.1.1.1.2.1.1.2.1.1.2.1.2.2.1.2.2.1.2.2.2.2.2.2.2.2.2.2.2.2.3.2.2.3.2.3.2.3.2.3.2.3.3.2.3.3.3.3.3.3.3.3.3.3.3.3.4.3.3.4 85~ 0' 84~ 50' 840 40' 84~ 30' 840 20' 84~ 10' 84~ 0' 830 50' 83~ 40' 830 30' 830 20' 830 10' 830 0' 82~ 50' 820 40' 820 30' 820 20' 820 10' 82~ 0' 810 50' 810 40' 810 30' 810 20' 810 10' 81~ 0' 800 50' 80~ 40' 80~ 30' 800 20' 800 10' 80~ 0' 790 50' 790 40' 790-30' 790 20' 790 10' 790 0' 780 50' 780 40' 78~ 30' 780 20' 780 10' 780 0' 770 50' 770 40' 770 30' 770 20' 770 10' 770 0' 760 50' 760 40' 76~ 30' 760 20' 760 10' 76~ 0' 750 50' 750 40' 750 30' 75~ 20' 750 10' 750 0' I - I I log 005 diff. iI log cost diOM.1 log tan log sin diff. 1' Angle 75~-850 I 14 TABLE II. LOGARITHMIC SINES m 15~-25o Angleog sin d log otan i lg c log co lg s I diff. 1' 15~ 0' 15~ 10' 15~ 20' 15~ 30' 15~ 40' 15~ 50' 160 0' 16~ 10' 16~ 20' 16~ 30' 16~ 40' 16~ 50' 17~ 0' 17~ 10' 17~ 20' 17~ 30' 17~ 40' 17~ 50' 18~ 0' 18~ 10' 18~ 20' 18~ 30' 18~ 40' 18~ 50' 19~ 0' 19~ 10' 19~ 20' 19~ 30' 19~ 40' 19~ 50' 20~ 0' 20~ 10' 20~ 20' 20~ 30' 20~ 40' 20~ 50' 21~ 0' 21~ 10' 21~ 20' 21~ 30' 21~ 40' 21~ 50' 22~ 0' 22~ 10' 22~ 20' 22~ 30' 22~ 40' 22~ 50' 23~ 0' 23~ 10' 23~ 20' 23~ 30' 23~ 40' 23~ 50' 24~ 0' 24~ 10' 24~ 20' 24~ 30' 24~ 40' 24~ 50' 25~ 0' 9.4130 9.4177 9.4223 9.4269 9.4314 9.4359 9.4403 9.4447 9.4491 9.4533 9.4576 9.4618 9.4659 9.4700 9.4741 9.4781 9.4821 9.4861 9.4900 9.4939 9.4977 9.5015 9.5052 9.5090 9.5126 9.5163 9.5199 9.5235 9.5270 9.5306 9.5341 9.5375 9.5409 9.5443 9.5477 9.5510 9.5543 9.5576 9.5609 9.5641 9.5673 9.5704 9.5736 9.5767 9.5798 9.5828 9.5859 9.5889 9.5919 9.5948 9.5978 9.6007 9.6036 9.6065 9.6093 9.6121 9.6149 9.6177 9.6205 9.6232 9.6259 4.7 4.6 4.6 4.5 4.5 4.4 4.4 4.4 4.2 4.3 4.2 4.1 4.1 4.1 4.0 4.0 4.0 3.9 3.9 3.8 3.8 3.7 3.8 3.6 3.7 3.6 3.6 3.5 3.6 3.5 3.4 3.4 3.4 3.4 3.3 3.3 3.3 3.3 3.2 3.2 3.1 3.2 3.1 3.1 3.0 3.1 3.0 3.0 2.9 3.0 2.9 2.9 2.9 2.8 2.8 2.8 2.8 2.8 2.7 2.7 9.4281 9.4331 9.4381 9.4430 9.4479 9.4527 9.4575 9.4622 9.4669 9.4716 9.4762 9.4808 9.4853 9.4898 9.4943 9.4987 9.5031 9.5075 9.5118 9.5161 9.5203 9.5245 9.5287 9.5329 9.5370 9.5411 9.5451 9.5491 9.5531 9.5571 9.5611 9.5650 9.5689 9.5727 9.5766 9.5804 9.5842 9.5879 9.5917 9.5954 9.5991 9.6028 9.6064 9.6100 9.6136 9.6172 9.6208 9.6243 9.6279 9.6314 9.6348 9.6383 9.6417 9.6452 9.6486 9.6520 9.6553 9.6587 9.6620 9.6654 9.6687 5.0 5.0 4.9 4.9 4.8 4.8 4.7 4.7 4.7 4.6 4.6 4.5 4.5 4.5 4.4 4.4 4.4 4.3 4.3 4.2 4.2 4.2 4.2 4.1 4.1 4.0 4.0 4.0 4.0 4.0 3.9 3.9 3.8 3.9 3.8 3.8 3.7 3.8 3.7 3.7 3.7 3.6 3.6 3.6 3.6 3.6 3.5 3.6 3.5 3.4 3.5 3.4 3.5 3.4 3.4 3.3 3.4 3.3 3.4 3.3 0.5719 0.5669 0.5619 0.5570 0.5521 0.5473 0.5425 0.5378 0.5331 0.5284 0.5238 0.5192 0.5147 0.5102 0.5057 0.5013 0.4969 0.4925 0.4882 0.4839 0.4797 0.4755 0.4713 0.4671 0.4630 0.4589 0.4549 0.4509 0.4469 0.4429 0.4389 0.4350 0.4311 0.4273 0.4234 0.4196 0.4158 0.4121 0.4083 0.4046 0.4009 0.3972 0.3936 0.3900 0.3864 0.3828 0.3792 0.3757 0.3721 0.3686 0.3652 0.3617 0.3583 0.3548 0.3514 0.3480 0.3447 0.3413 0.3380 0.3346 0.3313 9.9849 9.9846 9.9843 9.9839 9.9836 9.9832 9.9828 9.9825 9.9821 9.9817 9.9814 9.9810 9.9806 9.9802 9.9798 9.9794 9.9790 9.9786 9.9782 9.9778 9.9774 9.9770 9.9765 9.9761 9.9757 9.9752 9.9748 9.9743 9.9739 9.9734 9.9730 9.9725 9.9721 9.9716 9.9711 9.9706 9.9702 9.9697 9.9692 9.9687 9.9682 9.9677 9.9672 9.9667 9.9661 9.9656 9.9651 9.9646 9.9640 9.9635 9.9629 9.9624 9.9618 9.9613 9.9607 9.9602 9.9596 9.9590 9.9584 9.9579 9.9573.3.3.4.3.4.4.3.4.4.3.4.4.4.4.4.4.4.4.4.4.4.5.4.4.5.4.5.4.5.4.5.4.5.5.5.4.5.5.5.5.5.5.5.6.5.5.5.6.5.6.5.6.5.6.5.6.6.6.5.6 75~ 0' 74~ 50' 74~ 40' 740 30' 74~ 20' 740 10' 74~ 0' 73~ 50' 730 40' 73~ 30' 73~ 20' 73~ 10' 73~ 0' 72~ 50' 72~ 40' 72~ 30' 72~ 20' 72~ 10' 72~ 0' 71~ 50' 71~ 40' 71~ 30' 71~ 20' 71~ 10' 71~ 0' 70~ 50' 70~ 40' 70~ 30' 70~ 20' 70~ 10' 70~ 0' 69~ 50' 69~ 40' 69~ 30' 69~ 20' 69~ 10' 69~ 0' 68~ 50' 68~ 40' 68~ 30' 68~ 20' 68~ 10' 68~ 0' 67~ 50' 67~ 40' 67~ 30' 67~ 20' 67~ 10' 67~ 0' 66~ 50' 66~ 40' 66~ 30' 66~ 20' 66~ 10' 66~ 0' 65~ 50' 65~ 40' 65~ 30' 65~ 20' 65~ 10' 65~ 0' -I I I I I I- log cos gff '' log co log tan log sin diff. 1' Angle 65~-75~, [] COSINES, TANGENTS, AND COTANGENTS 15 25~-35o Angle |logsin diff.1llogtan-dcom/ log cot log cos diff.1'I1 ~~~~ I ~~~~ diff. 1' I 25~ 0' 25~ 10' 25~ 20' 25~ 30' 25~ 40' 25~ 50' 26~ 0' 26~ 10' 26~ 20' 26~ 30' 26~ 40' 26~ 50' 27~ 0' 27~ 10' 27~ 20' 27~ 30' 27~ 40' 27~ 50' 28~ 0' 28~ 10' 28~ 20' 28~ 30' 28~ 40' 28~ 50' 29~ 0' 29~ 10' 29~ 20' 29~ 30' 29~ 40' 29~ 50' 30~ 0' 30~ 10' 30~ 20' 30~ 30' 30~ 40' 30~ 50' 31~ 0' 310 10' 31~ 20' 31~ 30' 31~ 40' 31~ 50' 32~ 0' 32~ 10' 32~ 20' 32~ 30' 32~ 40' 32~ 50' 33~ 0' 33~ 10' 33~ 20' 33~ 30' 33~ 40' 33~ 50' 34~ 0' 34~ 10' 34~ 20' 34~ 30' 34~ 40' 34~ 50' 35~ 0' 9.6259 9.6286 9.6313 9.6340 9.6366 9.6392 9.6418 9.6444 9.6470 9.6495 9.6521 9.6546 9.6570 9.6595 9.6620 9.6644 9.6668 9.6692 9.6716 9.6740 9.6763 9.6787 9.6810 9.6833 9.6856 9.6878 9.6901 9.6923 9.6946 9.6968 9.6990 9.7012 9.7033 9.7055 9.7076 9.7097 9.7118 9.7139 9.7160 9.7181 9.7201 9.7222 9.7242 9.7262 9.7282 9.7302 9.7322 9.7342 9.7361 9.7380 9.7400 9.7419 9.7438 9.7457 9.7476 9.7494 9.7513 9.7531 9.7550 9.7568 9.7586 2.7 2.7 2.7 2.6 2.6 2.6 2.6 2.6 2.5 2.6 2.5 2.4 2.5 2.5 2.4 2.4 2.4 2.4 2.4 2.3 2.4 2.3 2.3 2.3 2.2 2.3 2.2 2.3 2.2 2.2 2.2 2.1 2.2 2.1 2.1 2.1 2.1 2.1 2.1 2.0 2.1 2.0 2.0 2.0 2.0 2.0 2.0 1.9 1.9 2.0 1.9 1.9 1.9 1.9 1.8 1.9 1.8 1.9 1.8 1.8 9.6687 9.6720 9.6752 9.6785 9.6817 9.6850 9.6882 9.6914 9.6946 9.6977 9.7009 9.7040 9.7072 9.7103 9.7134 9.7165 9.7196 9.7226 9.7257 9.7287 9.7317 9.7348 9.7378 9.7408 9.7438 9.7467 9.7497 9.7526 9.7556 9.7585 9.7614 9.7644 9.7673 9.7701 9.7730 9.7759 9.7788 9.7816 9.7845 9.7873 9.7902 9.7930 9.7958 9.7986 9.8014 9.8042 9.8070 9.8097 9.8125 9.8153 9.8180 9.8208 9.8235 9.8263 9.8290 9.8317 9.8344 9.8371 9.8398 9.8425 9.8452 3.3 3.2 3.3 3.2 3.3 3.2 3.2 3.2 3.1 3.2 3.1 3.2 3.1 3.1 3.1 3.1 3.0 3.1 3.0 3.0 3.1 3.0 3.0 3.0 2.9 3.0 2.9 3.0 2.9 2.9 2.9 2.9 2.8 2.9 2.9 2.9 2.8 2.9 2.8 2.9 2.8 2.8 2.8 2.8 2.8 2.8 2.7 2.8 2.8 2.7 2.8 2.7 2.8 2.7 2.7 2 7 2 7 2.7 2.7 2.7 0.3313 0.3280 0.3248 0.3215 0.3183 0.3150 0.3118 0.3086 0.3054 0.3023 0.2991 0.2960 0.2928 0.2897 0.2866 0.2835 0.2804 0.2774 0.2743 0.2713 0.2683 0.2652 0.2622 0.2592 0.2562 0.2533 0.2503 0.2474 0.2444 0.2415 0.2386 0.2356 0.2327 0.2299 0.2270 0.2241 0.2212 0.2184 0.2155 0.2127 0.2098 0.2070 0.2042 0.2014 0.1986 0.1958 0.1930 0.1903 0.1875 0.1847 0.1820 0.1792 0.1765 0.1737 0.1710 0.1683 0.1656 0.1629 0.1602 0.1575 0.1548 9.9573 9.9567 9.9561 9.9555 9.9549 9.9543 9.9537 9.9530 9.9524 9.9518 9.9512 9.9505 9.9499 9.9492 9.9486 9.9479 9.9473 9.9466 9.9459 9.9453 9.9446 9.9439 9.9432 9.9425 9.9418 9.9411 9.9404 9.9397 9.9390 9.9383 9.9375 9.9368 9.9361 9.9353 9.9346 9.9338 9.9331 9.9323 9.9315 9.9308 9.9300 9.9292 9.9284 9.9276 9.9268 9.9260 9.9252 9.9244 9.9236 9.9228 9.9219 9.9211 9.9203 9.9194 9.9186 9.9177 9.9169 9.9160 9.9151 9.9142 9.9134.6.6.6.6.6.6.7.6.6.6.7.6.7.6.7.6.7.7.6.7.7.7.7.7.7.7.7.7.7.8.7.7.8.7.8.7.8.8.7.8.8.8.8.8.8.8.8.8.8.9.8.8.9.8.9.8.9.9.9.8 65~ 0' 64~ 50' 64~ 40' 64~ 30' 64~ 20' 64~ 10' 64~ 0' 63~ 50' 63~ 40' 63~ 30' 63~ 20' 63~ 10' 63~. 0' 62~ 50' 62~ 40' 62~ 30' 62~ 20' 62~ 10' 62~ 0' 61~ 50' 61~ 40' 61~ 30' 61~ 20' 61~ 10' 61~ 0' 60~ 50' 60~ 40' 60~ 30' 60~ 20' 60~ 10' 60~ 0' 59~ 50' 59~ 40' 59~ 30' 59~ 20' 59~ 10' 59~ 0' 58~ 50' 58~ 40' 58~ 30' 58~ 20' 58~ 10' 58~ 0' 57~ 50' 57~ 40' 57~ 30' 57~ 20' 57~ 10' 57~ 0' 56~ 50' 56~ 40' 56~ 30' 56~ 20' 56~ 10' 56~ 0' 55~ 50' 55~ 40' 55~ 30' 55~ 20' 55~ 10' 55~ 0' __ log cos di. log cot csm', log tan log sin diff. 1' Angle 65diff. 1 550~-650 16 TABLE II. LOGARITHMIC SINES 35~-45~ 0 Angle log sin diff. logtan dii log cot log cos diff. 1'. - - - I I I 35~ 0' 350 10' 350 20' 350 30' 350 40' 350 50' 360 0' 36~ 10' 36~ 20' 36~ 30' 360 40' 36~ 50' 37~ 0' 370 10' 370 20' 370 30' 370 40' 370 50' 380 0' 38~ 10' 380 20' 380 30' 380 40' 380 50' 390 0' 390 10' 390 20' 390 30' 390 40' 390 50' 400 0' 400 10' 400 20' 400 30' 400 40' 40~ 50' 410 0' 410 10' 410 20' 41~ 30' 41~ 40' 41~ 50' 42~ 0' 42~ 10' 42~ 20' 420 30' 42~ 40' 42~ 50' 430 0' 430 10' 430 20' 430 30' 430 40' 430 50' 44~ 0' 440 10' 440 20' 440 30' 440 40' 440 50' 450 0' 9.7586 9.7604 9.7622 9.7640 9.7657 9.7675 9.7692 9.7710 9.7727 9.7744 9.7761 9.7778 9.7795 9.7811 9.7828 9.7844 9.7861 9.7877 9.7893 9.7910 9.7926 9.7941 9.7957 9.7973 9.7989 9.8004 9.8020 9.8035 9.8050 9.8066 9.8081 9.8096 9.8111 9.8125 9.8140 9.8155 9.8169 9.8184 9.8198 9.8213 9.8227 9.8241 9.8255 9.8269 9.8283 9.8297 9.8311 9.8324 9.8338 9.8351 9.8365 9.8378 9.8391 9.8405 9.8418 9.8431 9.8444 9.8457 9.8469 9.8482 9.8495 1.8 1.8 1.8 1.7 1.8 1.7 1.8 1.7 1.7 1.7 1.7 1.7 1.6 1.7 1.6 1.7 1.6 1.6 1.7 1.6 1.5 1.6 1.6 1.6 1.5 1.6 1.5 1.5 1.6 1.5 1.5 1.5 1.4 1.5 1.5 1.4 1.5 1.4 1.5 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.3 1.4 1.3 1.4 1.3 1.3 1.4 1.3 1.3 1.3 1.3 1.2 1.3 1.3 9.8452 9.8479 9.8506 9.8533 9.8559 9.8586 9.8613 9.8639 9.8666 9.8692 9.8718 9.8745 9.8771 9.8797 9.8824 9.8850 9.8876 9.8902 9.8928 9.8954 9.8980 9.9006 9.9032 9.9058 9.9084 9.9110 9.9135 9.9161 9.9187 9.9212 9.9238 9.9264 9.9289 9.9315 9.9341 9.9366 9.9392 9.9417 9.9443 9.9468 9.9494 9.9519 9.9544 9.9570 9.9595 9.9621 9.9646 9.9671 9.9697 9.9722 9.9747 9.9772 9.9798 9.9823 9.9848 9.9874 9.9899 9.9924 9.9949 9.9975 0.0000 2.7 2.7 2.7 2.6 2.7 2.7 2.6 2.7 2.6 2.6 2.7 2.6 2.6 2.7 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.5 2.6 2.6 2.5 2.6 2.6 2.5 2.6 2.6 2.5 2.6 2.5 2.6 2.5 2.6 2.5 2.5 2.6 2.5 2.6 2.5 2.5 2.6 2.5 2.5 2.5 2.6 2.5 2.5 2.6 2.5 2.5 2.5 2.6 2.5 0.1548 0.1521 0.1494 0.1467 0.1441 0.1414 0.1387 0.1361 0.1334 0.1308 0.1282 0.1255 0.1229 0.1203 0.1176 0.1150 0.1124 0.1098 0.1072 0.1046 0.1020 0.0994 0.0968 0.0942 0.0916 0.0890 0.0865 0.0839 0.0813 0.0788 0.0762 0.0736 0.0711 0.0685 0.0659 0.0634 0.0608 0.0583 0.0557 0.0532 0.0506 0.0481 0.0456 0.0430 0.0405 0.0379 0.0354 0.0329 0.0303 0.0278 0.0253 0.0228 0.0202 0.0177 0.0152 0.0126 0.0101 0.0076 0.0051 0.0025 0.0000 9.9134 9.9125 9.9116 9.9107 9.9098 9.9089 9.9080 9.9070 9.9061 9.9052 9.9042 9.9033 9.9023 9.9014 9.9004 9.8995 9.8985 9.8975 9.8965 9.8955 9.8945 9.8935 9.8925 9.8915 9.8905 9.8895 9.8884 9.8874 9.8864 9.8853 9.8843 9.8832 9.8821 9.8810 9.8800 9.8789 9.8778 9.8767 9.8756 9.8745 9.8733 9.8722 9.8711 9.8699 9.8688 9.8676 9.8665 9.8653 9.8641 9.8629 9.8618 9.8606 9.8594 9.8582 9.8569 9.8557 9.8545 9.8532 9.8520 9.8507 9.8495.9.9.9.9.9.9 1.0.9.9 1.0.9 1.0.9 1.0.9 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.1 1.0 1.0 1.1 1.0 1.1 1.1 1.1 1.0 1.1 1.1 1.1 1.1 1.1 1.2 1.1 1.1 1.2 1.1 1.2 1.1 1.2 1.2 1.2 1.1 1.2 1.2 1.2 1.3 1.2 1.2 1.3 1.2 1.3 1.2 550 0' 540 50' 540 40' 54~ 30' 540 20' 540 10' 540 0' 530 50' 530 40' 530 30' 530 20' 530 10' 530 0' 52~ 50' 520 40' 52~ 30' 520 20' 52~ 10' 520 0' 510 50' 510 40' 510 30' 510 20' 51~ 10' 51~ 0' 500 50' 500 40' 500 30' 500 20' 500 10' 500 0' 490 50' 490 40' 490 30' 490 20' 490 10' 490 0' 48~ 50' 48~ 40' 48~ 30' 480 20' 48~ 10' 48~ 0' 470 50' 470 40' 470 30' 470 20' 470 10' 47~ 0' 460 50' 46~ 40' 460 30' 46~ 20' 46~ 10' 46~ 0' 450 50' 450 40' 450 30' 45~ 20' 450 10' 450 0' log cos diff. 1' log cot cdif1j' log tan log sin diff. 1' Angle 45- 55~ l~ ~ ~ ~~~~~~~i v CONVERSION TABLES FOR ANGLES 17 To CHANGE FROM MINUTES AND SECONDS INTO THE DECIMAL PARTS OF A DEGREE OR INTO RADIANS From seconds From minutes From degrees into radians 1"=0.00028~ =0.0000048 Rad. 1'= 0.017~= 0.00029 Rad. 1~=0.01745 Rad. 2" =0.00056~=0.0000097 2' =0.033 = 0.00058 " 2 =0.03491 " 3" = 0.00083 = 0.0000145 " 3' = 0.050~ = 0.00087 " 3~ 0.05236" 4" =0.00111~=0.0000194 " 4'=0.067~=0.00116 4 = 0.06981 " 5"=0.00139~=0.0000242 " 5'=0.0830 =0.00145 " 5~=0.08727 " 6",=0.001670~0.0000291 ' 6'=0.1000 =0.00175 " 60=0.10472 ' 7"= 0.00194 =0.0000339 " 7'=0.117~=0.00204 " 70=0.12217 " 8"=0.00222 =0.0000388 " 8'=0.133~=0.00233 " 80=0.13963 9" =0.00250 =0.0000436 9'=0.150 =0.00262 90=0.15708 " 10" =0.002780 =0.0000485 " 10' =0.167=0.00291 " 10 =0.17453 " 20 =0.00556~= 0.0000970 " 20'-0.3330~=0.00582 " 20~=0.34907 " 30 =0.008330= 0.0001454 " 30'=0.5000=0.00873 " 300=0.52360, 40" =0.0111 1 = 0.0001939 " 40' =0.667 =0.01164 " 40~=0.69813, 50"= 0.01389~= 0.0002424 " 50'=0.833 =0.01454 " 50 =0.87266 " TO CHANGE FROM DECIMAL PARTS OF A DEGREE INTO MINUTES AND SECONDS 0.0000~ 0.000' = 0" 0.20~ = 12.0'= 12' 0.60~ = 36.0' = 36' 0.0001~= 0.006' = 0.36" 0.21~ = 12.6'= 12' 36" 0.61~ = 36.6' = 36' 36" 0.0002~ = 0.012' = 0.72" 0.22~ = 13.2' = 13' 12" 0.62~ = 37.2' = 37' 12" 0.0003~ = 0.018' = 1.08" 0.230 = 13.8' = 13' 48" 0.630 = 37.8' = 37' 48" 0.0004~ = 0.024' =, 1.44" 0.240 = 14.4' = 14' 24" 0.64~ = 38.4' = 38' 24" 0.00050 = 0.030' = 1.80" 0.25~ = 15.0' = 15' 0.650 = 39.0' = 39' 0.0006~ = 0.036' = 2.16" 0.260 = 15.6' = 15' 36" 0.660 =39.6' = 39' 36" 0.00070 = 0.042' = 2.52" 0.270 = 16.2' = 16' 12" 0.67~ = 40.2' = 40' 12" 0.00080 = 0.048' = 2.88" 0.28~ = 16.8' = 16' 48" 0.680 = 40.8' = 40' 48" 0.0009~ = 0.054' = 3.24" 0.29~ = 17.4' = 17' 24" 0.690 = 41.4' = 41' 24" 0.00100 = 0.060' = 3.60" 0.30~ = 18.0' = 18' 0.70~ = 42.0' =42' 0.0010 = 0.06' = 3.6" 0.31~ = 18.6' = 18' 36" 0.71~ = 42.6' = 42' 36" 0.0020 = 0.12' = 7.2" 0.32~ = 19.2'= 19' 12" 0.720 = 43.2' = 43' 12" 0.003~ = 0.18' = 10.8" 0.33~ = 19.8' = 19' 48" 0.730 = 43.8' = 43' 48" 0.004~ = 0.24' = 14.4" 0.34~ = 20.4' = 20' 24" 0.74~ = 44.4' = 44' 24" 0.005~ = 0.30' = 18.0" 0.35~ = 21.0'= 21' 0.75~ = 45.0' = 45' 0.006~ = 0.36' = 21.6" 0.36~ = 21.6' = 21' 36" 0.76~ = 45.6' = 45' 36" 0.007~ = 0.42' = 25.2" 0.37~ = 22.2'= 22' 12" 0.770 = 46.2' = 46' 12" 0.0080 = 0.48' = 28.8" 0.380 = 22.8' = 22' 48" 0.780 = 46.8' = 46' 48" 0.0090 = 0.54' = 32.4" 0.39~ = 23.4' = 23' 24" 0.790 = 47.4' = 47' 24" 0.010~ = 0.60' = 36.0" 0.40~ = 24.0' = 24' 0.80~ = 48.0' = 48' 0.01~ = 0.6' = 36" 0.410 = 24.6' = 24' 36" 0.810 = 48.6' = 48' 36" 0.02~ = 1.2' = 1' 12" 0.42~ = 25.2' = 25' 12" 0.82 = 49.2' = 49' 12" 0.03~ = 1.8' = 1' 48" 0.430 = 25.8' = 25' 48" 0.83~ = 49.8' = 49' 48" 0.04~ = 2.4' = 2' 24" 0.44~ = 26.4' = 26' 24" 0.84~ = 50.4' = 50' 24" 0.05~ = 3.0' = 3' 0.45 = 27.0'= 27' 0.850 = 51.0'= 51' 0.060 = 3.6' = 3' 36" 0.460 = 27.6' = 27' 36" 0.86~ = 51.6' = 51' 36" 0.07~ = 4.2' = 4' 12" 0.47~ = 28.2' =28' 12" 0.87~ = 52.2' = 52' 12" 0.08~ = 4.8' = 4' 48" 0.48~ = 28.8' = 28' 48" 0.880 = 52.8' = 52' 48" 0.090 = 5.4' = 5' 24" 0.49~ = 29.4'= 29' 24" 0.89~ = 53.4' = 53' 24" 0.100 = 6.0' = 6' 0.50~ = 30.0' = 30' 0.90~ = 54.0' = 54' 0.110 = 6.6' = 6' 36" 0.51~ = 30.6' = 30' 36" 0.91~ = 54.6' = 54' 36" 0.120 = 7.2' = 7' 12" 0.52~ = 31.2'= 31' 12" 0.92~ = 55.2' = 55' 12" 0.130 = 7.8' = 7' 48" 0.53~ = 31.8' = 31' 48" 0.93~ = 55.8' = 55' 48" 0.14~ = 8.4' = 8' 24" 0.54~ = 32.4' = 32' 24" 0.940 = 56.4' = 56' 24" 0.15~ = 9.0' = 9' 0.550 = 33.0' = 33' 0.950 = 57.0' = 57' 0.16~ = 9.6' = 9' 36" 0.56~ =.33.6' = 33' 36" 0.96~ = 57.6' = 57' 36" 0.170 = 10.2' = 10' 12" 0.570 = 34.2' = 34' 12" 0.970 = 58.2' = 58' 12" 0.18~ = 10.8' = 10' 48" 0.58~ = 34.8' = 34' 48" 0.980 = 58.8' = 58' 48" 0.19~ = 11.4' = 11' 24" 0.59~ = 35.4' = 35' 24" 0.99~ = 59.4' = 59' 24" 0.20~ = 12.0' = 12' 0.60~ = 36.0' = 36' 1.00~ = 60.0' = 60' TABLE III FOUR-PLACE LOGARITHMS OF TRIGONOMETRIC FUNCTIONS, THE ANGLE BEING EXPRESSED IN DEGREES AND THE DECIMAL PART OF A DEGREE This table gives the common logarithms (base 10) of the sines, cosines, tangents, and cotangents of all angles from 0~ to 5~, and from 85~ to 90~ for every hundredth part of a degree, and from 5~ to 85~ for every tenth of a degree, all calculated to four places of decimals. In order to avoid the printing of negative characteristics, the number 10 has been added to every logarithm in the first, second, and fourth columns (those having log sin, log tan, and log cos at the top). Hence in writing down any logarithm taken from these three columns - 10 should be written after it. Logarithms taken from the third column (having log cot at the top) should be used as printed. A logarithm found from this table by interpolation may be in error by one unit in the last decimal place, except for angles between 0~ and 0.3~ or between 89.7~ and 90~, when the error may be larger. In the latter cases the table refers the student to the formulas on page 6 for more accurate results. 19 20 TABLE III. LOGARITHMIC SINES 00 Angle log sin diff. log tan log cot log cos Prop. Parts 0.000 0.010 0.020 0.030 0.04~ 0.05~ 0.060 0.070 0.080 0.09~ 0.10~ 0.110 0.120 0.13~ 0.140 0.150 0.160 0.17~ 0.180 0.190 0.20~ 0.210 0.22~ 0.230 0.24~ 0.250 0.260 0.270 0.28" 0.29~ 0.30~ 0.31~ 0.320 0.330 0.34~ 0.35~ 0.36~ 0.37~ 0.380 0.39~ 0.40~ 0.41~ 0.42~ 0.430 0.440 0.45~ 0.46~ 0.470 0.48~ 0.490 0.50~ 6.2419 6.5429 6.7190 6.8439 6.9408 7.0200 7.0870 7.1450 7.1961 7.2419 7.2833 7.3211 7.3558 7.3880 7.4180 7.4460 7.4723 7.4971 7.5206 7.5429 7.5641 7.5843 7.6036 7.6221 7.6398 7.6568 7.6732 7.6890 7.7043 7.7190 7.7332 7.7470 7.7604 7.7734 7.7859 7.7982 7.8101 7.8217 7.8329 7.8439 7.8547 7.8651 7.8753 7.8853 7.8951 7.9046 7.9140 7.9231 7.9321 7.9408 142 k c3.119 110 108 10 10. 100 138 95 94 91 1.1 90 87 1 2._ d aD 1420 138 134 130 125 123 119 112 108 104 102 95 94 91 90 87 6.2419 6.5429 6.7190 6.8439 6.9408 7.0200 7.0870 7.1450 7.1961 7.2419 7.2833 7.3211 7.3558 7.3880 7.4180 7.4460 7.4723 7.4972 7.5206 7.5429 7.5641 7.5843 7.6036 7.6221 7.6398 7.6569 7.6732 7.6890 7.7043 7.7190 7.7332 7.7470 7.7604 7.7734 7.7860 7.7982 7.8101 7.8217 7.8329 7.8439 7.8547 7.8651 7.8754 7.8853 7.8951 7.9046 7.9140 7.9231 7.9321 7.9409 3 Oe 0 a 134 10 o 142 119 104 103 99 12 130 llG 112 110 108 104 103 94 91 90 88 3.7581 3.4571 3.2810 3.1561 3.0592 2.9800 2.9130 2.8550 2.8039 2.7581 2.7167 2.67~9 2.6442 2.6120 2.5820 2.5540 2.5277 2.5028 2.4794 2.4571 2.4359 2.4157 2.3964 2.3779 2.3602 2.3431 2.3268 2.3110 2.2957 2.2810 2.2668 2.2530 2.2396 2.2266 2.2140 2.2018 2.1899 2.1783 2.1671 2.1561 2.1453 2.1349 2.1246 2.1147 2.1049 2.0954 2.0860 2.0769 2.0678 2.0591 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 90.00~ 89.990 89.98~ 89.97~ 89.96~ 89.95~ 89.94~ 89.93~ 89.920 89.910 89.900 89.890 89.8SS 89.87~ 89.860 89.850 89.84~ 89.830 89.820 89.81~ 89.80~ 89.790 89.78~ 89.770 89.76~ 89.75~ 89.74~ 89.730 89.720 89.710 89.70~ 89.690 89.68~ 89.67~ 89.660 89.650 89.640 89.630 89.620 89.610 89.600 89.59~ 89.580 89.57~ 89.56~ 89.55~ 89.540 89.53~ 89.520 89.510 89.50~ ko be sq 4ca P4 Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 79 7.9 15.8 23.7 31.6 39.5 47.4 55.3 63.2 71.1 76 7.6 15.2 22.8 30.4 38.0 45.6 53.2 60.8 68.4 73 7.3 14.6 21.9 29.2 36.5 43.8 51.1 58.4 65.7 69 6.9 13.8 20.7 27.6 34.5 41.4 48.3 55.2 62.1 66 6.6 13.2 19.8 26.4 33.0 39.6 46.2 52.8 59.4 63 6.3 12.6 18.9 25.2 31.5 37.8 44.1 50.4 56.7 78 7.8 15.6 23.4 31.2 39.0 46.8 54.6 62.4 70.2 75 7.5 15.0 22.5 30.0 37.5 45.0 52.5 60.0 67.5 72 7.2 14.4 21.6 28.8 36.0 43.2 50.4 57.6 64.8 68 6.8 13.6 20.4 27.2 34.0 40.8 47.6 54.4 61.2 65 6.5 13.0 19.5 26.0 32.5 39.0 45.5 52.0 58.5 62 6.2 12.4 18.6 24.8 31.0 37.2 43.4 49.6 55.8 77 7.7 15.4 23.1 30.8 38.5 46.2 53.9 61.6 69.3 74 7.4 14.8 22.2 29.6 37.0 44.4 51.8 59.2 66.6 71 7.1 14.2 21.3 28.4 35.5 42.6 49.7 56.8 63.9 67 6.7 13.4 20.1 26.8 33.5 40.2 46.9 53.6 60.3 64 6.4 12.8 19.2 25.6 32.0 38.4 44.8 51.2 57.6 61 6.1 12.2 18.3 24.4 30.5 36.6 42.7 48.8 54.9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 I I I Ic r log cos diff. log cot com. logtan log sin Angle 89~ a COSINES, TANGENTS, AND COTANGENTS 21 corn. o Angle log sin diff. log tan com. log cot log cos Prop. Parts 0.500 0.510 0.520 0.530 0.540 0.550 0.56~ 0.57~ 0.58~ 0.590 0.60~ 0.61~ 0.620 0.630 0.640 0.650 0.66~ 0.670 0.6S8 0.690 0.70~ 0.710 0.720 0.73~ 0.74~ 0.75~ 0.760 0.770 0.78~ 0.790 0.800 0.81~ 0.820 0.83~ 0.840 0.850 0.86~ 0.870 0.88S 0.890 0.900 0.910 0.920 0.93~ 0.940 0.950 0.96~ 0.970 0.98~ 0.990 1.00~ 7.9408 7.9494 7.9579 7.9661 7.9743 7.9822 7.9901 7.9977 8.0053 8.0127 8.0200 8.0272 8.0343 8.0412 8.0480 8.0548 8.0614 8.0679 8.0744 8.0807 8.0870 8.0931 8.0992 8.1052 8.1111 8.1169 8.1227 8.1284 8.1340 8.1395 8.1450 8.1503 8.1557 8.1609 8.1661 8.1713 8.1.764 8.1814 8.1863 8.1912 8.1961 8.2009 8.2056 8.2103 8.2150 8.2196 8.2241 8.2286 8.2331 8.2375 8.2419 86 85 82 82 79 79 76 76 74 73 72 71 69 68 68 66 65 65 63 63 61 61 60 59 58 58 57 56 55 55 53 54 52 52 52 51 50 49 49 49 48 47 47 47 46 45 45 45 44 44 7.9409 7.9495 7.9579 7.9662 7.9743 7.9823 7.9901 7.9978 8.0053 8.0127 8.0200 8.0272 8.0343 8.0412 8.0481 8.0548 8.0614 8.0680 8.0744 8.0807 8.0870 8.0932 8.0992 8.1052 8.1111 8.1170 8.1227 8.1284 8.1340 8.1395 8.1450 8.1504 8.1557 8.1610 8.1662 8.1713 8.1764 8.1814 8.1864 8.1913 8.1962 8.2010 8.2057 8.2104 8.2150 8.2196 8.2242 8.2287 8.2331 8.2376 8.2419 86 84 83 81 80 78 77 75 74 73 72 71 69 69 67 66 66 64 63 63 62 60 60 59 59 57 57 56 55 55 54 53 53 52 51 51 50 50 49 49 48 47 47 46 46 46 45 44 45 43 2.0591 2.0505 2.0421 2.0338 2.0257 2.0177 2.0099 2.0022 1.9947 1.9873 1.9800 1.9728 1.9657 1.9588 1.9519 1.9452 1.9386 1.9320 1.9256 1.9193 1.9130 1.9068 1.9008 1.8948 1.8889 1.8830 1.8773 1.8716 1.8660 1.8605 1.8550 1.8496 1.8443 1.8390 1.8338 1.8287 1.8236 1.8186 1.8136 1.8087 1.8038 1.7990 1.7943 1.7896 1.7850 1.7804 1.7758 1.7713 1.7669 1.7624 1.7581 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 89.500 89.490 89.48~ 89.470 89.46~ 89.450 89.440 89.430 89.42~ 89.410 89.400 89.390 89.38~ 89.37~ 89.360 89.35~ 89.34~ 89.33~ 89.320 89.310 89.30~ 89.290 89.28~ 89.27~ 89.26~ 89.250 89.240 89.230 89.220 89.210 89.20~ 89.19~ 89.180 89.170 89.16~0 89.15~ 89.140 89.130 89.12~ 89.110 89.100 89.090 89.080 89.070 89.060 89.050 89.04~ 89.030 89.02~ 89.010 89.000 1 2 3 4 5 6 7 8 9 4-3,:4 'j i., 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 Difference 60 59 58 6.0 5.9 5.8 12.0 11.8 11.6 18.0 17.7 17.4 24.0 23.6 23.2 30.0 29.5 29.0 36.0 35.4 34.8 42.0 41.3 40.6 48.0 47.2 46.4 54.0 53.1 52.2 57 56 55 5.7 5.6 5.5 11.4 11.2 11.0 17.1 16.8 16.5 22.8 22.4 22.0 28.5 28.0 27.5 34.2 33.6 33.0 39.9 39.2 38.5 45.6 44.8 44.0 51.3 50.4 49.5 54 53 52 5.4 5.3 5.2 10.8 10.6 10.4 16.2 15.9 15.6 21.6 21.2 20.8 27.0 26.5 26.0 32.4 31.8 31.2 37.8 37.1 36.4 43.2 42.4 41.6 48.6 47.7 46.8 51 50 49 5.1 5.0 4.9 10.2 10.0 9.8 15.3 15.0 14.7 20.4 20.0 19.6 25.5 25.0 24.5 30.6 30.0 29.4 35.7 35.0 34.3 40.8 40.0 39.2 45.9 45.0 44.1 48 47 46 4.8 4.7 4.6 9.6 9.4 9.2 14.4 14.1 13.8 19.2 18.8 18.4 24.0 23.5 23.0 28.8 28.2 27.6 33.6 32.9 32.2 38.4 37.6 36.8 43.2 42.3 41.4 45 44 43 4.5 4.4 4.3 9.0 8.8 8.6 13.5 13.2 12.9 18.0 17.6 17.2 22.5 22.0 21.5 27.0 26.4 25.8 31.5 30.8 30.1 36.0 35.2 34.4 40.5 39.6 38.7 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 I -- - I I.I I ~~ I ----— I --- - -I — -- I log cos diff. log cot coi log tan log sin Angle d 89 89~ m 22 TABLE III. LOGARITHMIC SINES 1~ Angle logsin di. logtan ' logcot logcos | Prop. Parts Angle log si dif log tan cr log cot log cos Prop. Parts l l l l 1.00~ 1.010 1.02~ 1.03~ 1.04~ 1.05~ 1.06~ 1.07~ 1.08~ 1.09~ 1.100 1.11~ 1.12~ 1.13~ 1.14~ 1.15~ 1.16~ 1.17~ 1.18~ 1.19~ 1.20~ 1.21~ 1.22~ 1.23~ 1.24~ 1.25~ 1.26~ 1.27~ 1.28~ 1.29~ 1.30~ 1.31~ 1.32~ 1.33~ 1.34~ 1.35~ 1.36~ 1.37~ 1.38~ 1.39~ 1.40~ 1.41~ 1.42~ 1.43~ 1.44~ 1.45~ 1.46~ 1.47~ 1.48~ 1.49~ 1.50~ 8.2419 8.2462 8.2505 8.2547 8.2589 8.2630 8.2672 8.2712 8.2753 8.2793 8.2832 8.2872 8.2911 8.2949 8.2988 8.3025 8.3063 8.3100 8.3137 8.3174 8.3210 8.3246 8.3282 8.3317 8.3353 8.3388 8.3422 8.3456 8.3491 8.3524 8.3558 8.3591 8.3624 8.3657 8.3689 8.3722 8.3754 8.3786 8.3817 8.3848 8.3880 8.3911 8.3941 8.3972 8.4002 8.4032 8.4062 8.4091 8.4121 8.4150 8.4179 log cos 43 43 42 42 41 42 40 41 40 39 40 39 38 39 37 38 37 37 37 36 36 36 35 36 35 34 34 35 33 34 33 33 33 32 33 32 32 31 31 32 31 30 31 30 30 30 29 30 29 29 diff. 8.2419 8.2462 8.2505 8.2548 8.2590 8.2631 8.2672 8.2713 8.2754 8.2794 8.2833 8.2873 8.2912 8.2950 8.2988 8.3026 8.3064 8.3101 8.3138 8.3175 8.3211 8.3247 8.3283 8.3318 8.3354 8.3389 8.3423 8.3458 8.3492 8.3525 8.3559 8.3592 8.3625 8.3658 8.3691 8.3723 8.3755 8.3787 8.3818 8.3850 8.3881 8.3912 8.3943 8.3973 8.4003 8.4033 8.4063 8.4093 8.4122 8.4152 8.4181 log cot 43 43 43 42 41 41 41 41 40 39 40 39 38 38 38 38 37 37 37 36 36 36 35 36 35 34 35 34 33 34 33 33 33 33 32 32 32 31 32 31 31 31 30 30 30 30 30 29 30 29 corn. diff. 1.7581 1.7538 1.7495 1.7452 1.7410 1.7369 1.7328 1.7287 1.7246 1.7206 1.7167 1.7127 1.7088 1.7050 1.7012 1.6974 1.6936 1.6899 1.6862 1.6825 1.6789 1.6753 1.6717 1.6682 1.6646 1.6611 1.6577 1.6542 1.6508 1.6475 1.6441 1.6408 1.6375 1.6342 1.6309 1.6277 1.6245 1.6213 1.6182 1.6150 1.6119 1.6088 1.6057 1.6027 1.5997 1.5967 1.5937 1.5907 1.5878 1.5848 1.5819 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 — 89.00~ 88.99~ -88.98~ 88.97~ 88.96~ 88.95~ 88.94~ 88.93~ 88.92~ 88.91~ 88.90~ 88.89~ 88.88~ 88.87~ 88.86~ 88.85~ 88.84~ 88.83~ 88.82~ 88.81~ 88.800 88.79~ 88.78~ 88.77~ 88.76~ 88.75~ 88.74~ 88.73~ 88.72~ 88.71~ 88.70~ 88.69~ 88.68~ 88.67~ 88.66~ 88.65~ 88.64~ 88.63~ 88.62~ 88.61~ 88.60~ 88.59~ 88.58~ 88.57~ 88.560 88.55~ 88.54~ 88.53~ 88.52~ 88.51~ 88.50~ I 4-.,l ba.ql 03 FIl Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 43 4.3 8.6 12.9 17.2 21.5 25.8 30.1 34.4 38.7 41 4.1 8.2 12.3 16.4 20.5 24.6 28.7 32.8 36.9 39 3.9 7.8 11.7 15.6 19.5 23.4 27.3 31.2 35.1 37 3.7 7.4 11.1 14.8 18.5 22.2 25.9 29.6 33.3 34 3.4 6.8 10.2 13.6 17.0 20.4 23.8 27.2 30.6 31 3.1 6.2 9.3 12.4 15.5 18.6 21.7 24.8 27.9 42 4.2 8.4 12.6 16.8 21.0 25.2 29.4 33.6 37.8 40 4.0 8.0 12.0 16.0 20.0 24.0 28.0 32.0 36.0 38 3.8 7.6 11.4 15.2 19.0 22.8 26.6 30.4 34.2 36 3.6 7.2 10.8 14.4 18.0 21.6 25.2 28.8 32.4 33 3.3 6.6 9.9 13.2 16.5 19.8 23.1 26.4 29.7 30 3.0 6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 I I 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 35 3.5 7.0 10.5 14.0 17.5 21.0 24.5 28.0 31.5 32 3.2 6.4 9.6 12.8 16.0 19.2 22.4 25.6 28.8 29 2.9 5.8 8.7 11.6 14.5 17.4 20.3 23.2 26.1 1 2 3 4 5 6 7 8 9 log tan log sin Angle 88~ COSINES, TANGENTS, AND COTANGENTS 23 Anglelogsin lo1~cot Angle log sin iff log tan log cot log cos Prop. Parts 1.500 1.510 1.520 1.530 1.540 1.550 1.560 1.570 1.58~ 1.59~ 1.60~ 1.61~ 1.62~ 1.63~ 1.640 1.65~ 1.66~ 1.670 1.680 1.69~ 1.700 1.71~ 1.72~ 1.73~ 1.740 1.75~ 1.760 1.770 1.780 1.790 1.800 1.810 1.82~ 1.83~ 1.840 1.850 1.860 1.870 1.880 1.89~ 1.900 1.91~ 1.920 1.930 1.940 1.950 1.96~ 1.97~ 1.980 1.990 2.000 I 8.4179 8.4208 8.4237 8.4265 8.4293 8.4322 8.4349 8.4377 8.4405 8.4432 8.4459 8.4486 8.4513 8.4540 8.4567 8.4593 8.4619 8.4645 8.4671 8.4697 8.4723 8.4748 8.4773 8.4799 8.4824 8.4848 8.4873 8.4898 8.4922 8.4947 8.4971 8.4995 8.5019 8.5043 8.5066 8.5090 8.5113 8.5136 8.5160 8.5183 8.5206 8.5228 8.5251 8.5274 8.5296 8.5318 8.5340 8.5363 8.5385 8.5406 8.5428 29 29 28 28 29 27 28 28 27 27 27 27 27 27 26 26 26 26 26 26 25 25 26 25 24 25 25 24 25 24 24 24 24 23 24 23 23 24 23 23 22 23 23 22 22 22 23 22 21 I I 8.4181 8.4210 8.4238 8.4267 8.4295 8.4323 8.4351 8.4379 8.4406 8.4434 8.4461 8.4488 8.4515 8.4542 8.4568 8.4595 8.4621 8.4647 8.4673 8.4699 8.4725 8.4750 8.4775 8.4801 8.4826 8.4851 8.4875 8.4900 8.4924 8.4949 8.4973 8.4997 8.5021 8.5045 8.5068 8.5092 8.5115 8.5139 8.5162 8.5185 8.5208 8.5231 8.5253 8.5276 8.5298 8.5321 8.5343 8.5365 8.5387 8.5409 29 28 29 28 28 28 28 27 28 27 27 27 27 26 27 26 26 26 26 26 25 25 26 25 25 24 25 24 25 24 24 24 24 23 24 23 24 23 23 23 23 22 23 22 23 22 22 22 22 22 I I 1.5819 1.5790 1.5762 1.5733 1.5705 1.5677 1.5649 1.5621 1.5594 1.5566 1.5539 1.5512 1.5485 1.5458 1.5432 1.5405 1.5379 1.5353 1.5327 1.5301 1.5275 1.5250 1.5225 1.5199 1.5174 1.5149 1.5125 1.5100 1.5076 1.5051 1.5027 1.5003 1.4979 1.4955 1.4932 1.4908 1.4885 1.4861 1.4838 1.4815 1.4792 1.4769 1.4747 1.4724 1.4702 1.4679 1.4657 1.4635 1.4613 1.4591 1.4569 9.9999 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9998 9.9997 9.9997 9.9997 9.9997 9.9997 9,9997 I 88.50~ 88.490 88.48~ 88.47~ 88.460 88.450 88.44~ 88.430 88.42~ 88.410 88.400 88.39~ 88.38~ 88.370 88.360 88.350 88.340 88.330 88.320 88.31~ 88.300 88.290 88.280 88.270 88.260 88.250 88.240 88.230 88.220 88.210 88.200 88.19~ 88.18~ 88.17~ 88.160 88.15~ 88.140 88.130 88.120 88.110 88.100 88.090 88.080 88.07~ 88.060 88.05~ 88.04~ 88.030 88.020 88.010 88.00~ 4-3 bo C [-. 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 Difference 29 2.9 5.8 8.7 11.6 14.5 17.4 20.3 23.2 26.1 27 2.7 5.4 8.1 10.8 13.5 16.2 18.9 21.6 24.3 25 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 23 2.3 4.6 6.9 9.2 11.5 13.8 16.1 18.4 20.7 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 28 2.8 5.6 8.4 11.2 14.0 16.8 19.6 22.4 25.2 26 2.6 5.2 7.8 10.4 13.0 15.6 18.2 20.8 23.4 24 2.4 4.8 7.2 9.6 12.0 14.4 16.8 19.2 21.6 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 1 2 3 4 5 6 7 8 9 0 22 8.5431 a I I I I I I log cos diff. log log cot tan log sin Angle di88ff. 88~..,~~~~ 24 TABLE III. LOGARITHMIC SINES 2~ Angle log sin diff. log tan com, log cot log cos Prop. Parts diff. I 2.000 2.01~ 2.020 2.03~ 2.040 2.05~ 2.060 2.070 2.080 2.09~ 2.100 2.110 2.12~ 2.13~ 2.140 2.150 2.16~ 2.170 2.18~ 2.190 2.20~ 2.210 2.220 2.230 2.240 2.250 2.260 2.270 2.280 2.290 2.300 2.310 2.320 2.330 2.340 2.350 2.360 2.37~ 2.38~ 2.390 2.40~ 2.410 2.420 2.430 2.440 2.45~ 2.460 2.470 2.480 2.49~ 2.500 I 8.5428 8.5450 8.5471 8.5493 8.5514 8.5535 8.5557 8.5578 8.5598 8.5619 8.5640 8.5661 8.5681 8.5702 8.5722 8.5742 8.5762 8.5782 8.5802 8.5822 8.5842 8.5862 8.5881 8.5901 8.5920 8.5939 8.5959 8.5978 8.5997 8.6016 8.6035 8.6054 8.6072 8.6091 8.6110 8.6128 8.6147 8.6165 8.6183 8.6201 8.6220 8.6238 8.6256 8.6274 8.6291 8.6309 8.6327 8.6344 8.6362 8.6379 8.6397 22 21 22 21 21 22 21 20 21 21 21 20 21 20 20 20 20 20 20 20 20 19 20 19 19 20 19 19 19 19 19 18 19 19 18 19 18 18 18 19 8.5431 8.5453 8.5474 8.5496 8.5517 8.5538 8.5559 8.5580 8.5601 8.5622 8.5643 8.5664 8.5684 8.5705 8.5725 8.5745 8.5765 8.5785 8.5805 8.5825 8.5845 8.5865 8.5884 8.5904 8.5923 8.5943 8.5962 8.5981 8.6000 8.6019 8.6038 8.6057 8.6076 8.6095 8.6113 8.6132 8.6150 8.6169 8.6187 8.6205 8.6223 8.6242 8.6260 8.6277 8.6295 8.6313 8.6331 8.6348 8.6366 8.6384 8.6401 22 21 22 21 21 21 21 21 21 21 21 20 21 20 20 20 20 20 20 20 20 19 20 19 20 19 19 19 19 19 19 19 19 18 19 18 19 18 18 18 19 18 17 18 18 18 17 18 18 17 1.4569 1.4547 1.4526 1.4504 1.4483 1.4462 1.4441 1.4420 1.4399 1.4378 1.4357 1.4336 1.4316 1.4295 1.4275 1.4255 1.4235 1.4215 1.4195 1.4175 1.4155 1.4135 1.4116 1.4096 1.4077 1.4057 1.4038 1.4019 1.4000 1.3981 1.3962 1.3943 1.3924 1.3905 1.3887 1.3868 1.3850 1.3831 1.3813 1.3795 1.3777 1.3758 1.3740 1.3723 1.3705 1.3687 1.3669 1.3652 1.3634 1.3616 1.3599 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.99-97 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 I 88.000 87.99~ 87.980 87.97~ 87.96~ 87.950 87.94~ 87.930 87.92~ 87.91~ 87.900 87.890 87.880 87.870 87.860 87.850 87.840 87.830 87.82~ 87.810 87.800 87.790 87.780 87.77~ 87.760 87.750 87.740 87.730 87.72~ 87.710 87.700 87.690 87.68~ 87.670 87.660 87.65~ 87.640 87.63~ 87.62~ 87.61~ 87.600 87.590 87.580 87.570 87.56~ 87.55~ 87.54~ 87.530 87.520 87.510 87.50~.,4 c3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 9 a) o Cl.L 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 20 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 19 1.9 3.8 5.7 7.6 9.5 11.4 13.3 15.2 17.1 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 18 18 18 17 18 18 17 18 17 18 i I I log cos difi. log cot com. log tan log sin Angle 87~ PESO~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~i 0 COSINES, TANGENTS, AND COTANGENTS 25 2~ Angle log sin dif. log tan co' log cot log cos Prop. Parts I 2.500 2.510 2.520 2.530 2.54~ 2.550 2.560 2.570 2.580 2.59~ 2.600 2.61~ 2.62~ 2.63~ 2.640 2.650 2.66~ 2.670 2.680 2.690 2.700 2.710 2.720 2.73~ 2.740 2.750 2.760 2.770 2.780 2.790 2.800 2.81~ 2.82~ 2.830 2.840 2.850 2.860 2.870 2.880 2.890 2.900 2.910 2.920 2.930 2.940 2.950 2.960 2.97~ 2.980 2.990 3.00~ I 8.6397 8.6414 8.6431 8.6449 8.6466 8.6483 8.6500 8.6517 8.6534 8.6550 8.6567 8.6584 8.6600 8.6617 8.6633 8.6650 8.6666 8.6682 8.6699 8.6715 8.6731 8.6747 8.6763 8.6779 8.6795 8.6810. 8.6826 8.6842 8.6858 8.6873 8.6889 8.6904 8.6920 8.6935 8.6950 8.6965 8.6981 8.6996 8.7011 8.7026 8.7041 8.7056 8.7071 8.7086 8.7100 8.7115 8.7130 8.7144 8.7159 8.7174 8.7188 17 17 18 17 17 17 17 17 I 16 17 17 16 17 16 17 16 16 17 16 16 16 16 16 16 15 16 16 16 15 16 15 16 15 15 15 16 15 15 15 15 15 15 15 14 15 15 14 15 15 14 8.6401 8.6418 8.6436 8.6453 8.6470 8.6487 8.6504 8.6521 8.6538 8.6555 8.6571 8.6588 8.6605 8.6621 8.6638 8.6654 8.6671 8.6687 8.6703 8.6719 8.6736 8.6752 8.6768 8.6784 8.6800 8.6815 8.6831 8.6847 8.6863 8.6878 8.6894 8.6909 8.6925 8.6940 8.6956 8.6971 8.6986 8.7001 8.7016 8.7031 8.7046 8.7061 8.7076 8.7091 8.7106 8.7121 8.7136 8.7150 8.7165 8.7179 8.7194 17 18 17 17 17 17 17 17 17 16 17 17 16 17 16 17 16 16 16 17 16 16 16 16 15 16 16 16 15 16 15 16 15 16 15 15 15 15 15 15 15 15 15 15 15 15 14 15 14 15 1.3599 1.3582 1.3564 1.3547 1.3530 1.3513 1.3496 1.3479 1.3462 1.3445 1.3429 1.3412 1.3395 1.3379 1.3362 1.3346 1.3329 1.3313 1.3297 1.3281 1.3264 1.3248 1.3232 1.3216 1.3200 1.3185 1.3169 1.3153 1.3137 1.3122 1.3106 1.3091 1.3075 1.3060 1.3044 1.3029 1.3014 1.2999 1.2984 1.2969 1.2954 1.2939 1.2924 1.2909 1.2894 1.2879 1.2864 1.2850 1.2835 1.2821 1.2806 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 87.500 87.490 87.480 87.47~ 87.460 87.45~ 87.440 87.430 87.42~ 87.410 87.400 87.390 87.38~ 87.370 87.360 87.350 87.340 87.330 87.320 87.310 87.300 87.290 87.280 87.27~ 87.26~ 87.250 87.240 87.23~ 87.22~ 87.210 87.200 87.190 87.18~ 87.17~ 87.160 87.150 87.140 87.130 87.120 87.110 87.10~ 87.090 87.080 87.070 87.060 87.05~ 87.040 87.030 87.020 87.010 87.00~ I I kbo 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 0> 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 14 1.4 2.8 4.2 5.6 7.0 8.4 9.8 11.2 12.6 I I I log cos di. log c coom log tan log sin Angle 87di I I 26 TABLE III. LOGARITHMIC SINES I I,i 3~ Angle log log tan com log cot log cos Prop. Parts 3.00~ 3.01~ 3.02~ 3.03~ 3.04~ 3.05~ 3.06~ 3.07~ 3.08~ 3.09~ 3.10~ 3.11~ 3.12~ 3.13~ 3.14~ 3.15~ 3.16~ 3.17~ 3.18~ 3.19~ 3.20~ 3.21~ 3.22~ 3.23~ 3.24~ 3.25~ 3.26~ 3.27~ 3.28~ 3.29~ 3.30~ 3.31~ 3.32~ 3.330 3.34~ 3.350 3.36~ 3.37~ 3.38~ 3.390 3.40~ 3-41~ 3.42~ 3.43~ 3.44~ 3.450 3.46~ 3.47~ 3.48~ 3.490 3.50~ 8.7188 8.7202 8.7217 8.7231 8.7245 8.7260 8.7274 8.7288 8.7302 8.7316 8.7330 8.7344 8.7358 8.7372 8.7386 8.7400 8.7413 8.7427 8.7441 8.7454 8.7468 8.7482 8.7495 8.7508 8.7522 8.7535 8.7549 8.7562 8.7575 8.7588 8.7602 8.7615 8.7628 8.7641 8.7654 8.7667 8.7680 8.7693 8.7705 8.7718 8.7731 8.7744 8.7756 8.7769 8.7782 8.7794 8.7807 8.7819 8.7832 8.7844 8.7857 14 8.7194 _ _ 8.7208 15 14 14 15 14 14 14 14 14 14 14 14 14 14 13 14 14 13 14 14 13 13 14 13 14 13 13 13 14 13 13 13 13 13 13 13 12 13 13 13 12 13 13 12 13 12 13 12 13 8.7223 8.7237 8.7252 8.7266 8.7280 8.7294 8.7308 8.7323 8.7337 8.7351 8.7365 8.7379 8.7392 8.7406 8.7420 8.7434 8.7448 8.7461 8.7475 8.7488 8.7502 8.7515 8.7529 8.7542 8.7556 8.7569 8.7582 8.7596 8.7609 8.7622 8.7635 8.7648 8.7661 8.7674 8.7687 8.7700 8.7713 8.7726 8.7739 8.7751 8.7764 8.7777 8.7790 8.7802 8.7815 8.7827 8.7840 8.7852 8.7865 14 15 14 15 14 14 14 14 15 14 14 14 14 13 14 14 14 14 13 14 13 14 13 14 13 14 13 13 14 13 13 13 13 13 13 13 13 13 13 13 12 13 13 13 12 13 12 13 12 13 1.2806 1.2792 1.2777 1.2763 1.2748 1.2734 1.2720 1.2706 1.2692 1.2677 1.2663 1.2649 1.2635 1.2621 1.2608 1.2594 1.2580 1.2566 1.2552 1.2539 1.2525 1.2512 1.2498 1.2485 1.2471 1.2458 1.2444 1.2431 1.2418 1.2404 1.2391 1.2378 1.2365 1.2352 1.2339 1.2326 1.2313 1.2300 1.2287 1.2274 1.2261 1.2249 1.2236 1.2223 1.2210 1.2198 1.2185 1.2173 1.2160 1.2148 1.2135 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 87.000 86.99~ 86.98~ 86.97~ 86.96~ 86.95~ 86.94~ 86.93~ 86.92~ 86.91~ 86.90~ 86.89~ 86.88~ 86.87~ 86.86~ 86.85~ 86.84~ 86.83~ 86.82? 86.81~ 86.80~ 86.79~ 86.78~ 86.77~ 86.76~ 86.75~ 86.74~ 86.73~ 86.72~ 86.71~ 86.700 86.69~ 86.68~ 86.67~ 86.66~ 86.65~ 86.64~ 86.63~ 86.62~ 86.61~ 86.60~ 86.59~ 86.58~ 86.57~ 86.56~ 86.55~ 86.54~ 86.53~ 86.52~ 86.51~ 86.50~ 4a.r1 2 3 4 5 6 7 8 9 o 3.0.I 1.5 12.0 3.5 4.5 6.0 7.5 9.0 10.5 12.0 13.5 14 1 1.4 2 2.8 3 4.2 4 5.6 5 7.0 6 8.4 7 9.8 8 11.2 9 12.6 13 1 1.3 2 2.6 3 3.9 4 5.2 5 6.5 6 7.8 7 9.1 8 10.4 9 11.7 _12 1 1.2 2 2.4 3 3.6 4 4.8 5 6.0 6 7.2 7 8.4 8 9.6 9 10.8 0 log cos diff. log cot com t l og sin Angle 86~ COSINES, TANGENTS, AND COTANGENTS 27 30 Angle log sin di. log tan ' log cot log cos Prop Parts Angle 1 log sin diff. log tan log cot log cos Prop. Parts diff. 3.50~ 3.51~ 3.52~ 3.530 3.540 3.550 3.56~ 3.570 3.58~ 3.590 3.600 3.61~ 3.62~ 3.63~ 3.64~ 3.65~ 3.66~ 3.67~ 3.68~ 3.69~ 3.70~ 3.71~ 3.72~ 3.730 3.740 3.750 3.76~ 3.770 3.78~ 3.79~ 3.80~ 3.81~ 3.82~ 3.83~ 3.84~ 3.85~ 3.86~ 3.87~ 3.88~ 3.89~ 3.90~ 3.91~ 3.92~ 3.93~ 3.940 3.950 3.96~ 3.970 3.98~ 3.990 4.00~ 8.7857 8.7869 8.7881 8.7894 8.7906 8.7918 8.7930 8.7943 8.7955 8.7967 8.7979 8.7991 8.8003 8.8015 8.8027 8.8039 8.8051 8.8062 8.8074 8.8086 8.8098 8.8109 8.8121 8.8133 8.8144 8.8156 8.8168 8.8179 8.8191 8.8202 8.8213 8.8225 8.8236 8.8248 8.8259 8.8270 8.8281 8.8293 8.8304 8.8315 8.8326 8.8337 8.8348 8.8359 8.8370 8.8381 8.8392 8.8403 8.8414 8.8425 8.8436 12 12 13 12 12 12 13 12 12 12 12 12 12 12 12 12 11 12 12 12 11 12 12 11 12 12 11 12 11 11 12 11 12 11 11 11 12 11 11 11 11 11 11 11 11 11 11 11 11 11 8.7865 8.7877 8.7890 8.7902 8.7914 8.7927 8.7939 8.7951 8.7963 8.7975 8.7988 8.8000 8.8012 8.8024 8.8036 8.8048 8.8059 8.8071 8.8083 8.8095 8.8107 8.8119 8.8130 8.8142 8.8154 8.8165 8.8177 8.8188 8.8200 8.8212 8.8223 8.8234 8.8246 8.8257 8.8269 8.8280 8.8291 8.8302 8.8314 8.8325 8.8336 8.8347 8.8358 8.8370 8.8381 8.8392 8.8403 8.8414 8.8425 8.8436 8.8446 12 13 12 12 13 12 12 12 12 13 12 12 12 12 12 11 12 12 12 12 12 11 12 12 11 12 11 12 12 11 11 12 11 12 11 11 11 12 11 11 11 11 12 11 11 11 11 11 11 10 1.2135 1.2123 1.2110 1.2098 1.2086 1.2073 1.2061 1.2049 1.2037 1.2025 1.2012 1.2000 1.1988 1.1976 1.1964 1.1952 1.1941 1.1929 1.1917 1.1905 1.1893 1.1881 1.1870 1.1858 1.1846 1.1835 1.1823 1.1812 1.1800 1.1788 1.1777 1.1766 1.1]754 1.1743 1.1731 1.1720 1.1709 1.1698 1.1686 1.1675 1.1664 1.1653 1.1642 1.1630 1.1619 1.1608 1.1597 1.1586 1.1575 1.1564 1.1554 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9989 9.9989 86.50~ 86.49~ 86.48~ 86.47~ 86.46~ 86.45~ 86.44~ 86.43~ 86.42~ 86.41~ 86.40~ 86.39~ 86.38~ 86.37~ 86.36~ 86.35~ 86.34~ 86.33~ 86.32~ 86.31~ 86.30~ 86.29~ 86.28~ 86.27~ 86.26~ 86.25~ 86.24~ 86.23~ 86.22~ 86.21~ 86.20~ 86.19~ 86.18~ 86.17~ 86.16~ 86.15~ 86.14~ 86.13~ 86.12~ 86.11~ 86.10~ 86.09~ 86.08~ 86.07~ 86.06~ 86.05~ 86.04~ 86.03~ 86.02~ 86.01~ 86.00~.4-i 1 2 3 4 5 6 7 8 9 0 0f p 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 12 1 1.2 2 2.4 3 3.6 4 4.8 5 6.0 6 7.2 7 8.4 8 9.6 9 10.8 11 1 1.1 2 2.2 3 3.3 4 4.4 5 5.5 6 6.6 7 7.7 8 8.8 9 9.9 10 1 1.0 2 2.0 3 3.0 4 4.0 5 5.0 6 6.0 7 7.0 8 8.0 9 9.0 log cos diff. log cot coi. log tan log sin Angle 86~ 28 TABLE III. LOGARITHMIC SINES 4~ Angleog sin log tanco log cos Prop. Parts Angleog sin diffl log cot log cos Prop. Parts -- I 4.000 4;01~ 4.020 4.030 4.04~ 4.05~ 4.06~ 4.07~ 4.08~ 4.090 4.100 4.11~ 4.120 4.130 4.140 4.15~ 4.16~ 4.170 4.180 4.190 4.200 4.210 4.220 4.23~ 4.240 4.25~ 4.26~ 4.270 4.280 4.290 4.300 4.31~ 4.320 4.330 4.340 4.350 4.360 4.370 4.380 4.390 4.400 4.410 4.420 4.430 4.440 4.450 4.46~ 4.470 4.48~ 4.490 4.50~ 8.8436 8.8447 8.8457 8.8468 8.8479 8.8490 8.8500 8.8511 8.8522 8.8532 8.8543 8.8553 8.8564 8.8575 8.8585 8.8595 8.8606 8.8616 8.8627 8.8637 8.8647 8.8658 8.8668 8.8678 8.8688 8.8699 8.8709 8.8719 8.8729 8.8739 8.8749 8.8759 8.8769 8.8780 8.8790 8.8799 8.8809 8.8819 8.8829 8.8839 8.8849 8.8859 8.8869 8.8878 8.8888 8.8898 8.8908 8.8917 8.8927 8.8937 8.8946 11 10 11 11 11 10 11 11 10 11 10 11 11 10 10 11 10 11 10 10 11 10 10 10 11 10 10 10 10 10 10 10 11 10 9 10 10 10 10 10 10 10 9 10 10 10 9 10 10 9 8.8446 8.8457 8.8468 8.8479 8.8490 8.8501 8.8511 8.8522 8.8533 8.8543 8.8554 8.8565 8.8575 8.8586 8.8596 8.8607 8.8617 8.8628 8.8638 8.8649 8.8659 8.8669 8.8680 8.8690 8.8700 8.8711 8.8721 8.8731 8.8741 8.8751 8.8762 8.8772 8.8782 8.8792 8.8802 8.8812 8.8822 8.8832 8.8842 8.8852 8.8862 8.8872 8.8882 8.8891 8.8901 8.8911 8.8921 8.8931 8.8940 8.8950 8.8960 11 11 11 11 11 10 11 11 10 11 11 10 11 10 11 10 11 10 11 10 10 11 10 10 11 10 10 10 10 11 10 10 10 10 10 10 10 10 10 10 10 10 9 10 10 10 10 9 10 10 lO 1.1554 1.1543 1.1532 1.1521 1.1510 1.1499 1.1489 1.1478 1.1467 1.1457 1.1446 1.1435 1.1425 1.1414 1.1404 1.1393 1.1383 1.1372 1.1362 1.1351 1.1341 1.1331 1.1320 1.1310 1.1300 1.1289 1.1279 1.1269 1.1259 1.1249 1.1238 1.1228 1.1218 1.1208 1.1198 1.1188 1.1178 1.1168 1.1158 1.1148 1.1138 1.1128 1.1118 1.1109 1.1099 1.1089 1.1079 1.1069 1.1060 1.1050 1.1040 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 86.000 85.990 85.98~ 85.970 85.96~ 85.95~ 85.94~ 85.930 85.920 85.91~ 85.90~ 85.89~ 85.880 85.87~ 85.86~ 85.850 85.840 85.83~ 85.82~ 85.810 85.80~ 85.790 85.780 85.770 85.76~ 85.750 85.740 85.730 85.72~ 85.710 85.700 85.690 85.68~ 85.67~ 85.660 85.65~ 85.64~ 85.630 85.62~ 85.61~ 85.600 85.59~ 85.58~ 85.570 85.56~ 85.550 85.54~ 85.53~ 85.520 85.510 85.50~ I bn I., 1 2 3 4 5 6 7 8 9 o a) 0) 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 10 1 1.0 2 2.0 3 3.0 4 4.0 5 5.0 6 6.0 7 7.0 8 8.0 9 9.0 9 1 0.9 2 1.8 3 2.7 4 3.6 5 4.5 6 5.4 7 6.3 8 7.2 9 8.1 0 log cos dif. log cot com. log tan log sin Angle 85~ I COSINES, TANGENTS, AND COTANGENTS 29 Angle.4~0 i.n Angle log sin if log tan co log cot log cosProp. Parts duff, log diff. _____ _____ 4.50~ 4.51~ 4.520 4.530 4.540 4.550 4.56~ 4.570 4.58~ 4.590 4.600 4.61~ 4.62~ 4.63~ 4.64~ 4.650 4.660 4.67~ 4.68~ 4.69~ 4.70~ 4.71~ 4.720 4.730 4.740 4.750 4.760 4.77~ 4.780 4.790 4.80~ 4.810 4.82~ 4.83~ 4.840 4.85~ 4.86~ 4.870 4.88~ 4.890 4.90~ 4.910 4.92~ 4.930 4.940 4.950 4.96~ 4.970 4.980 4.990 5.00~ 8.8946 8.8956 8.8966 8.8975 8.8985 8.8994 8.9004 8.9013 8.9023 8.9032 8.9042 8.9051 8.9060 8.9070 8.9079 8.9089 8.9098 8.9107 8.9116 8.9126 8.9135 8.9144 8.9153 8.9162 8.9172 8.9181 8.9190 8.9199 8.9208 8.9217 8.9226 8.9235 8.9244 8.9253 8.9262 8.9271 8.9280 8.9289 8.9298 8.9307 8.9315 8.9324 8.9333 8.9342 8.9351 8.9359 8.9368 8.9377 8.9386 8.9394 8.9403 10 10 9 10 9 10 9 10 9 10 9 9 10 9 10 9 9 9 10 9 9 9 9 10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 9 9 9 9 8 9 9 9 8 9 8.8960 8.8970 8.8979 8.8989 8.8998 8.9008 8.9018 8.9027 8.9037 8.9046 8.9056 8.9065 8.9075 8.9084 8.9093 8.9103 8.9112 8.9122 8.9131 8.9140 8.9150 8.9159 8.9168 8.9177 8.9186 8.9196 8.9205 8.9214 8.9223 8.9232 8.9241 8.9250 8.9260 8.9269 8.9278 8.9287 8.9296 8.9305 8.9313 8.9322 8.9331 8.9340 8.9349 8.9358 8.9367 8.9376 8.9384 8.9393 8.9402 8.9411 8.9420 10 9 io 10 9 10 10 9 10 9 10 9 10 9 9 10 9 10 9 9 10 9 9 9 9 10 9 9 9 9 9 9 10 9 9 9 9 9 8 9 9 9 9 9 8 9 9 9 9 1.1040 1.1030 1.1021 1.1011 1.1002 1.0992 1.0982 1.0973 1.0963 1.0954 1.0944 1.0935 1.0925 1.0916 1.0907 1.0897 1.0888 1.0878 1.0869 1.0860 1.0850 1.0841 1.0832 1.0823 1.0814 1.0804 1.0795 1.0786 1.0777 1.0768 1.0759 1.0750 1.0740 1.0731 1.0722 1.0713 1.0704 1.0695 1.0687 1.0678 1.0669 1.0660 1.0651 1.0642 1.0633 1.0624 1.0616 1.0607 1.0598 -1.0589 1.0580 9.9987 9.9987 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9983 85.500 85.49~ 85.48~ 85.47~ 85.46~ 85.45~ 85.44~ 85.430 85.42~ 85.41~ 85.400 85.39~ 85.380 85.37~ 85.36~ 85.35~ 85.34~ 85.330 85.320 85.310 85.30~ 85.29~ 85.280 85.27~ 85.26~ 85.25~ 85.24~ 85.23~ 85.220 85.210 85.20~ 85.190 85.180 85.17~ 85.160 85.15~ 85.14~ 85.130 85.12~ 85.110 85.100 85.090 85.080 85.070 85.060 85.050 85.04~ 85.030 85.020 85.01~ 85.00~ 4-3 1 2 3 4 5 6 7 8 9 C) 0 10 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 9 1 0.9 2 1.8 3 2.7 4 3.6 5 4.5 6 5.4 7 6.3 8 7.2 9 8.1 8 1 0.8 2 1.6 3 2.4 4 3.2 5 4.0 6 4.8 7 5.6 8 6.4 9 7.2 log sinf co log cos ff. log cot co an log sin Angle 8diff. 85~ ago~~~~~ 30 TABLE III. LOGARITHMIC SINES 5~0100 Angle l iog nsin diff. om. Angle log sin di log tan com.ff, log cot log cos diff. Prop. Parts _ _ _ _ _ _ _ _ _ _. _ ___diff. 5.00 5.1~ 5.2~ 5.30 5.40 5.50 5.60 5.70 5.80 5.90 6.0~ 6.1~ 6.20 6.30 6.4~ 6.5~ 6.60 6.70 6.8~ 6.90 7.00 7.1~ 7.2~ 7.30 7.40 7.50 7.60 7.70 7.80 7.9~ 8.0~ 8.10 8.20 8.30 8.40 8.5~ 8.60 8.70 8.80 8.90 9.00 9.10 9.2~ 9.30 9.4~ 9.50 9.6~ 9.7~ 9.8~ 9.90 10.0~ 8.9403 8.9489 8.9573 8.9655 8.9736 8.9816 8.9894 8.9970 9.0046 9.0120 9.0192 9.0264 9.0334 9.0403 9.0472 9.0539 9.0605 9.0670 9.0734 9.0797 9.0859 9.0920 9.0981 9.1040 9.1099 9.1157 9.1214 9.1271 9.1326 9.1381 9.1436 9.1489 9.1542 9.1594 9.1646 9.1697 9.1747 9.1797 9.1847 9.1895 9.1943 9.1991 9.2038 9.2085 9.2131 9.2176 9.2221 9.2266 9.2310 9.2353 9.2397 86 84 82 81 80 78 76 76 74 72 72 70 69 69 67 66 65 64 63 62 61 61 59 59 58 57 57 55 55 55 53 53 52 52 51 50 50 50 48 48 48 47 47 46 45 45 45 44 43 44 8.9420 8.9506 8.9591 8.9674 8.9756 8.9836 8.9915 8.9992 9.0068 9.0143 9.0216 9.0289 9.0360 9.0430 9.0499 9.0567 9.0633 9.0699 9.0764 9.0828 9.0891 9.0954 9.1015 9.1076 9.1135 9.1194 9.1252 9.1310 9.1367 9.1423 9.1478 9.1533 9.1587 9.1640 9.1693 9.1745 9.1797 9.1848 9.1898 9.1948 9.1997 9.2046 9.2094 9.2142 9.2189 9.2236 9.2282 9.2328 9.2374 9.2419 9.2463 86 85 83 82 80 79 77 76 75 73 73 71 70 69 68 66 66 65 64 63 63 61 61 59 59 58 58 57 56 55 55 54 53 53 52 52 51 50 50 49 49 48 48 47 47 46 46 46 45 44 1.0580 1.0494 1.0409 1.0326 1.0244 1.0164 1.0085 1.0008 0.9932 0.9857 0.9784 0.9711 0.9640 0.9570 0.9501 0.9433 0.9367 0.9301 0.9236 0.9172 0.9109 0.9046 0.8985 0.8924 0.8865 0.8806 0.8748 0.8690 0.8633 0.8577 0.8522 0.8467 0.8413 0.8360 0.8307 0.8255 0.8203 0.8152 0.8102 0.8052 0.8003 0.7954 0.7906 0.7858 0.7811 0.7764 0.7718 0.7672 0.7626 0.7581 0.7537 9.9983 9.9983 9.9982 9.9981 9.9981 9.9980 9.9979 9.9978 9.9978 9.9977 9.9976 9.9975 9.9975 9.9974 9.9973 9.9972 9.9971 9.9970 9.9969 9.9968 9.9968 9.9967 9.9966 9.9965 9.9964 9.9963 9.9962 9.9961 9.9960 9.9959 9.9958 9.9956 9.9955 9.9954 9.9953 9.9952 9.9951 9.9950 9.9949 9.9947 9.9946 9.9945 9.9944 9.9943 9.9941 9.9940 9.9939 9.9937 9.9936 9.9935 9.9934 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 2 1 1 2 1 1 1 85.00 84.90 84.8~ 84.70 84.60 84.50 84.40 84.30 84.20 84.10 84.00 83.90 83.80 83.7~ 83.60 83.50 83.40 83.3~ 83.20 83.10 83.0~ 82.90 82.80 82.70 82.60 82.50 82.4~ 82.30 82.20 82.10 82.00 81.90 81.80 81.70 81.60 81.50 81.40 81.3~ 81.2~ 81.10 81.00 80.90 80.8~ 80.70 80.60 80.5~ 80.4~ 80.30 80.2~ 80.10 80.0~ 4-o *l) *.3 F1 Ca Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 62 6.2 12.4 18.6 24.8 31.0 37.2 43.4 49.6 55.8 59 5.9 11.8 17.7 23.6 29.5 35.4 41.3 47.2 53.1 56 5.6 11.2 16.8 22.4 28.0 33.6 39.2 44.8 50.4 53 5.3 10.6 15.9 21.2 26.5 31.8 37.1 42.4 47.7 50 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 47 4.7 9.4 14.1 18.8 23.5 28.2 32.9 37.6 42.3 61 60 6.1 6.0 12.2 12.0 18.3 18.0 24.4 24.0 30.5 30.0 36.6 36.0 42.7 42.0 48.8 48.0 54.9 54.0 58 57 5.8 5.7 11.6 11.4 17.4 17.1 23.2 22.8 29.0 28.5 34.8 34.2 40.6 39.9 46.4 45.6 52.2 51.3 55 54 5.5 5.4 11.0 10.8 16.5 16.2 22.0 21.6 27.5 27.0 33.0 32.4 38.5 37.8 44.0 43.2 49.5 48.6 52 51 5.2 5.1 10.4 10.2 15.6 15.3 20.8 20.4 26.0 25.5 31.2 30.6 36.4 35.7 41.6 40.8 46.8 45.9 49 48 4.9 4.8 9.8 9.6 14.7 14.4 19.6 19.2 24.5 24.0 29.4 28.8 34.3 33.6 39.2 38.4 44.1 43.2 46 45 4.6 4.5 9.2 9.0 13.8 13.5 18.4 18.0 23.0 22.5 27.6 27.0 32.2 31.5 36.8 36.0 41.4 40.5 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 log cos diff. log cot com. log tan log sin diff. Angle 8 diff 800-85~ j COSINES, TANGENTS, AND COTANGENTS 31 10~-15~ Angle log sin diff. log tan codn. log cot log cos diff. Prop. Parts 0,,.I I - 10.0~ 10.1~ 10.2~ 10.3~ 10.4~ 10.5~ 10.6~ 10.7~ 10.8~ 10.9~ 11.0~ 11.1~ 11.2~ 11.3~ 11.4~ 11.5~ 11.6~ 11.7~ 11.8~ 11.9~ 12.0~ 12.1~ 12.2~ 12.3~ 12.4~ 12.5~ 12.6~ 12.7~ 12.8~ 12.9~ 13.0~ 13.1~ 13.2~ 13.3~ 13.4~ 13.5~ 13.6~ 13.7~ 13.8~ 13.9~ 14.0~ 14.1~ 14.2~ 14.3~ 14.4~ 14.5~ 14.6~ 14.7~ 14.8~ 14.9~ 15.0~ 9.2397 9.2439 9.2482 9.2524 9.2565 9.2606 9.2647 9.2687 9.2727 9.2767 9.2806 9.2845 9.2883 9.2921 9.2959 9.2997 9.3034 9.3070 9.3107 9.3143 9.3179 9.3214 9.3250 9.3284 9.3319 9.3353 9.3387 9.3421 9.3455 9.3488 9.3521 9.3554 9.3586 9.3618 9.3650 9.3682 9.3713 9.3745 9.3775 9.3806 9.3837 9.3867 9.3897 9.3927 9.3957 9.3986 9.4015 9.4044 9.4073 9.4102 9.4130 42 43 42 41 41 41 40 40 40 39 39 38 38 38 38 37 36 37 36 36 35 36 34 35 34 34 34 34 33 33 33 32 32 32 32 31 32 30 31 31 30 30 30 30 29 29 29 29 29 28 9.2463 9.2507 9.2551 9.2594 9.2637 9.2680 2.2722 9.2764 9.2805 9.2846 9.2887 9.2927 9.2967 9.3006 9.3046 9.3085 9.3123 9.3162 9.3200 9.3237 9.3275 9.3312 9.3349 9.3385 9.3422 9.3458 9.3493 9.3529 9.3564 9.3599 9.3634 9.3668 9.3702 9.3736 9.3770 9.3804 9.3837 9.3870 9.3903 9.3935 9.3968 9.4000 9.4032 9.4064 9.4095 9.4127 9.4158 9.4189 9.4220 9.4250 9.4281 44 44 43 43 43 42 42 41 41 41 40 40 39 40 39 38 39 38 37 38 37 37 36 37 36 35 36 35 35 35 34 34 34 34 34 33 33 33 32 33 32 32 32 31 32 31 31 31 30 31 0.7537 0.7493 0.7449 0.7406 0.7363 0.7320 0.7278 0.7236 0.7195 0.7154 0.7113 0.7073 0.7033 0.6994 0.6954 0.6915 0.6877 0.6838 0.6800 0.6763 0.6725 0.6688 0.6651 0.6615 0.6578 0.6542 0.6507 0.6471 0.6436 0.6401 0.6366 0.6332 0.6298 0.6264 0.6230 0.6196 0.6163 0.6130 0.6097 0.6065 0.6032 0.6000 0.5968 0.5936 0.5905 0.5873 0.5842 0.5811 0.5780 0.5750 0.5719, I _, 9.9934 9.9932 9.9931 9.9929 9.9928 9.9927 9.9925 9.9924 9.9922 9.9921 9.9919 9.9918 9.9916 9.9915 9.9913 9.9912 9.9910 9.9909 9.9907 9.9906 9.9904 9.9902 9.9901 9.9899 9.9897 9.9896 9.9894 9.9892 9.9891 9.9889 9.9887 9.9885 9.9884 9.9882 9.9880 9.9878 9.9876 9.9875 9.9873 9.9871 9.9869 9.9867 9.9865 9.9863 9.9861 9.9859 9.9857 9.9855 9.9853 9.9851 9.9849 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 80.0~ 79.90 79.8~ 79.70 79.6~ 79.50 79.40 79.30 79.2~ 79.1~ 79.00 78.9~ 78.8~ 78.7~ 78.6~ 78.50 78.4~ 78.3~ 78.2~ 78.1~ 78.0~ 77.9~ 77.8~ 77.7~ 77.6~ 77.5~ 77.4~ 77.30 77.2~ 77.1~ 77.00 76.9~ 76.8~ 76.7~ 76.6~ 76.5~ 76.4~ 76.3~ 76.2~ 76.1~ 76.00 75.9~ 75.8~ 75.7~ 75.6~ 75.50 75.4~ 75.3~ 75.2~ 75.1~ 75.0~ I 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 -4 Id ea *C q 4-3 m 1 2 3 4 5 6 7 8 9 Difference 44 43 42 4.4 4.3 4.2 8.8 8.6 8.4 13.2 12.9 12.6 17.6 17.2 16.8 22.0 21.5 21.0 26.4 25.8 25.2 30.8 30.1 29.4 35.2 34.5 33.6 39.6 38.8 37.8 41 40 39 4.1 4.0 3.9 8.2 8.0 7.8 12.3 12.0 11.7 16.4 16.0 15.6 20.5 20.0 19.5 24.6 24.0 23.4 28.7 28.0 27.3 32.8 32.0 31.2 36.9 36.0 35.1 38 37 36 3.8 3.7 3.6 7.6 7.4 7.2 11.4 11.1 10.8 15.2 14.8 14.4 19.0 18.5 18.0 22.8 22.2 21.6 26.6 25.9 25.2 30.4 29.6 28.8 34.2 33.3 32.4 35 34 33 3.5 3.4 3.3 7.0 6.8 6.6 10.5 10.2 9.9 14.0 13.6 13.2 17.5 17.0 16.5 21.0 20.4 19.8 24.5 23.8 23.1 28.0 27.2 26.4 31.5 30.6 29.7 32 31 30 3.2 3.1 3.0 6.4 6.2 6.0 9.6 9.3 9.0 12.8 12.4 12.0 16.0 15.5 15.0 19.2 18.6 18.0 22.4 21.7 21.0 25.6 24.8 24.0 28.8 27.9 27.0 29 28 2 2.9 2.8 0.2 5.8 5.6 0.4 8.7 8.4 0.6 11.6 11.2 0.8 14.5 14.0 1.0 17.4 16.8 1.2 20.3 19.6 1.4 23.2 22.4 1.6 26.1 25.2 1.8 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 --- — --- 1 I- I. - - I - - 1 I -. log cos diff. log cot com, log tan log sin diff. Angle diff. 75~-80o I 32 TABLE III. LOGARITHMIC SINES I 150-200 Angle log sin diff. Jlog tan codiff log cot log cos diff. Prop. Parts ~-~~I~p~- ~- I COI~diff 15.0~ 15.10 15.20 15.30 15.4~ 15.50 15.60 15.70 15.80 15.90 16.0~ 16.1~ 16.2~ 16.30 16.40 16.50 16.60 16.70 16.80 16.90 17.00 17.10 17.20 17.30 17.40 17.50 17.60 17.70 17.80 17.90 18.00 18.10 18.20 18.30 18.40 18.50 18.60 18.70 18.8~ 18.90 19.00 19.10 19.20 19.30 19.40 19.50 19.60 19.7~ 19.8~ 19.90 20.00 9.4130 9.4158 9.4186 9.4214 9.4242 9.4269 9.4296 9.4323 9.4350 9.4377 9.4403 9.4430 9.4456 9.4482 9.4508 9.4533 9.4559 9.4584 9.4609 9.4634 9.4659 9.4684 9.4709 9.4733 9.4757 9.4781 9.4805 9.4829 9.4853 9.4876 9.4900 9.4923 9.4946 9.4969 9.4992 9.5015 9.5037 9.5060 9.5082 9.5104 9.5126 9.5148 9.5170 9.5192 9.5213 9.5235 9.5256 9.5278 9.5299 9.5320 9.5341 28 28 28 28 27 27 27 27 27 26 27 26 26 26 25 26 25 25 25 25 25 25 24 24 24 24 24 24 23 24 23 23 23 23 23 22 23 22 22 22 22 22 22 21 22 21 22 21 21 21 9.4281 9.4311 9.4341 9.4371 9.4400 9.4430 9.4459 9.4488 9.4517 9.4546 9.4575 9.4603 9.4632 9.4660 9.4688 9.4716 9.4744 9.4771 9.4799 9.4826 9.4853 9.4880 9.4907 9.4934 9.4961 9.4987 9.5014 9.5040 9.5066 9.5092 9.5118 9.5143 9.5169 9.5195 9.5220 9.5245 9.5270 9.5295 9.5320 9.5345 9.5370 9.5394 9.5419 9.5443 9.5467 9.5491 9.5516 9.5539 9.5563 9.5587 9.5611 30 30 30 29 30 29 29 29 29 29 28 29 28 28 28 28 27 28 27 27 27 27 27 27 26 27 26 26 26 26 25 26 26 25 25 25 25 25 25 25 24 25 24 24 24 25 23 24 24 24 0.5719 0.5689 0.5659 0.5629 0.5600 0.5570 0.5541 0.5512 0.5483 0.5454 0.5425 0.5397 0.5368 0.5340 0.5312 0.5284 0.5256 0.5229 0.5201 0.5174 0.5147 0.5120 0.5093 0.5066 0.5039 0.5013 0.4986 0.4960 0.4934 0.4908 0.4882 0.4857 0.4831 0.4805 0.4780 0.4755 0.4730 0.4705 0.4680 0.4655 0.4630 0.4606 0.4581 0.4557 0.4533 0.4509 0.4484 0.4461 0.4437 0.4413 0.4389 9.9849 9.9847 9.9845 9.9843 9.9841 9.9839 9.9837 9.9835 9.9833 9.9831 9.9828 9.9826 9.9824 9.9822 9.9820 9.9817 9.9815 9.9813 9.9811 9.9808 9.9806 9.9804 9.9801 9.9799 9.9797 9.9794 9.9792 9.9789 9.9787 9.9785 9.9782 9.9780 9.9777 9.9775 9.9772 9.9770 9.9767 9.9764 9.9762 9.9759 9.9757 9.9754 9.9751 9.9749 9.9746 9.9743 9.9741 9.9738 9.9735 9.9733 9.9730 2 2 2 2 2 2 2 2 2 3 2 2 2 2 3 2 2 2 3 2 2 3 2 2 3 2 3 2 3 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 2 3 3 2 3 75.0~ 74.90 74.8~ 74.70 74.60 74.50 74.40 74.30 74.2~ 74.10 74.00 73.90 73.80 73.70 73.60 73.50 73.40 73.30 73.20 73.10 73.00 72.9~ 72.80 72.70 72.60 72.5~ 72.40 72.30 72.20 72.10 72.00 71.90 71.8~ 71.70 71.60 71.5~ 71.40 71.30 71.20 71.10 71.00 70.90 70.8~ 70.70 70.60 70.50 70.40 70.30 70.20 70.1~ 70.00 a 0'3 2 3.4 5 6 7 8 9 Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 30 3.0 6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 28 2.8 5.6 8.4 11.2 14.0 16.8 19.6 22.4 25.2 26 2.6 5.2 7.8 10.4 13.0 15.6 18.2 20.8 23.4 24 2.4 4.8 7.2 9.6 12.0 14.4 16.8 19.2 21.6 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 29 2.9 5.8 8.7 11.6 14.5 17.4 20.3 23.2 26.1 27 2.7 5.4 8.1 10.8 13.5 16.2 18.9 21.6 24.3 25 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 23 2.3 4.6 6.9 9.2 11.5 13.8 16.1 18.4 20.7 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 I I log cos diff. log cot com. logtan log sin diff. Angle diff 70 75 0 COSINES, TANGENTS, AND COTANGENTS 33 20~-250 Angle log sin diff, log tan comi log cot log cos diff. Prop. Parts lo_ sin diff. log tan diff. I I I 20.0~ 20.10 20.20 20.30 20.40 20.5~ 20.6~ 20.70 20.80 20.90 21.00 21.10 21.2~ 21.30 21.4~ 21.50 21.60 21.7~ 21.8S 21.90 22.00 22.1~ 22.20 22.3~ 22.40 22.50 22.60 22.70 22.8~ 22.90 23.00 23.10 23.20 23.3~ 23.4~ 23.50 23.60 23.70 23.8~ 23.90 24.00 24.10 24.20 24.30 24.4~ 24.5~ 24.60 24.70 24.80 24.9~ 25.00 9.5341 9.5361 9.5382 9.5402 9.5423 9.5443 9.5463 9.5484 9.5504 9.5523 9.5543 9.5563 9.5583 9.5602 9.5621 9.5641 9.5660 9.5679 9.5698 9.5717 9.5736 9.5754 9.5773 9.5792 9.5810 9.5828 9.5847 9.5865 9.5883 9.5901 9.5919 9.5937 9.5954 9.5972 9.5990 9.6007 9.6024 9.6042 9.6059 9.6076 9.6093 9.6110 9.6127 9.6144 9.6161 9.61.77 9.6194 9.6210 9.6227 9.6243 9.6259 I 20 21 20 21 20 20 21 20 19 20 20 20 19 19 20 19 19 19 19 19 18 19 19 18 18 19 18 18 18 18 18 17 18 18 17 17 18 17 17 17 17 17 17 17 16 17 16 17 16 16 9.5611 9.5634 9.5658 9.5681 9.5704 9.5727 9.5750 9.5773 9.5796 9.5819 9.5842 9.5864 9.5887 9.5909 9.5932 9.5954 9.5976 9.5998 9.6020 9.6042 9.6064 9.6086 9.6108 9.6129 9.6151 9.6172 9.6194 9.6215 9.6236 9.6257 9.6279 9.6300 9.6321 9.6341 9.6362 9.6383 9.6404 9.6424 9.6445 9.6465 9.6486 9.6506 9.6527 9.6547 9.6567 9.6587 9.6607 9.6627 9.6647 9.6667 9.6687 23 24 23 23 23 23 23 23 23 23 22 23 22 23 22 22 22 22 22 22 22 22 21 22 21 22 21 21 21 22 21 21 20 21 21 21 20 21 20 21 20 21 20 20 20 20 20 20 20 20 0.4389 0.4366 0.4342 0.4319 0.4296 0.4273 0.4250 0.4227 0.4204 0.4181 0.4158 0.4136 0.4113 0.4091 0.4068 0.4046 0.4024 0.4002 0.3980 0.3958 0.3936 0.3914 0.3892 0.3871 0.3849 0.3828 0.3806 0.3785 0.3764 0.3743 0.3721 0.3700 0.3679 0.3659 0.3638 0.3617 0.3596 0.3576 0.3555 0.3535 0.3514 0.3494 0.3473 0.3453 0.3433 0.3413 0.3393 0.3373 0.3353 0.3333 0.3313 I 9.9730 9.9727 9.9724 9.9722 9.9719 9.9716 9.9713 9.9710 9.9707 9.9704 9.9702 9.9699 9.9696 9.9693 9.9690 9.9687 9.9684 9.9681 9.9678 9.9675 9.9672 9.9669 9.9666 9.9662 9.9659 9.9656 9.9653 9.9650 9.9647 9.9643 9.9640 9.9637 9.9634 9.9631 9.9627 9.9624 9.9621 9.9617 9.9614 9.9611 9.9607 9.9604 9.9601 9.9597 9.9594 9.9590 9.9587 9.9583 9.9580 9.9576 9.9573 3 3 2 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 3 3 3 4 3 3 3 3 3 4 3 3 3 3 4 3 3 4 3 3 4 3 3 4 3 4 3 4 3 4 3 70.00 69.9~ 69.8~ 69.7~ 69.6~ 69.50 69.4~ 69.30 69.20 69.1~ 69.00 68.90 68.80 68.70 68.6~ 68.50 68.4~ 68.30 68.20 68.10 68.0~ 67.90 67.8~ 67.70 67.60 67.50 67.40 67.3~ 67.2~ 67.1~ 67.00 66.90 66.80 66.70 66.6~ 66.50 66.4~ 66.3~ 66.20 66.10 66.00 65.90 65.80 65.7~ 65.60 65.50 65.4~ 65.30 65.2~ 65.1~ 65.00 I 4.! f. 4 g -4 I3 33 L4.2 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 Difference 23 22 2.3 2.2 4.6 4.4 6.9 6.6 9.2 8.8 11.5 11.0 13.8 13.2 16.1 15.4 18.4 17.6 20.7 19.8 21 20 2.1 2.0 4.2 4.0 6.3 6.0 8.4 8.0 10.5 10.0 12.6 12.0 14.7 14.0 16.8 16.0 18.9 18.0 19 18 1.9 1.8 3.8 3.6 5.7 5.4 7.6 7.2 9.5 9.0 11.4 10.8 13.3 12.6 15.2 14.4 17.1 16.2 17 16 1.7 1.6 3.4 3.2 5.1 4.8 6.8 6.4 8.5 8.0 10.2 9.6 11.9 11.2 13.6 12.8 15.3 14.4 2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 3 4 0.3 0.4 0.6 0.8 0.9 1.2 1.2 1.6 1.5 2.0 1.8 2.4 2.1 2.8 2.4 3.2 2.7 3.6 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 I I I I I I I I~ ~ I I~ I ~ log cos diff. log cot codif log tan log sin diff. Angle 65~-70~ I 34 TABLE III. LOGARITHMIC SINES 25~-30~ Angle log sin diff. log tan com. log cot log cos diff. Prop. Parts 25.0~ 25.1~ 25.2~ 25.3~ 25.4~ 25.5~ 25.6~ 25.7~ 25.8~ 25.9~ 26.00 26.1~ 26.2~ 26.3~ 26.4~ 26.5~ 26.6~ 26.7~ 26.8~ 26.9~ 27.0~ 27.1~ 27.2~ 27.3~ 27.4~ 27.5~ 27.6~ 27.7~ 27.8~ 27.9~ 28.00 28.1~ 28.2~ 28.3~ 28.4~ 28.5~ 28.6~ 28.7~ 28.8~ 28.9~ 29.00 29.1~ 29.2~ 29.3~ 29.4~ 29.5~ 29.6~ 29.7~ 29.8~ 29.9~ 30.0~ 9.6259 9.6276 9.6292 9.6308 9.6324 9.6340 9.6356 9.6371 9.6387 9.6403 9.6418 9.6434 9.6449 9.6465 9.6480 9.6495 9.6510 9.6526 9.6541 9.6556 9.6570 9.6585 9.6600 9.6615 9.6629 9.6644 9.6659 9.6673 9.6687 9.6702 9.6716 9.6730 9.6744 9.6759 9.6773 9.6787 9.6801 9.6814 9.6828 9.6842 9.6856 9.6869 9.6883 9.6896 9.6910 9.6923 9.6937 9.6950 9.6963 9.6977 9.6990 17 16 16 16 16 16 15 16 16 15 16' 15 16 15 15 15 16 15 15 14 15 15 15 14 15 15 14 14 15 14 14 14 15 14 14 14 13 14 14 14 13 14 13 14 13 14 13 13 14 13 9.6687 9.6706 9.6726 9.6746 9.6765 9.6785 9.6804 9.6824 9.6843 9.6863 9.6882 9.6901 9.6920 9.6939 9.6958 9.6977 9.6996 9.7015 9.7034 9.7053 9.7072 9.7090 9.7109 9.7128 9.7146 9.7165 9.7183 9.7202 9.7220 9.7238 9.7257 9.7275 9.7293 9.7311 9.7330 9.7348 9.7366 9.7384 9.7402 9.7420 9.7438 9.7455 9.7473 9.7491 9.7509 9.7526 9.7544 9.7562 9.7579 9.7597 9.7614 19 20 20 19 20 19 20 19 20 19 19 19 19 19 19 19 19 19 19 19 18 19 19 18 19 18 19 18 18 19 18 18 18 19 18 18 18 18 18 18 17 18 18 18 17 18 18 17 18 17 0.3313 0.3294 0.3274 0.3254 0.3235 0.3215 0.3196 0.3176 0.3157 0.3137 0.3118 0.3099 0.3080 0.3061 0.3042 0.3023 0.3004 0.2985 0.2966 0.2947 0.2928 0.2910 0.2891 0.2872 0.2854 0.2835 0.2817 0.2798 0.2780 0.2762 0.2743 0.2725 0.2707 0.2689 0.2670 0.2652 0.2634 0.2616 0.2598 0.2580 0.2562 0.2545 0.2527 0.2509 0.2491 0.2474 0.2456 0.2438 0.2421 0.2403 0.2386 9.9573 9.9569 9.9566 9.9562 9.9558 9.9555 9.9551 9.9548 9.9544 9.9540 9.9537 9.9533 9.9529 9.9525 9.9522 9.9518 9.9514 9.9510 9.9506 9.9503 9.9499 9.9495 9.9491 9.9487 9.9483 9.9479 9.9475 9.9471 9.9467 9.9463 9.9459 9.9455 9.9451 9.9447 9.9443 9.9439 9.9435 9.9431 9.9427 9.9422 9.9418 9.9414 9.9410 9.9406 9.9401 9.9397 9.9393 9.9388 9.9384 9.9380 9.9375 4 3 4 4 3 4 3 4 4 3 4 4 4 3 4 4 4 4 3 4 4 ' 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 4 4 4 4 5 4 4 5 4 4 5 65.0~ 64.9~ 64.8~ 64.7~ 64.6~ 64.5~ 64.4~ 64.3~ 64.2~ 64.1~ 64.0~ 63.9~ 63.8~ 63.7~ 63.6~ 63.5~ 63.4~ 63.3~ 63.2~ 63.1~ 63.0~ 62.9~ 62.8~ 62.7~ 62.6~ 62.5~ 62.4~ 62.3~ 62.2~ 62.1~ 62.0~ 61.9~ 61.8~ 61.7~ 61.6~ 61.5~ 61.4~ 61.3~ 61.2~ 61.1~ 61.0~ 60.9~ 60.8~ 60.7~ 60.6~ 60.5~ 60.4~ 60.3~ 60.2~ 60.1~ 60.0~.rl -14 *so C M4 - Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 20 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 14 1.4 2.8 4.2 5.6 7.0 8.4 9.8 11.2 12.6 3 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 19 1.9 3.8 5.7 7.6 9.5 11.4 13.3 15.2 17.1 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 4 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 log cos diff. log cot com. log tan log sin diff. Angle 60~-65~ COSINES, TANGENTS, AND COTANGENTS 35 0 30~-35~ Angle log sin diff. log tan com' log cot log cos, diff. Prop. Parts 30.0~ 30.1~ 30.2~ 30.30 30.40 30.50 30.6~ 30.7~ 30.8~ 30.90 31.0~ 31.1~ 31.2~ 31.3~ 31.4~ 31.5~ 31.6~ 31.7~ 31.8~ 31.9~ 32.0~ 32.1~ 32.2~ 32.3~ 32.4~ 32.5~ 32.6~ 32.7~ 32.8~ 32.9~ 33.0~ 33.1~ 33.2~ 33.30' 33.40 33.50 33.6~ 33.7~ 33.8~ 33.9~ 34.0~ 34.1~ 34.2~ 34.3~ 34.4~ 34.50 34.6~ 34.7~ 34.8~ 34.90 35.0~ 9.6990 9.7003 9.7016 9.7029 9.7042 9.7055 9.7068 9.7080 9.7093 9.7106 9.7118 9.7131 9.7144 9.7156 9.7168 9.7181 9.7193 9.7205 9.7218 9.7230 9.7242 9.7254 9.7266 9.7278 9.7290 9.7302 9.7314 9.7326 9.7338 9.7349 9.7361 9.7373 9.7384v 9.7396 9.7407 9.7419 9.7430 9.7442 9.7453 9.7464 9.7476 9.7487 9.7498 9.7509 9.7520 9.7531 9.7542 9.7553 9.7564 9.7575 9.7586 13 13 13 13 13 13 12 13 13 12 13 13 12 12 13 12 12 13 12 12 12 12 12 12 12 12 12 12 11 12 12 11 12 11 12 11 12 11 11 12 11 11 11 11 11 11 11 11 11 11 9.7614 9.7632 9.7649 9.7667 9.7684 9.7701 9.7719 9.7736 9.7753 9.7771 9.7788 9.7805 9.7822 9.7839 9.7856 9.7873 9.7890 9.7907 9.7924 9.7941 9.7958 9.7975 9.7992 9.8008 9.8025 9.8042 9.8059 9.8075 9.8092 9.8109 9.8125 9.8142 9.8158 9.8175 9.8191 9.8208 9.8224 9.8241 9.8257 9.8274 9.8290 9.8306 9.8323 9.8339 9.8355 9.8371 9.8388 9.8404 9.8420 9.8436 9.8452 18 17 18 17 17 18 17 17 18 17 17 17 17 17 17 17 17 17 17 17 17 17 16 17 17 17 16 17 17 16 17 16 17 16 17 16 17 16 17 16 16 17 16 16 16 17 16 16 16 16 0.2386 0.2368 0.2351 0.2333 0.2316 0.2299 0.2281 0.2264 0.2247 0.2229 0.2212 0.2195 0.2178 0.2161 0.2144 0.2127 0.2110 0.2093 0.2076 0.2059 0.2042 0.2025 0.2008 0.1992 0.1975 0.1958 0.1941 0.1925 0.1908 0.1891 0.1875 0.1858 0.1842 0.1825 0.1809 0.1792 0.1776 0.1759 0.1743 0.1726 0.1710 0.1694 0.1677 0.1661 0.1645 0.1629 0.1612 0.1596 0.1580 0.1564 0.1548 9.9375 9.9371 9.9367 9.9362 9.9358 9.9353 9.9349 9.9344 9.9340 9.9335 9.9331 9.9326 9.9322 9.9317 9.9312 9.9308 9.9303 9.9298 9.9294 9.9289 9.9284 9.9279 9.9275 9.9270 9.9265 9.9260 9.9255 9.9251 9.9246 9.9241 9.9236 9.9231 9.9226 9.9221 9.9216 9.9211 9.9206 9.9201 9.9196 9.9191 9.9186 9.9181 9.9175 9.9170 9.9165 9.9160 9.9155 9.9149 9.9144 9.9139 9.9134 4 4 5 4 5 4 5 4 5 4 5 4 5 5 4 5 5 4 5 5 5 4 5 5 5 5 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 5 5 5 5 6 5 5 5 60.0~ 59.90 59.8~ 59.70 59.6~ 59.5~ 59.40 59.30 59.2~ 59.1~ 59.0~ 58.9~ 58.8~ 58.7~ 58.6~ 58.5~ 58.4~ 58.3~ 58.2~ 58.1~ 58.0~ 57.90 57.8~ 57.70 57.6~ 57.5~ 57.4~ 57.3~ 57.2~ 57.1~ 57.0~ 56.9~ 56.8~ 56.7~ 56.6~ 56.5~ 56.4~ 56.3~ 56.2~ 56.1~ 56.0~ 55.90 55.8~ 55.7~ 55.6~ 55.50 55.40 55.3~ 55.2~ 55.1~ 55.0~.lIC -4. 4-. mr~ Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 1 2 3 4 5 6 7 8 9 12 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4 log cos diff. log cot com. log tan log sin diff. Angle 55diff. 550-60~ 36 TABLE III. LOGARITHMIC SINES 0 35~-40~ Angle log sin diff. log tan dif' log cot log cos diff. Prop. Parts I I-~- com, I_ diff. at 35.0~ 35.10 35.2~ 35.30 35.4~ 35.50 35.6~ 35.70 35.8~ 35.90 36.0~ 36.1~ 36.20 36.3~ 36.4~ 36.5~ 36.60 36.70 36.8~ 36.9~ 37.0~ 37.10 37.20 37.3~ 37.4~ 37.50 37.6~ 37.70 37.8~ 37.9~ 38.0~ 38.1~ 38.2~ 38.3~ 38.40 38.50 38.6~ 38.70 38.8~ 38.9~ 39.00 39.10 39.2~ 39.30 39.40 39.50 39.6~ 39.70 39.80 39.90 40.0~ I 9.7586 9.7597 9.7607 9.7618 9.7629 9.7640 9.7650 9.7661 9.7671 9.7682 9.7692 9.7703 9.7713 9.7723 9.7734 9.7744 9.7754 9.7764 9.7774 9.7785 9.7795 9.7805 9.7815 9.7825 9.7835 9.7844 9.7854 9.7864 9.7874 9.7884 9.7893 9.7903 9.7913 9.7922 9.7932 9.7941 9.7951 9.7960 9.7970 9.7979 9.7989 9.7998 9.8007 9.8017 9.8026 9.8035 9.8044 9.8053 9.8063 9.8072 9.8081 11 10 11 11 11 10 11 10 11 10 11 10 10 11 10 10 10 10 11 10 10 10 10 10 9 10 10 10 10 9 10 10 9 10 9 10 9 10 lO 9.8452 9.8468 9.8484 9.8501 9.8517 9.8533 9.8549 9.8565 9.8581 9.8597 9.8613 9.8629 9.8644 9.8660 9.8676 9.8692 9.8708 9.8724 9.8740 9.8755 9.8771 9.8787 9.8803 9.8818 9.8834 9.8850 9.8865 9.8881 9.8897 9.8912 9.8928 9.8944 9.8959 9.8975 9.8990 9.9006 9.9022 9.9037 9.9053 9.9068 9.9084 9.9099 9.9115 9.9130 9.9146 9.9161 9.9176 9.9192 9.9207 9.9223 9.9238 16 16 17 16 16 16 16 16 16 16 16 15 16 16 16 16 16 16 15 16 16 16 15 16 16 15 16 16 15 16 16 15 16 15 16 16 15 16 15 16 15 16 15 16 15 15 16 15 16 15 0.1548 0.1532 0.1516 0.1499 0.1483 0.1467 0.1451 0.1435 0.1419 0.1403 0.1387 0.1371 0.1356 0.1340 0.1324 0.1308 0.1292 0.1276 0.1260 0.1245 0.1229 0.1213 0.1197 0.1182 0.1166 0.1150 0.1135 0.1119 0.1103 0.1088 0.1072 0.1056 0.1041 0.1025 0.1010 0.0994 0.0978 0.0963 0.0947 0.0932 0.0916 0.0901 0.0885 0.0870 0.0854 0.0839 0.0824 0.0808 0.0793 0.0777 0.0762 9.9134 9.9128 9.9123 9.9118 9.9112 9.9107 9.9101 9.9096 9.9091 9.9085 9.9080 9.9074 9.9069 9.9063 9.9057 9.9052 9.9046 9.9041 9.9035 9.9029 9.9023 9.9018 9.9012 9.9006 9.9000 9.8995 9.8989 9.8983 9.8977 9.8971 9.8965 9.8959 9.8953 9.8947 9.8941 9.8935 9.8929 9.8923 9.8917 9.8911 9.8905 9.8899 9.8893 9.8887 9.8880 9.8874 9.8868 9.8862 9.8855 9.8849 9.8843 6 5 5 6 5 6 5 5 6 5 6 5 6 6 5 6 5 6 6 6 5 6 6 6 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 6 6 6 7 6 6 55.0~ 54.90 54.80 54.7~ 54.6~ 54.50 54.4~ 54.3~ 54.20 54.1~ 54.0~ 53.90 53.80 53.70 53.6~ 53.50 53.40 53.30 53.2~ 53.10 53.0~ 52.9~ 52.8~ 52.7~ 52.6~ 52.50 52.4~ 52.3~ 52.2~ 52.1~ 52.00 51.90 51.80 51.7~ 51.60 51.50 51.4~ 51.30 51.20 51.10 51.0~ 50.9~ 50.80 50.7~ 50.60 50.50 50.40 50.30 50.2~ 50.10 50.00 ba,.o [. Difference 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 9 0.9 1.8 2.7 3.6 4.5 5.4 6.3 7.2 8.1 5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 10 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 1 2 3 4 5 6 7 8 9 9 10 9 9 10 9 9 9 9 10 9 9 1 2 3 4 5 6 7 8 9 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4 log cos diff. log cot com log tan log sin diff. Angle diff.0 55 50~-55~ - I I..... II COSINES, TANGENTS, AND COTANGENTS 37 I 40~-45o Angle log sin diff. log tan 40.0~ 40.1~ 40.2~ 40.3~ 40.4~ 40.50 40.6~ 40.7~ 40.8~ 40.9~ 41.0~ 41.1~ 41.2~ 41.3~ 41.4~ 41.5~ 41.6~ 41.7~ 41.8~ 41.9~ 42.0~ 42.1~ 42.2~ 42.3~ 42.4~ 42.5~ 42.6~ 42.7~ 42.8~ 42.9~ 43.0~ 43.1~ 43.2~ 43.3~ 43.40 43.5~ 43.6~ 43.7~ 43.8~ 43.90 44.0~ 44.1~ 44.2~ 44.3~ 44.40 44.50 44.6~ 44.70 44.8~ 44.90 45.0~ 9.8081 9.8090 9.8099 9.8108 9.8117 9.8125 9.8134 9.8143 9.8152 9.8161 9.8169 9.8178 9.8187 9.8195 9.8204 9.8213 9.8221 9.8230 9.8238 9.8247 9.8255 9.8264 9.8272 9.8280 9.8289 9.8297 9.8305 9.8313 9.8322 9.8330 9.8338 9.8346 9.8354 9.8362 9.8370 9.8378 9.8386 9.8394 9.8402 9.8410 9.8418 9.8426 9.8433 9.8441 9.8449 9.8457 9.8464 9.8472 9.8480 9.8487 9.8495 9 9. 9 9 8 9 9 9 9 8 9 9 8 9 9 8 9 8 9 8 9 8 8 9 8 8 8 9 8 8 8 8 8 8 8 8 8 8 8 8 8 7 8 8 8 7 8 8 7 8 1 9.9238 9.9254 9.9269 9.9284 9.9300 9.9315 9.9330 9.9346 9.9361 9.9376 9.9392 9.9407 9.9422 9.9438 9.9453 9.9468 9.9483 9.9499 9.9514 9.9529 9.9544 9.9560 9.9575 9.9590 9.9605 9.9621 9.9636 9.9651 9.9666 9.9681 9.9697 9.9712 9.9727 9.9742 9.9757 9.9772 9.9788 9.9803 9.9818 9.9833 9.9848 9.9864 9.9879 9.9894 9.9909 9.9924 9.9939 9.9955 9.9970 9.9985 10.0000 corn. diff. 16 15 15 16 15 15 16 15 15 16 15 15 16 15 15 15 16 15 15 15 16 15 15 15 16 15 15 15 15 16 15 15 15 15 15 16 15 15 15 15 16 15 15 15 15 15 16 15 15 15 log cot log cos diff. 0.0762 0.0746 0.0731 0.0716 0.0700 0.0685 0.0670 0.0654 0.0639 0.0624 0.0608 0.0593 0.0578 0.0562 0.0547 0.0532 0.0517 0.0501 0.0486 0.0471 0.0456 0.0440 0.0425 0.0410 0.0395 0.0379 0.0364 0.0349 0.0334 0.0319 0.0303 0.0288 0.0273 0.0258 0.0243 0.0228 0.0212 0.0197 0.0182 0.0167 0.0152 0.0136 0.0121 0.0106 0.0091 0.0076 0.0061 0.0045 0.0030 0.0015 0.0000 9.8843 9.8836 9.8830 9.8823 9.8817 9.8810 9.8804 9.8797 9.8791 9.8784 9.8778 9.8771 9.8765 9.8758 9.8751 9.8745 9.8738 9.8731 9.8724 9.8718 9.8711 9.8704 9.8697 9.8690 9.8683 9.8676 9.8669 9.8662 9.8655 9.8648 9.8641 9.8634 9.8627 9.8620 9.8613 9.8606 9.8598 9.8591 9.8584 9.8577 9.8569 9.8562 9.8555 9.8547 9.8540 9.8532 9.8525 9.8517 9.8510 9.8502 9.8495 7 6 7 6 7 6 7 6 7 6 7 6 7 7 6 7 7 7 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 7 7 7 8 7 7 8 7 8 7 8 7 8 7 Prop. Parts 50.00 49.9~ 49.8~ 49.7~ 49.6~ 49.50 49.40 49.3~ 49.2~ 49.1~ 49.00 48.9~ 48.8~ 48.7~ 48.6~ 48.5~ 48.4~ 48.3~ 48.2~ 48.1~ 48.00 47.9~ 47.8~ 47.7~ 47.6~ 47.50 47.4~ 47.3~ 47.2~ 47.1~ 47.0~ 46.9~ 46.8~ 46.7~ 46.6~ 46.5~ 46.4~ 46.3~ 46.2~ 46.1~ 46.00 45.9~ 45.8~ 45.70 45.6~ 45.50 45.4~ 45.30 45.2~ 45.1~ 45.0~ 4-2 bn cd 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 a16 o 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 9 0.9 1.8 2.7 3.6 4.5 5.4 6.3 7.2 8.1 8 0.8 1.6 2.4 3.2 4.0 4.8 5.6 6.4 7.2 7 0.7 1.4 2.1 2.8 3.5 4.2 4.9 5.6 6.3 6 0.6 1.2 1.8 2.4 3.(f 3.6 4.2 4.8 5.4 I log cos diff. log cot corn log tan log sin diff. Angle 450-50~ I -- I,..L. t 38 TABLE OF TRIGONOMETRIC FUNCTIONS TABLE OF NATURAL VALUES OF THE TRIGONOMETRIC FUNCTIONS Angle sin cos tan cot sec sc 00 1~ 2~ 30 40 50 60 70 8~ 9~ 100 110 12~ 13~ 14~ 15~ 16~ 17~ 18~ 19~ 20~ 21~ 22~ 23~ 24~ 25~ 26~ 27~ 280 29~ 300 31~ 32~ 330 340 350 36~ 370 38~ 390 40~ 41~ 42~ 430 440 450 I.0000.0175.0349.0523.0698.0872.1045.1219.1392.1564.1736.1908.2079.2250.2419.2588.2756.2924.3090.3256.3420.3584.3746.3907.4067.4226.4384.4540.4695.4848.5000.5150.5299.5446.5592.5736.5878.6018.6157.6293.6428.6561.6691.6820.6947.7071 1.0000.9998.9994.9986.9976.9962.9945.9925.9903.9877.9848.9816.9781.9744.9703.9659.9613.9563.9511.9455.9397.9336.9272.9205.9135.9063.8988..8910.8829.8746.8660.8572.8480.8387.8290.8192.8090.7986.7880.7771.7660.7547.7431.7314.7193.7071 I I I I I -.0000.0175.0349.0524.0699.0875.1051.1228.1405.1584.1763.1944.2126.2309.2493.2679.2867.3057.3249.3443.3640.3839.4040.4245.4452.4663.4877.5095.5317.5543.5774.6009.6249.6494.6745.7002.7265.7536.7813.8098.8391.8693.9004.9325.9657 1.0000 00 57.290 28.636 19.081 14.300 11.430 9.5144 8.1443 7.1154 6.3138 5.6713 5.1446 4.7046 4.3315 4.0108 3.7321 3.4874 3.2709 3.0777 2.9042 2.7475 2.6051 2.4751 2.3559 2.2460 2.1445 2.0503 1.9626 1.8807 1.8040 1.7321 1.6643 1.6003 1.5399 1.4826 1.4281 1.3764 1.3270 1.2799 1.2349 1.1918 1.1504 1.1106 1.0724 1.0355 1.0000 1.0000 1.0002 1.0006 1.0014 1.0024 1.0038 1.0055 1.0075 1.0098 1.0125 1.0154 1.0187 1.0223 1.0263 1.0306 1.0353 1.0403 1.0457 1.0515 '1.0576 1.0642 1.0711 1.0785 1.0864 1.0946 1.1034 1.1126 1.1223 1.1326 1.1434 1.1547 1.1666 1.1792 1.1924 1.2062 1.2208 1.2361 1.2521 1.2690 1.2868 1.3054 1.3250 1.3456 1.3673 1.3902 1.4142 00 57.299 28.654 19.107 14.336 11.474 9.5668 8.2055 7.1853 6.3925 5.7588 5.2408 4.8097 4.4454 4.1336, 3.8637 3.6280 3.4203 3.2361 3.0716 2.9238 2.7904 2.6695 2.5593 2.4586 2.3662 2.2812 2.2027 2.1301 2.0627 2.0000 1.9416 1.8871 1.8361 1.7883 1.7434 1.7013 1.6616 1.6243 1.5890 1.5557 1.5243 1.4945 1.4663 1.4396 1.4142 90~ 89~ 88~ 87~ 86~ 85~ 84~ 83~ 82~ 81~ 80~ 790 780 770 76~, 750 740 730 72~ 71~ 700 696 68~ 67~ 66~ 65~ 64~ 63~ 62~ 61~ 60s 590 58~ 57~ 56~ 550 540 530 52~ 51~ 500 490 48~ 470 46~ 45~ 0 cos sin cot tan cso sec Angle