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THE ELEMENTS PLANE AND SPHERICAL TRIGONOMETRY. EUGENE L™0HAEDS, B. A., ASSISTANT FB07ESSOR OF UATBEMATICS IN YAIE COLLEGE. NEW YORK: D. APPLETON AND COMPANY, 549 AND 651 BROADWAT. 1879. COPYRIGHT BY D. APPLETON AND COMPACT. 18T9. PEEFACE. In the following pages the author has aimed to make the subject of Trigonometry plain to beginners, and has, therefore, devoted a great deal of space to the elementary definitions and to the application of them. Spherical Trigonometry is a subject so difficult to graisp, that any treatise on it designed for beginners should, in the opinion of the writer, make frequent use of diagrams, to convey to thfe student a clear idea of the relations of the magiiitudes under treatment. Con- sequently, this kind of illustration has been applied wherever it could be of any possible assistance. The references are to " Todhunter's Euclid " and to " Ohauvenet's Geometry." The answers to examples have been, obtained by means of six-place tables. OOl^TEI^f TS. CHAPTER I. „.,„ PAGE Some Instruments used in Drawing 'j CHAPTEE II. Plane Trigonometry. — Definitions . 14 CHAPTEK m. Trigonometrical Eatios of an Angle of 30°, of 46°, and of 60" . 23 CHAPTEE IV., The Use of Trigonometrical Tables. — Solution of Eight-angled Triangles 28 CHAPTEE V. Trigonometric Eatios of a Eight Angle, of 0, and of Obtuse Angles 30 CHAPTEE VI. Oblique-angled Triangles 49 CHAPTER VII. Relations of the Trigonometrical Eatios to each other. — ^Trigono- metrical Eatios of Two Angles 69 6 CONTENTS. CHAPTER VIII. InTeree Trigonometric Functions. — Circular Measure of an Angle . 85 CHAPTER IX. Angles having Corresponding Trigonometric Functions of the Same Numerical Value. — Table of Important Trigonometric Func- tions. — Angles Greater than tour Right Angles. — Positive and Negative Angles 93 CHAPTER X. Trigonometric Functions Represented by Lines. — Outline of a Method of Constructing Trigonometric Tables . . .106 CHAPTER XI. Definitions. — Theorems of Right-angled Triangles, . . . 113 CHAPTER XII. Solution of Right-angled Triangles, 127 CHAPTER Xm. Quadrantal Triangles, 138 CHAPTER XrV. Theorems and Formulas of Oblique-angled Triangles, . . 143 CHAPTER XV. Solution of Oblique-angled Triangles, 186 THE ELEMENTS OF PLANE TRIGONOMETRY. CHAPTEE I. SOME mSTKUMENTS USED IN DBAWING. Foe tlie right representation of tlie magnitudes of which Trigonometry treats, the following instruments will be found useful : 1. A pair of dividers or conh- passes, having one leg to which can be adjusted a pen- cil, or, if necessary, an ink-point ; 2. A plane-scale. One of the simplest scales, and one which is suffi- cient for all ordinary drawing purposes, is six inches in length, and is made of ivory or of box-wood. It con- tains a line or lines of chords, and also lines of equal parts, of which the units are of different lengths. (a) A lims of chords is used for setting off angles. It consists of a line, marked from 1 0° to 90°, and con- tains the chords of all arcs from 1° up to 90°. As the chord of 60° is equal to radius (Euc. 16, IV. Cor. Ch. 5, Y.), when we wish to lay down an angle of a given 8 THE ELEMENTS OF PLANE TKI60N0METRY. number of degrees we describe a circle, from a given point, with a radius equal to the length of the line measured from the beginning of the line of chords to the division marked 60 ; then taking the chord of the number of degrees whose angle is required as a radius, we place it in the circle. If the extremities of the chord be joined to the centre of the circle, the chord will subtend the angle required. Thus, suppose at a given point in a given line we wish to make an angle of 25°. Let A be the given point in the given line A C, at which it is required to make an angle of 25°. From the line of chords we take a chord of 60° as radius, and with it, from A as a centre, describe an arc B D. Then from ^ as a centre, with the line jBi^, as a radius, equal to a chord of 25°, taken from the sa/me line of chords, we describe an arc intersecting the former arc at F. Join BF &iiA A F. Then FA B is an angle of 25°, since it is at the centre of the circle, and subtends an arc of 25°. If it is required, at a given point, to make an ob- tuse angle of a given number of degrees, we produce the line through the given point, and make an angle on the line produced equal to the supplement of the given angle. Thus let it be required to make an angle of 100° at the point A in the given line A B. Produce the line A B through A. Make A C equal to a chord of 60° taken from the line of chords. From SOME INSTRUMENTS TOED IN DRAWING. 9 A as centre, witli AC a,B radius, describe an arc CE. From O as centre, with a radius OE, equal to a chord of 80°, taken from the same line of chords, describe an arc cutting the former arc at E. Then C AE\b an an- gle of 80°. But as the angles C J- ^and EA B are to- gether equal to two right angles, or 180°, EA B is equal to 100°. Also an obtuse angle of a certain number of degrees can be made by making two adjacent acute angles, the sum of which is equal to the given obtuse angle. (Angles of a certain number of degrees and mmvies, or degrees, minutes, and seconds, can be made only approximately correct by the line of chords.) (b) A Utw of equal pa/rts is used for drawing lines of a given length, one of the equal parts being taken as the unit, or as ten units, or as a hundred units, if the line is measured on the decimal system. Lines of equal parts are also given, by means of which feet and inches can be represented. The unit is then divided into twelve equal parts. To obtain hundredths of the unit on the plane- 10 THE ELEMENTS OF PLANE TRIGONOMETRY. scale a device is adopted which will be understood by the following figure : U Rx2 y4 6 s r M4 JEl'JLL In the above figure, which represents a part of a plane-scale in which the unit of measurement is an inch, let liSYte a square of which each side is an inch. Let li The divided into ten equal parts, Hx, x2, etc. j also let JES and ST he divided into ten equal parts. Through the points of division of ST let lines be drawn parallel to BT ov ES. Let a straight line be drawn from JR to the first point {F) of division in the line ES. Let xhhe drawn from the first point of H Tto the second point of ES, another line be drawn from the second point oi B T to the third point of ES, and so on. Since Hx, x2, 2y, etc., are equal and parallel to FK, EL, etc., RF, xK, 2 Z, etc., are parallel (Euc. 33, I. Ch. 32, I.). Therefore, the distances on the parallels to R T, through 1 and 3, etc., intercepted be- tween any two consecutive lines, are equal to the dis- 'tances on the same parallels between any other con- secutive two (Euc. 34, I. Ch.'SO, I.), and each of these distances is equal to -^^ of an inch. Again, in the triangle REF, since a 6 is parallel to SOME INSTRUMENTS USED IN DRAWING. H 7-77-, 1 » .., ... ai HO' 1 iLJ^, we have, trom similar triangles, -r-— = ■^r-^= — ; * ' EF RE 10' EF therefore aS = — — ; but EF= ^V of an inch; there- fore a J = ^ij- of ^ijj- of an inch, or ^-^ of an inch. In a similar manner it can be proved that CD equals Y^-g- of an inch, OP equals -jw ^f an inch, etc. If, on the left of the figure, we take TJ V equal to ^ inch and divide it into ten equal parts, and divide TF4 into ten equal parts, and draw a line from Y to the first point M of W4:, another line from the first point of UVto the second point of W4:, and so on, we can prove gh equal to -j-^tt of 2 inch or ^-^ of an incb, and r s equal to jf^ of ^ inch or -^-^ of an inch, etc. Suppose, now, we are required to represent a straight line of 125 feet, on the scale of 1 00 feet to the inch. We draw any line AB ; then placing one point of the dividers at O, on the plane-scale, and bringing the other to the point A, ^ ■ ^ the intersection of the ''' two lines 5 5 and 2 Z, we lay off on ^ ^ a line A equal to O A. Then J. C is the required line, as it is equal to 1 inch and ^-^-^ inch, or to if|- of an inch. The two instruments described above are all that are fiecessary for drawing lines of a given length, or laying down angles of a given number of degrees. The following, however, are very convenient instru- ments : A. protractor, for laying down angles, and & parallel ruler for drawing parallel lines. A protractor is a semicircle, usually - of brass or some other metal, a part of the semicircle being cut 13 THE ELEMENTS OF PLANE TRIGONOMETRT. out, as the part B CD. The outer rim left is divided into degrees, and sometimes into half degrees or even smaller parts, the lines of division all converging to a notch (J.) on the di- ameter of the semi- circle. The degrees are numbered from both left to right and right to left, from 10° to 170°, or from 10° to 90°, as in the figure. At a given point in a given line, to lay down an angle of a given number of degrees by means of the pro- tractor, place the notch. A, at the given point, and the line FE coincident with the given line. Then put a dot, or point, opposite the given number of degrees, on the outer or inner edge of the protractor. A line, drawn from this point to the given point, will make with the given straight line the required angle. B- A parallel rule consists of two rules joined to- gether by two pieces of metal of equal length, at equal distances apart. To draw a line through a given point parallel to a given line, place one edge of one of the rulers coinci- dent with the given line, and one edge of the other SOME INSTRUMENTS USED m DRAWING. 13 ruler on the point, and draw the line tlirough the given point. It will be parallel to the given line. Thus, suppose A to be the given point and B C the given straight line. Make the edge of the ruler, D, cpincide with jB 0, and place the edge of the rule, ^, on the point A. A line drawn through A will be parallel to B C. The parallel line might also have been drawn by placing the upper edge of ^ on the point, or by placing the lower edge of D coincident with the given line B O. Example 1. Draw the right-angled triangle of which the hypotenuse is 236 feet, and one a,cute angle is 30°, and find by measurement the base and perpendicular. Ans. Perpendicular = 118 ; base = 204, nearly. 2. Draw the triangle whose two sides are, respectively, 120 and 80, and the included angle 42°, and find by measurement the third side. Ans. 81, almost. 3. Draw the right-angled triangle whose sides about the right angle are each 34S, and find by measurement the hypotenuse. 4. Draw the triangle of which one side is 421, and the adjacent an- gles are 35° and 72°, and find by measurement the other sides. CHAPTER II. PLANE TEIGONOMETET. ^DEFINITIONS. Aeticle 1. Teigonometey is that brancli of mathe- matics whidh treats of angles, the relations of different angles to each other, and the relations of angles to lines, surfaces, and solids. 2. In Plane Teigonometey the angles considered are such as are contained by straight Unes, and the sur- faces considered axe plane surfaces. 3. The Teigonometeical Ratios of an angle are ratios between the sides of a right-angled triangle, in which, or with respect to which, the angle has a certain position. The right-angled t/riangle is called the triangle of reference. It is formed by the lines containing the angle and a perpendicular to one of the lines, or to one of the lines produced, drawn from any point in the other line. The trigonometrical ratios are six in number. 4. We will first consider the trigonomst/rical ratios of an acute angle. Let A be any acute angle, formed by the two lines A E and A F, meet- ing at the point A. \ts.AE take any point PLANE TRIGONOMETRY.— DEFINITIONS. 15 C, and from C draw G B at right angles to A F. Then there is formed the right-angled triangle ABC, the triangle of reference. In this triangle, calling C B, the side opjposite the given angle, the perpendicula/r, and AB, the side adjao&nt to the given angle, the (1.) The Sine of an angle is the ratio of the per- pendicular to the hypotenuse. Thus, Sine of A (written sin. A) = — ^ (2.) The Tai^qent of an angle is the ratid of the perpendicular to the iase. Thus, Tangent of A (written tan. A) = y-^. (3.) The Secant of an angle is the ratio of the hy- votenuse to the hase. AC Thus, Secant of A (written sec. A) — -j-j,- (4.) The CosmE of an angle is the ratio of the hase to the hypotenuse. Thus, Cosine of A (written cos. A) = — ^. (5.) The Cotangent of an angle is the ratio of the iase to the perpendicular. Thus, Cotangent of ^ (written cot. A, or cotan. A)' ^AB BC (6.) The Cosecant of an angle is the ratio of the hypotenuse to the perpendicular. Thus, Cosecant of A (written cosec. A) = -g— . 5. If we take C as the acute angle, A B as the per- 16 THE ELEMENTS OF PLANE TRIGONOMETRY. pendicular, and B G as tlie base, we shall have sin. C AB ^ ^ AB ^ ^ AC ^^ =■ -r—y, tan. a = -^-py, and sec. C = -^-^ JN ow, C is the complement of A, as their sum equals 90°. Com- paring the COS. A with the sin. O, it will be seen that AB they are identical, both being —r-py- Also, tan. is A (j identical with cotan. A, and sec. O with cosec. A. The cosine of an angle is, therefore, the sine of its comple- ment , the cotcmgent of an angle the tcmgent of its com- jpleTnent / and the cosecami of an angle the secant of its convplement f so that the six trigonometrical ratios of an acute angle are really composed of three ratios be- longing to thait angle, and three ratios belonging to its complement. 6. The Teigonometkical Ratios, for each acute am- gle, are constant ; i. e., are the same for the same or equal angle, in whatever right-angled triangle it is situated. Let A represent any acute angle formed by the two lines A E and,^ F, meeting at A. Make ABO the triangle of reference. From B draw BD perpendicular to A C, and from JD, where BD meets A G, draw a perpendicular DHioAB. In tn&n^eADH, in triangle AB G, sin. A = -r-y,/ and A (y sin. A = AD' PLANE TRIGONOMETRY.— DEFINITIONS. 17 in triangle ABD, sin. A = — — ((1) Art. 4). But DE BC BD Td = J^CE^c-^'"^!- Ch.4,III.) = 1^ (Euc. 8, VI. Ch. 13, III.). The angle A^DBO (Euc. 8, VI. Ch. 13, III.) = BDJI{Euc. 29, 1. Ch. 13, 1.). Sin. DBC=-- — = — ' ^ BO AG • AM ' T^nrr SB BD BO = sin. A. Also, Bin. BDII= -:=-^ = -—=, = ' BD AB AO = sin. A. T) W In triangle AD H, tan. A= y^/ i° triangle B ABC, tan. A = — — / and in triangle A B D, AB . BD „ DH BO BD tan.^ = — . But-^= — = --. T\ fi Ti Ti Also, tan. D B O — -^—=: = -— ;r = tan. A ; DB AD ' ^ ^^^ HB BD HD ^ . i.n.HDB = ^^ = -^ = ^=i.n.A. AD In triangle ADH, sec. A = -riEr> ^-^^ C, sec. A = -— -= ; and, in ABD, sec. A = — -^. But -— =. AB' AD AB AO AB ~ AB~ AD' Also, sec. D B = ^z-^ = -r-^ = sec. A; and BD AB ^,, „ BD AB . sec. HDB =rp:rr= — -r^ = sec. A. ED AD 18 THE ELEMENTS OP PLANE TRIGONOMETRT. In a similar manner it 'may be proved that the ratios for the cosine, cotangent, and cosecant of the same or equal angle are constant. T. The Sine and the Cosine, of an acute angle, are al/wwys LESS than 1. The 8vne being equal to the ratio of the perpendicu- lar to the hypotenuse of the triangle of reference, in the fraction expressing that ratio the numerator is always less than the denominator, and therefore the value of the fraction is less than 1. The cosine being equal to the ratio of the base to the hypotenuse, in the fraction expressing that ratio the numerator is always less than the denominator, and therefore the value of the fraction is less than 1. Thus let ABO, ALB, A ME, and ANF, be a series of right- angled triangles, all having an hypotenuse of the same length. Let the hypotenuse of each triangle represent 82 units of length. Let OB = 6 units ; DL =12 units ; ME = 22 units ; NF= 31 units. The numerical measures of these lines are, therefore, 32, 6, 12, 22, etc. OB =2^=((l)Art.4). Sin. OAB-- = — (Oh. Art. 43, IL) ; sin. BAL BL 12 EM = — — ; sin. EA M= BA 32 AE 22 EN 31 = ^ : sin. FAN'=--—^- — ; 32 FA 32 all less than 1. Also, let there be a series of right-angled triangles OAH, ABO, etc., all having an hypotenuse of the same length, viz., 16. Let AH-= 15, ^ .B = 12, AL = 8, AM= 4, and AN— 2, in terms of the unit of length. PLANE TRIGONOMETRY.— DEFINITIONS. 19 12 Cos. GAH=-^{i4:)kTt. 4)= -; COS. OAJB^ — = ^=z — - = - = If ; AD AG Sec. GAF- -— = 14 _ 20 AF~ 10 Each of these secants is greater than 1, and, as above, the secant of an acute angle is always greater than 1. Again, the cosecant being equal to the ratio of the hypotenuse to the perpendicular, the fraction express- ing that ratio is always greater than 1, as its numerator is always greater than its denominator. 33 THE ELEMENTS OF PLANE TRIGX)NOMETEY. Let ABO, ABE, and AFQ, be triangles right angled at B, D, and F, respectively. Let the numeri- cal measures of AC, A E, and A O, each equal 20 ; let the numerical measure oi B C equal 8, of D E equal 14, and of F G \.&, respectively. AC 20 Cosec. CAB= — = - = CB 8 = 2J AE 20 Cosec. EA B = — — - DE 14 = lf A a 20 Cosec. OAF=~r-=— = FG 16 = li. Each of those is greater than 1, and, therefore, the cosecant of an acute angle is always greater than 1. 13. Cor. As the acfute angle increases, its seoamt wv- oreases, but its coseocmt decreases. CHAPTEE III. TEIGONOMETEICAIi EATIOS OF AN Al^GLE OF 30°, OF 45% AND OF 60°- Aet, 14. To find the numerical values of the teigo- NOMETEicAL EATIOS of On angle of 30°. Suppose DAFio be an angle of 30°. lu AD take any point, O, and from Cdraw OB perpendicular to AF, forming tlie tri- angle of reference, OA B. At the point A, in the line A B, make the angle ^J.^ equal to BAG— that- is, to 30° — and pro- duce CB to meet -4 ^ at E. Then ABC&ndi ABE are equal triangJes (Euc. 26, I. Ch. 21, I.), and GB is equal to B E. Now, in the triangle AGE, the angle A is equal to 60°, and the angle E is also equal to 60°. There- fore GA is equal to GE (Euc. 6, 1. Ch. 27, I.), and GB, which is half of CE, is equal to half of G A. Now, AG^^AB^-^BG"" (Euc. 47, I. Ch. 14, \\\>)=AB''^IAG^; Therefore AB^ = AG^ -\AG'' = iAG^\ and 4/3 2 AB = ■AG. 24 THE ELEMENTS OF PLANE TRIGONOMETBT. CB 1 Now, sin. 30° = sin. GAB Tan. 30° = tan. CAB CB AC 2' CB AB 2CB jAC ^ 1 V3 |/3 AC = i4/3; Sec. 30° = sec. CA B = IV3; Cos. 30° = COS. CAB = Cot. 30° = cot. CAB = Cosec. 30° = cosec. CA B AC AG AB 1/3 AB 4/3 AC AG AC" ~ AC AB ^:ac BC ^AC ^ AC AC _2_ V3 i/3; 4/3; = 2; Versin. 30° BC iAC versin. CAB=1 — cos. 30° (Art. 9) 15. To find the rmmsrical vahies of the teigonomet- jaoAL EATios of OM angle of 45°. Suppose CAE to be an an- gle of 45°. Ivl AG take any point, C, and from G draw CB perpendicular to A E, meeting ^^at^. In the right-angled triangle ~:e AB C, as the angle A is equal TRIGONOMETRICAL RATIOS. 25 to 45°, tlie angle O is also equal to 45° (Euc. 32, I. Ch. 18, 1.). Consequently O is equal to A, and the side ABIqB C(Euc. 6, 1. Ch. 27, 1.). Is^ow, J. 67* = ^^« 4- ^ C^ = 2 ^ C« = 2 J. .5^ ; therefore AB = — = B O. V2 '■AC JNow, sin. 45° = sm. A = -7-7^= ^ ^ = — AC AG 1^^ ■■W2; B C B G Tan.45° = tan.^ = — = — = 1; AG AC Sec. 45° = sec. A = -7-= = = 4/2 ; ]^AG Cos.45° = cos.^=^ = -^- = -^=iV2. Cot.45° = cot.^ = f| = f| = l; ^ ..o ^ AC AC Cosec. 45° = cosec. A = -^-^ = -— = |/2 ; ^^ -^AG Yersin. 45° = versin. A = l — — — = 1 — ^ |/2. 4/2 16, To j^Jid:? #Ae numerioal values of the teigono- METEICAL EATIOS of (W, (Mhgle of 60°. Suppose D AS to be an angle of 60°. In A I) take any point, G, and from G draw GB perpendicular 2 26 THE ELEMENTS OF PLANE TBIGONOMETET. to AE, meeting AE in B. From B lay oflE BE, equal to AB, and join OH. Then tlie triangles OA B and B OH are equal (Euc. 4, I. Oh. 20, 1.), and the angle OH A is equal P to OA J?— that is, to 60°. / Therefore the angle A OH ,/ is also equal to 60°. In the triangle A OH the side AH is equal to A O, and AB, which is one-half of AH, is equal to one-half oiAO. Now, J.<7^ = AB^-^ BO'; therefore B O' = lAO' A 0^-AB^ = AO' _ |/ "3 = ^Aa Now, sin. 60° sin. A = BO AO lAO' and BG -^AO -j-c- = i^3; V 3 Tan. 60° = tan. A = AO BO AB iAO = Vs; Sec. 60° = sec. A = Cos. 60° = COS. A Cot. 60° = cot. A = AO _ AO AB AB iAO IAO = 2; 1 AO AB BO AO jAO 1 = *V3; TRIGONOMETRICAL RATIOS. 27 Cosec. 60° = cosec. A = ^= -4^ = — = Versin. 60° = versin. A = l — ^ = -- 2 "We might also derive the trigonometrical ratios of an angle of 60° from those of an angle of 30°, accord- ing to Art. 5. Thus, sin. 60° = cos. 30° = i 4/3 ; Tan. 60° = cot. 30° = /3; Sec. 60° = cosec. 30° — 2; Cos. 60° = sin. 30° = i; Cot. 60° = tan. 30° = ^ 4/3 . Cosec. 60° = sec. 30° = | ^3 . Yersin. 60° = 1 — cos. 60° = 1 - sin. 30° = -• 2 Example 1. If the hypotenuse of a right-angled triangle be 5, and the perpendicular be 4, required the trigonometrical ratios for the angle at the base. Required also the trigonometrical ratios for the angle at the perpendicular. 2. If the base and perpendicular of a right-angled triangle be Y and 8, find the trigonometrical ratios of the angles at the base and perpen- dicular. 8. Calculate to four decimal places the numerical values of the trigo- nometrical ratios of an angle of 30°, and of 60°. 4. Calculate to four decimal places the numerical values of the trigo- nometrical ratios of an angle of 46°. CHAPTEE IV. THE USE OF TEIGOITOMETEIOAL TABLES. SOLXPTION OF EIGHT-ANGLED TELiNGLES. Aet. 17. The trigonomdrie ratios for all angles between 0° and 90°, beginning at an angle of 1' and increasing in size by successive additions of V, have been calculated and have been arranged in tables. Ta- bles have also been calculated for angles beginning at an angle of 10", and increasing in size by successive additions of 10". Smaller tables, in which the interval between angles is 15', have also been arranged for cal- culations where great accuracy is not required. Such tables are called in general Trigonometric(d Tables, and also " Tables of Natural Sines and Cosines," " Tables of Natural Tangents and Cotangents," etc. 18. The loga/riihmie values of the trigonometric ra- tios have also been arranged in tables, called " Tables of Logarithmic Sines," etc., for use in calculations by logarithms. 19. The values of ratios, intermediate between the ratios of the tables, are obtained from those of the tables on the theory that "for small intervals, the differences of the ratios a/re proportional to the differences of the amgles." Thus suppose we are using a table of natural sines, in which the sines are calculated for intervals of 1', and we are required to find the THE USE OP TRIGONOMETRICAL TABLES. 39 sine of an angle of 10° 1' 1". This sine falls between the sine of 10° l' and the sine of 10° 2'.^ In the table of sines, under 10° and opposite 1 in the column of Min. (minutes), we find the sine of 10° 1' to be .173935, and, directly under it, and opposite 2 in the column headed Min., we find the sine of 10° 2' to be .174221. (The decimal point is not prefixed in the table, but is always understood.) The difference of the two sines is 286 , or .000286, and the angles differ by 1' or 60". Therefore 1,000,000 ^ since when the angle increases 60" the sine increases .000286, on the prin- ciple just enunciated, when-the angle increases 1", or j'utli of the former increase, the sine will also iacrease ^Vth of Its former increase, or will .000286 47 increase — = ^ — .000005, nearly. So that the sine of 60 10,000,000 10° 1' 1" is obtained by adding .000005 to .173935. The sme of 10° 1' 1" is therefore .173940. If we wished to obtain the sine of 10° 1' 2" we should add twice the increase for one second, or -(fjjths of the increase for 1' — ^that is, .000009, nearly — and so on. The increase for 1" is generally calculated at short intervals, and put in the table under the column of the ratios to which it belongs, with the name of "propor- tional paHs." To find the increase for' any number of seconds, we multiply the proportional parts for 1" by the number of seconds. To obtain from the tables a ratio for an angle be- tween two angles of the table, it is best in general to take the ratio belonging to the smaller of the two angles, between which it falls, and apply the correction accord- ing to the principle above given, or directly from the " Corrections " -or " Proportional Parts " of such tables, being careful to add such Gorrection, according as the ratio required is a sine, tangent, or secant; and to sub- tract the correction, if the reqwi/red ratio is a cosine, cotangent, or cosecant (Arts. 8, 11, 13). 30 THE ELEMENTS OF PLANE TRIGONOMETRY. 20. Conversely, to obtain the degrees, minutes, amd seconds, answering to a given ratio intermediate between two ratios of the table, take the degrees amd minutes he- longing to the smaUer of the two angles, between whose ratios it falls, amd, for the seconds, di/oide the difference between this ratio amd the gimen ratio by the jpropor- tional parts for 1", fownd under the colmnwt in which the two ratios appear. Thu3 suppose we are using a five-place table, and we have a sine given as .50060, and are required to find the number of degrees, minutes, and seconds. In the angle to which it belongs. Looking in the table of sines we find it falls between .60050 and . 50076; that is, that it belongs to an angle between 30° 2' and 30° 3'. The difference between the sine belonging to the smaller angle and the 10 given sine is 10 — that is, — and the proportional part for 1 is 100,000 26 — = 0.43 (considering 26 as a whole number). Dividing 10 by .43 we have for the seconds 23. Therefore the angle whose sine is .50060, is an angle of 30° 2' 23". 21. The logarithm of a ratio intermediate between the logarithms of two ratios is obtained on the same general principle as were the ratios themselves, it being assumed that, for small intervals, the difEerence of the logarithms of the ratios is proportional to the difference of the angles. The principle here assumed is not strictly true, nor is the principle assumed in the previous article strictly true ; but, employing it with the limitation of small intervals, we are not in danger of error, except in cer- tain angles near 0°, and near 90°, for which angles separate tables are provided. 22. In the tables of logarithmic sines, cosines, tan- SOLUTION OF RIGHT-ANGLED TRIANGLES. 31 gents, etc., the logarithms are generally calculated for intervals of ten seconds, and the corrections or propor- tional parts are given for the seconds from 1" to 9". Most of the trigonometric ratios, whose logarithms are given, are less than 1 (Arts. 7 and 10), and, there- fore, the characteristics of their logarithms would be negative (Art. iOi, Loomis's Alg.). To avoid the use of negative characteristics, however, 10 is always added to the logarithm. Consequently, in order to obtain the correct logarithm of a result, in a calculation in which trigonometric ratios home Tyeen represented lyy their log- a/rithms, from the resulting logarithm 10 rrvust ie sub- tracted for every trigonometric ratio used as a mmlti- vli^r, and, to the resulting logarithm, 10 must be added for every 1/rigonomet/ric ratio used as a dwisor. SOLUTION OF EIGHT-ANGLED TRIANGLES. Art. 23. The jparts of any triangle are the sides and the angles. To sohe a triangle is to find the unknown parts from certain known parts. Trigonometry was, primarily, the science by which triangles are solved, and was originally so defined. 24. The parts of amy triangle are six in number. In a right-angled triangle, as one of the parts is always a right angle, one part is always known. As the two acute angles together make a right angle, when one acute angle is known, the other, being its complement, is also known. So that to soIaic a right-amgled triangle it is only necessary to consider fov/r pwrts; viz., the 33 THE ELEMENTS OE PLANE TRIGONOMETKT. three sides and one acute angle. Any two of these pa/rts of a right-amgled triangle hemg knoivh, we a/re able, ly the use of trigonometric tables, to sol/oe the triangle. 25. Suppose two sides are known. These may be the hypotenuse and perpendicular / the hypotemuse and base y or the perpendicula/r and base. As the perpendicular is " the side opposite the given angle " (Art. 4), either of the sides about the right an- gle may be made the perpendicular, and the other side the base, or " the side adjacent to the given angle," according as we take one or the other of the acute angles as the given angle. "When two sides are given we really have, therefore, two cases only, viz. : 1. When the hypotenuse and a side are given ; 2. When the two sides about the right angle are given. 26. The hypotenuse and a side being known, to sdl/oe the t/riangle. Let AB C he a right-angled triangle, having its right angle at B. Denote the sides opposite the angle C by small letters of the same name as the capital letters de- noting the angles. _ J, (Generally, in all triangles, the no- tation will be adopted of capitals for the angles, and small letters of the same name for the opposite sides.) In the triangle AB G suppose we have given the hypoteniise b, and the side a. (1) Sin. .A = ^((1) Art. 4). ft (2) Also 7 = cos. A ((4) Art. 4) ; or, c = 5 cos. A. V SOLUTION OF EIGHT-ANGLED TRIANGLES. 33 (3) c also equals Vl* — a'^= V (b-^a) {b — a); a formula convenieBt for the use of logarithms, as then log. c = i {log. (6 + a) + log. (5 - a). \ From (1) we see that, when the hypotenuse and a side are given, the rnigle opposite the given side is an angle whose sine is equal to the ratio of the side to the hypotenuse. It is, therefore, to be found from the table of sines. When this angle is found, the third side can be found by formula (2). 27. If the hypotenuse, 5, and the side, e, were given, we could find the angle C, as we found the angle A / or, if we desired to find the angle A, we have - = COS. A ; that is, when the hypotenuse and a side are given, the .angle adjacent to the given side is an angle whose cosine is the ratio of the given side to the hypotenuse. It is, therefore, to be found from a table of cosines. In this case the other side, « = 5 sin. A ; for - = sin. A\ . ■ . a = 5 sin. J. ; or, « = 1/52 - c« = '/(5 + c) (5 - c). Example 1. If the hypotenuse and perpendicular of a right-angled triangle be 192 and 130, respectively, required the other parts of the tri- angle. Am. Angles =.47° 23' 2" and 42° 36' 58" ; side = 141.29. 2. The hypotenuse and base of a right-angled triangle being 71 and 64, respectively, required the other parts of the triangle. Am. Angles =' 25° 39' 22" and 64° 20' 38" ; side = 30.74. 3. If the hypotenuse of a right-angled triangle be 71, and the base 60, required the other parts. Am. Angles = 32° 19' 14" and 57° 40' 46" ; side = 37.96. 34 THE ELEMENTS OP PLANE TRIGONOMETRY. 4. If the hypotenuse and a side of a right-angled triangle be 140 and 84, respectively, what are the other parts ? 5. If the hypotenuse and a side of a right-angled triangle be 130 and 66, respectively, what are the other parts ? 6. The hypotenuse and perpendicular of a right-angled triangle are 200 and 100, respectively ; what are the other parts ? Art. 28. The two sides about the right angle iemg Tmown, to solve the triangle. In the figure, suppose the two sides, a and c, are f, known, and it is required to find the hypotenuse and an- a gles. 5 = v«^ + cHEuc. 47,1. -B Oh. 14, III.). (1) Tan. ^=-((2) Art. 4). After the angle A is found, we can also find the hypotenuse, thus : (2) - = sin.^; 5 = sm. A (3) - = sec. A; . • . h=--G sec. A. G From (1) we see that when the two sides of a right- angled tria/ngle are given, the amgle opposite either of the sides is an angle whose ta/ngent is the ratio of that side to the other side. It is, therefore, to be found from a table of tangents. Example 1. If the base of a right-angled triangle be 141, and the perpendicular be 193, required the angles and the hypotenuse. Am. 63° 60' 67", 36° 9' 3" ; 239.02, nearly. , SOLUTION OP RIGHT-ANGLED TRIANGLES. 35 2. If the two sides about the right angle of a right-angled triangle be 2.1 and 2, what are the angles and the hypotenuse ? Am. 43° 36' 10", 46° 23' 50" ; 2.9. 3. If the two sides of a right-angled triangle be 123.6 and 10.16, what are the angles and the hypotenuse ? Ans. 4° 42' 11", 86° 17' 49"; 123.92. 4. If the two sides of a right-angled triangle be 39 and 30, what arc the other parts ? 5. If the two sides of a right-angled triangle be 81.48 and 108.64, what are the other parts ? 6. If the two sides of a right-angled triangle are 131 and 13.1, find the other parts. 1. If the sides of a right-angled triangle are each equal to a, what is the hypotenuse and what are the angles ? 29. Now, suppose a side and an acute angle are known. These may be either ^^C the hypotenuse and an acute angle, or one of the sides about the right angle and an acute angle. ^^"° tJ 30. The hypotenuse and an acute angle being hnown, to solve the triangle. In the right-angled triangle, A B G, suppose the hypotenuse 5, and the angle A, are known. (1) Sin. A = -; .-. a = 5 sin. J.. (2) Cos. J. = - ; - • . c= 5 cos. A. In the same triangle suppose the hypotenuse h, and the angle C, are known. (3) Sin. <7 = ^ ; . • . c = 5 sin. C. B (4) Cos. ^ = 7 ; • ■ ■ C' = ^ COS. O. 36 THE ELEMENTS OF PLANE TBIGONOMETKT. From (1) and (3) it will be seen that, in a -right- angled triangle, the side opposite cm acute angle is equal to the product of the hypotenuse hy the sine of tlie am,gle. From (2) and (4) it will be seen that, in a right-an- gled triangle, the side adjacent to am, acute angle is equal to the product of the hypotenuse hy the cosine of the angle. ExiMPtE 1. If the hypotenuse of a right-angled triangle be 4.958, and one of the angles be 54° 44', find the other sides. Ans. 4.048, 2.8626. 2. The hypotenuse of a right-angled triangle is 25, and one of the acute angles is 16° 15' 31". Bequired the sides. Ans. T and 24. 3. The hypotenuse of a right-angled triangle is 8T.36, and one of the acute angles is 12° 30'. Required the sides. Am. 8.0862, 36.474. 4. The hypotenuse of a right-angled triangle is 120, and one of the acute angles is 36° 14' 15". Required the sides. Ans. 70.936, 96.789. 5. The hypotenuse of a right-angled triangle is 316.6, and one of the acute angles is 55° 30' 17" Required the sides. 6. The hypotenuse of a right-angled triangle is 656.16, and one of the acute angles is 75° 10' 5". Required the sides. 7. The hypotenuse of a right-angled triangle is 100, and one of the acute angles is 60°. Required the sides. 8. The hypotenuse of a right-angled triangle is 667, and one of the acute angles is 45°. Required the sides. 9. The diagonal of a square is 400. Required the length of a side. .10. The side of a rhombus is 48 feet long, and one of its angles is 68°. Required the length of its diagonals. 31. One of the sides about the right angle and one of the acute cmgles being hnown, to solve the triangle. In the right-angled trian- SOLUTION OF RIGHT-AKGLED TRIANGLES. 37 gle ABO, suppose the side a, and the angle A, oppo- site the side a, are known. (1) Sin..l = f; .-.l "' sin. A' (2)Tan.^ = ^; .■ . c= "^ '0' ' ' tan. J." (3) Or cosec. A = --; . • . b = a cosec. A. €v /I (4) Cot.J. = -; .-. G = a cot. A. Formula (1) and formula (2) are more convenient for general use than formula (3) and formula (4). From (1) and (2) it will be seen that when a side about the right angle of a right-angled triangle, and an angle opposite it, are given, the hypotermse is equal to the quotient of the gi/oen side divided hy the sine of the given angle, and the other side is equal to the quotient of the gi/oen side divided hy the tangent of tJie given angle. 32. If, instead of a side and the opposite angle be- ing known, we know a side and the adjacent angle, we can find the opposite angle — as each acute angle is the complement of the other — and then we can solve the triangle, as in Art. 31. Or we can solve the triangle di- rectly, thus : Suppose, in the right-angled triangle ABC, the side G and the angle A are given, to solve the triangle. (1) Co8.^=-!; .-. 1=-^. ^ ^ h COS. A 38 THE ELEMENTS OF PLANE TRIGONOMETKY. (2) Tan. A = -; . • . a = c tan. A. c (3) Also, sec. A^-; . • . 5 = c sec. A. From (2) and (3) it will be seen that, in a right- -|f'' angled triangle, ihe side oppo- site an acute angle is equal to the ^product of tlie side ad- jaoent to that angle iy the imigent of the angle ^ and that the hypotenuse is equal to the product of a side by the secant of the angle adjacent to the side. Example 1. The side of a right-angled triangle is 141, and the angle opposite this side is 33° 41' 6". Kequired the hypotenuse and the other side. Am. 254.225, 211.64. 2. The side of a right-angled triangle is 124.6, and the angle oppo- site it is 64° 20'. What Is the hypotenuse and the other side ? Am. 138.24, 59.8766. 3. The side of a right-angled triangle is 19.67, and the angle oppo- site to it is 18° 31' 4". Kequired the hypotenuse and the other side. 4. The side of a right-angled triangle is 111.11, and the angle adja- cent to this side is 73° 49'. Required the hypotenuse and the other side. 5. If one side of a rectangle is 50 feet, and the diagonal makes an angle of 30° with this side, what is the length of the diagonal, and what is the length of the other sides of the rectangle ? 6. The diagonal of a rhombus is 30 feet, and the angle through which the diagonal passes is 120°. What is the length of a side of the rhombus, and what is the length of the other diagonal ? 33. From the preceding articles it will be seen that to solve a right-angled .triangle we simply apply the definitions of trigonometric ratios. To find a required part we select, in each case, the definition in which occur the required part and the two known parts. OHAPTEE V. TEIGONOMETBIC EATIOS OF A EIGHT ANGLE, OF 0, AND OF OBTUSE ANGLES. Aet. 34. According to Art. 3, the tria/ngle of refer- ence for tlie trigonometric ratios of an angle is formed by dropping a perpendicular upon one side, or side produced containing the angle, from any point in the other side. In the case of a right angle, the perpen- dicular coincides with one of the sides of the angle and no triangle is formed. "We shall, therefore, obtain the trigonometric ratios of a right angle by the method of limits, considei'ing the case of a right-angled trianglej whose hypotenuse remains constant, while one of the acute angles, taken as a variable, " approaches indefinite- ly " to a right angle as its limit (Ch. Art. 28, Bk. V.). Art. 35. The sine of a right cmgle, or the sine of an angle of 90° is 1. Let AB G \)Q a trian- gle right - angled at B. Suppose, while the hypot- eause remains the same, the acute angle 5^ C in- creases.- At the same time B O increases. B B Also the sirie oi B AO increases (Art. 8). 40 THE ELEMENTS OF PLANE TRIGONOMETRY. Now, as the angle BAG increases, it approaches a right angle as its limit, and its sine approaches the sine of a right angle, as its limit. Also as the angle BAG increases and ap- proaches a right angle as its limit, the perpendicu- lar, B G, approaches the hypotenuse, A G, as its limit (Euc. 6, I. Ch. 27, B G I.) ; and, therefore, — — t^ A But sin. BAG AG approaches -j-p,, or 1, as its limit. always equals BG AG' therefore, at the limits, sin. 90°, or the sine of a right angle, equals 1 (Ch. Art. 29, Bk. V.). 36. The cosine of a right angle, or the cosine of am angle of 90°, is 0. When the angle A (see iigure) increases toward its limit (a right angle), its cosine decreases and approach- es the cosine of a right angle as its limit (Art. 8). At the same time that A increases, J. B decreases, and ap- AB proaches as its limit, so that the limit of -7-7^ is AG or 0. But COS. A always equals AB there- ^^, .. .. .. .. „.„„^„ .^_. ^^, fore, at the limits, the cosine of a right angle, or COS. 90°, equals 0. 37. The tangent of a right angle, or the ta/ngent of cm am.gle of 90°, equals 00. TKIGONOMETEIC RATIOS OF A EIGHT ANGLE. 41 When the angle A increases {see figure, page 40), the tangent increases (Art. 11). "When the angle A approaches a right angle as its limit, its tangent ap- proaches the tangent of a right angle as its limit. Also as the angle A increases, at the same time C B in- creases toward A C, as its limit, while A B decreases ...... ^ B C ^ AC toward as its hmit, so that -r-^:, approaches — — - or co A B B O as a limit. But tan. A = -—= ; therefore, at the limits, AB the tangent of a right angle, or tangent of 90°, equals oo. 38. By a similar method of proof, it can be shown that the' cotcmgent of a right angle is ; that the secant of a right a/ngle is oo ; and that the cosecant of a right angle is 1. 39. To find the trigonometric ratios of 0°. As 0° is the complement of 90°, we can determine the trig- onometric ratios of 0° from those of a right angle, ac- cording to Art. 5. Thus sin. 0° = cos. 90° = ; Cos. 0° = sin. 90° = 1 ; Tan. 0° — cot. 90° = ; Cot. 0° = tan. 90° = 00 ; Sec. 0° = cosec. 90° = 1 ; Cosec. 0° = sec. 90° = oo. 40. We can also find the trigonometric ratios of 0° by the method of limits, considering 0° as the limit to which an acute angle approaches indefinitely when de- creasing. Thus, take the case of the sine of 0°. When A approaches 0° {see figure, page 40), as its 43 THE ELEMENTS OF PLANE TRIGONOMETRY. limit, sin. -A approaclies sin. 0° as its limit, and C B ap- O B proaches as its limit. Therefore the limit of -— ^ = •— -^ = 0. But sin. A always equals -77^. There- fore, at the limits, we shall have sin. 0°= (Ch. Bk. V., Art. 29). In a similar manner the other trigonometric ratios of 0° can be found. TEIGONOMETEICAL RATIOS OF AN OBTUSE ANGLE. Art. 41. Before proceeding to the trigonometrical ratios of an obtuse angle, we will define the sign of a line as distinguished from its value, or numerical measure. The sign of a line generally denotes the direction in which it is measured. To lines measured in one direc- tion there are given, positive signs (not always prefixed), while to lines measured in the opposite direction there are given negative signs. Thus, taking two lines, D E and G H (which we will name initial lines), intersecting at right angles at A, we call all lines going to the right of GH, or from A toward E, positive, and we call all lines to the left -B oi GM, or going from A toward I), negative. So we call all pei-pendiculars upon ^ DE, from above D E, posi- time lines, and we call all perpendiculars upon JDJE, from below, negative. V TRIGONOMETRICAL RATIOS OF AN OBTUSE ANGLE. 43 We shall also consider the hypotenuse of a right- angled triangle always positive. Now, if we put the vertex of the angle, whose trig- onometric ratios we are to consider, at A, and one of its sides coincident with A JE, it will be appareiit that (in the sense of our definitions with regard to the signs of lines), the trigonometric ratios of all angles thus far considered are positive, because the sides of the triangle of reference are all positive. 42. Take now an obtuse angle, and apply its vertex to the point A (see figure), and let one of its sides, as A B, he coincident with A E. To construct the triam^gle of reference (Art. 3), we drop a perpendicular, CD, from any point, C, in A (7, upon AB produced, meeting A B pro- duced in D. In this triangle, AD, the side adjacent to the obtuse angle is neg- ative, because we consider all lines running from A, toward the right of O H, as positive ; and those from A, toward the left of Q H, negati/oe (Art. 41). 43. The trigonometrical ratios of an obtuse angle have the same name^ as the corresponding ratios of an acute angle, but have not always the same sign. For, as A D, the iase of the triangle of reference, or the side adjacent to the obtuse angle, is always negative, and as the hypotenuse, A C, and the perpendicular, 44 THE ELEMENTS OF PLANE TBIGONOMETEY. CD, are always positive (Art. 41), every ratio in whicli A D occurs must be negative. 44. Thus (1), the sine of an obtuse angle is tlie ratio of the perjpendiGular to the hypotenuse of the triangle of reference, and is positive (Art. 41). fIT) = positive quantity. (2.) The tangent of an obtuse angle is the ratio of the jper- pendicular to the base, and is negatvoe. Tan CA B -^^ = P"^^^'^^ quantity _ ^ negative AD negative quantity (quantity. (3.) The secant of an obtuse angle is the ratio of the hypotenuse to the base, and is negative. Sec OA B= — = positive quanti ty ^ r negative AD negative quantity (quantity. (4.) The cosine of an obtuse angle is the ratio of the base to the hypotenuse, and is negative. Cos. OA B = — — = negative quantity. (5.) The cotangent of an obtuse angle is the ratio of the base to \he perpendicular, and is negative. Cot. CA B = -=-— = negative quantity. (6.) The cosecant of an obtuse a/ngle is the ratio of the hypotenuse to the perpendicular of the triangle of reference, and is positive. TRIGONOMETRICAL RATIOS OF AN OBTUSE ANGLE. 45 A. O Cosec. 0A£ = — ^ = positive quantity. 45. It can be proved that the sine and eosme of cm obi/use angle are always less than 1 in numerical value, in the same manner that the sine and cosine of an acute angle have been proved less than 1 (Art. 7) ; also, that when an Muse a/ngle increases its si/ne decreases, but that its cosine increases (negatively) in numerical value. It may also be proved that the tangent and cota/n^ gent of an obtuse angle may he equal to — 1, greater than — 1, or less tha/n, — 1, as it was proved that the tangent and cotangent of an acute angle might be equal to 1, greater than 1, or less than 1 (Art. 10) ; also, that as the obtuse a/ngle increases its tangent decreases, but its cotangent increases, both negatively. It may further be proved that the secant and cose- cami of an obtuse angle are always greater thorn, 1 nu- merically (the secamt being greater than — 1, the cose- cant being greater than -|- 1), iil the same manner that the secant and cosecant of an acute angle were proved to be always greater than 1 (Art. 12) ; also, that as the obtuse angle irhcreases, the secant decreases {fiegati/oel/y), but its cosecant increases. Def. The versed sine of an obtuse angle is equal to the algebraic difference between 1 and the cosine of the angle. Thus {see figure, page 43), versin. J. = 1 — cos. A. But, as cos. J. is a negative quantity (4, Art. 44), versin. A is greater than 1. (In the case of an acute angle, the versed sine is less than 1.) 46. Any trigonometrioal ratio of an acute or obi/use 46 THE ELEMENTS OF PLANE TRIGONOMETRY. angle has the same numerical value as the correspond- ing ratio of its swppl&ment. The SINE amd cosecant of am acute cmgle, and the SINE a/nd COSECANT of its sujp^lementam/ obtuse cmgle, are all of the same sign, and are positive. The TANGENT, COTANGENT, COSINE, and SECANT of an acute amgle a/re positive, while the corresponding ratios of the supplemental) obtuse a/ngle are negatme. In the figure CAB is an acute angle. Produce B A through A, beyond the perpendicular A Y. At A, in the straight line ji E, make the angle PA ^ equal to GAB. Then ieP AB the sup- plement of P J. ^ (Oh. Bk. I., Art. 19), and, conse- quently, equal to the supplement of OA B, the equal oiDAE. Construct the triangle of reference, GA B. ln.AP take AD equal to A C, and from D draw DE per- pendicular to A E. Then is DAE, the triangle of reference for the obtuse angle DAB, equal to the tri- angle OA B (Euc. 26, 1. Oh. 23, 1.). In the triangles A GB and ADE the hypotenuses A G and A D, and the perpendiculars G B and D E, and the base, A B, are all positive, while the base, A E, is negative (Arts. 41 and 42). As the triangles of reference are equal in all re- spects, and therefore similar, the ratio between any two of the sides of one is equal to the ratio between the cor- TRIGONOMETRICAL RATIOS OF AN OBTUSE ANGLE. 47 responding sides of the other (Euc. 4, YI. Oh. 4, III.) ; that is, the mmierical value of any trigonometric ratio of the acute angle, A B, is equal to the nwnerical value of the corresponding ratio of the obtuse angle, DAB. O Ti Ti W Now, sin. O A B ^=—r-^\ also, sin. D A B = -:=--.. ' AG DA But -7—?^^='^ir~,-, as the triangles CAB and D AE2iXQ AG D A equal. Also as the lines CB, DE, AC, and A D, are 1. . . , . OB ^ DE aU positive, the ratios -j-^ and yr-j are positive, and therefore the sine of the angle, O A B, and the sine of its supplement, DAB, are of the same sign, and are positive. AG Again, co&qc. C A B = -^•, also, cosec. DAB DA ^ AC DA , ^, ^ „ ,, = rr—rp,. But 77^ = 17^) and, as the terms 01 these D E C B D JtL ratios are positive, the ratios are positive ; and, there- fore, the cosecant of an angle and the cosecant of its supplement are of the same sign, and &■£& positive.. C B Tan. GAB = -r-^, = a positive quantity, as the A -B lines CB and A B are both positive. But tan. DAB DE (the supplement of OA B) = l-g.= * negative quan- tity, as J. -E" is negative (Arts. 41 and 42). AB Cotan. CAB= ^j, = a positive quantity ; AE Cotan. DAB= -77^= a negative quantity. JJ Ji 48 THE ELEMENTS OF PLANE TRIGONOMETRY. A JB Cos. CAJB = -T-p^ = a positive quantity ; AE Cos. D AB ^= 'A~h ^^ ^ negatiye quantity. A C Sec. C AB =^ "T~D ^= ^ positive quantity ; Sec. D AB ^^ ~A~lf ^^ ^ negative quantity. 47. It follows from the preceding article that trig, onometrical tables for acute angles can be used for obtuse angles, as the numerical values of the ratios of the one are the same as those of the corresponding ra- tios of the other. To find, therefore, the trigonometric ratio of any obtuse angle, we subtract the given angle from 180° and find the corresponding ratio of the remainder. Where it is important — as in certain computations — to distinguish the obtuse angle from its supplementary acute angle, we retain the sign of the ratio. Example 1. Find from the tables the trigonometric ratios of an an- gle of 100°. 2. Required from the tables the trigonometric ratios of an angle of 120° 13' 46". 3. Required the trigonometric ratios of an angle of 136°. 4. Required the trigonometric ratios of an angle of 150°. 6. Required the trigonometric ratios of an angle of 120°. 6. If the cosine of an angle is — .81346'?, what is the angle? Am. 144° 26' 10". Neglecting the sign, required the obtuse angle whose 1. Cotan. is 2.31821. Am. 156° 39' 68.4"- 8. Sine is .63416V. Am. 140° 38' 30.3". - 9. Tangent is 1.81611. Am. 118° 51' 7.4". CHAPTER VI. OBLIQUE-AKTGLED TKIANGLES. Def. An oJMque-cmgled tricmgle is one wliich does not contain a right angle. It is, therefore, either an acute-angled or an obtuse-angled triangle. Art. 48. In omy tricmgle, the sine of any angle is to the sine of a second angle as the side opposite the first angle is to the side opposite the second angle. This principle is sometimes stated in another form, thus : " The sines of the angles of a triangle are proportional to the opposite sides." In the given triangle, ABO, it is required to prove Sin. B AC Sin. C~J.J?' Suppose that J. ^ (7 be a tri- angle, in which a perpendicular •S', from one of the angles upon the opposite side falls within the triangle, as the perpendicular, AD, from A upon B G. Sin. B — -r-^ ((1) Art. 4) ; also, sin. C- AB therefore AD sin. B AB AO AC sm. O' AB AO AB 50 THE ELEMENTS OF PLANE TRIGONOMETRY. By drawing a perpendicular from O upon A B, it can also be proved, in the same way, that Sin. A BO ^m.B~Ad Next, let the triangle AB CbeB, triangle in which the perpendicular falls with- out the triangle, on the side prdduced. In the figure, A D, the perpendicular, falls without the triangle. The angle, A OB, is therefore an obtuse angle. A D Now, sin. J. 5 C = • also, sin. B OA AD ' AO AB' ((1) Art. 44); therefore sin. ABO AD AB AO mi..AOB AD AB AO By drawing a perpendicular from B, upon A pro- duced, it may be proved in the same way that Sin. ^^(7 BO Sin. BOA ~AB' Lastly, let JL -S <7be a right- angled triangle ; then also Sin. B AO Sin. 0~ AB' For, since C is a right angle, ^, its sine is 1 (Art. 35). Also sin. ^ = AO . sin. 5 AO sin. B AO sin. G~~AB' AB'^-\ 1 ~AB or OBLIQUE-ANGLED TEIANGLES. 51 49. In amy triangle, the stjm of amy two sides is to their DiFFEKENCE US the TAifGENT of HALF the sum of tlie OPPOSITE ANGLES is tO the TANGENT of HALF th^iv DIF- FEEENCE. O^^'K Let A CB be any triangle. Then BC^GA _ tan. \ {A-\-B) BG-GA tan. i(^-^) Produce G A to D, making G D equal to G B. Produce B Gto E, making G E equal to C J.. Join D and B, by the straight line I) B. Join E and A, by the straight line EA, and produce EA to meet 2)^ at K. By the construction, B E=B G-\- GA and AD = BG-GA. Now the sum of the angles GA B and GBA equals the sum of the angles B and GB B, because each sum is the supplement of the angle A GB (Euc. 32,1. Ch. 18,1.). BuiGDB-\-GBD = 2B (Euc. 5, 1. Ch. 25, 1.). Therefore I) ^l {GAB + GBA) = ^{A-\-B). Also ^{A + B)-\-i{A-B) = A; 52 THE ELEMENTS OP PLANE TRIGONOMETRY. That is, J)-\-i {A-B)=CAB = J) + ABD (Euc. 32, 1. Oh. 18, 1. Cor. 1) ; therefore, ABI> = l{A—B). Again, the angle CEA^ OA E= DAK; also CBK=ADK', therefore, B EK+ EB K= DAK + ADK; consequently EKB^=EKD, and each is a right an- gle, according to the definition of a right angle. Also the triangles KB E and ADK m^ similar (Euc. 4, I. Ch.4, III.). (1) Now ^^ = tan. D = tan. i{A-\-B); (2) also ^^^= tan. A BK— tan. \{A-B\ 4errfo™,ai^di.g(l)bj(2),^ = ^iM±|. But, from the similarity of the triangles, KBE z.-a.^ADK KB ^BE ^ BC-\-CA , KD~AD~ BG-OA' therefore ^^^^^ - taD-i(^+^ ) therefore ^ G - GA- iB.^.^{A - B)' SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 50. To solve an oblique-angled triangles i* ^ neces- sary to know tJm'ee parts, of which one at least imust be a side. The three angles of a triangle determine only its form. The same three angles may belong to a number of triangles, of different sides and of different areas, but all similar. We shall then have different cases of the solution of SOLUTION OF OBLIQUE-ANGLED TBIANGLES. 53 oblique-angled triangles, whicli may be classified as fol- lows: 1. When two angles and a side are given ; 2. When two sides and an omgle are given ; 3. When the three sides are given. 51. Two angles and a side of an dbUque-cmgled tria/ngle ieing known, to solve the i/ria/ngle. The third angle is found by subtracting the sum of the given angles from 180°. The sides are found by the theorem of Art. 48. Thus in the triangle A C B, suppose the side a is given and any two angles as B and A. Then 0=1^° -{A-^B). .• , . ,„ sin. J. _ a ^ a sin. B Also, by Art. 48, -: — — — ;.•. J = — ; -. ' "^ 'sm.B h sm. A . sin. C c asm. C " sm. A a sm. A Solve the triangle when there are given : Example 1. A aide = 121, and the adjacent angles = 16°, and 65° 31'. Am. Sides = 83.219 and 105.79T. 2. A side = 14.3, an adjacent angle = 43° 10' 15", an opposite angle = '73° 13' 4". Ans. Sides = 10.219 and 13.3'798. 3. A side = 31. 5*7, angle opposite = 65°, and the other sides equal. 4. a = 15.189, B = 76° 10', (7= 33° 14' T'. 6. c = 668.9, 5 = 3 ^, and C = 6 ^. 6. S = 7.93, A-Z B,a.TiAB=%C. 52. Two sides and an angle of am dbUque-angled triangle being known, to sol/oe the tria/agle. Under this head we have two cases : 1. When the 54 THE ELEMENTS OF PLANE TRIGONOMETRY. given angle is inchided by the given sides ; 2. When the given angle is opposite to one of the given sides. 53. Two sides, am,d the mcluded angle, of am, oblique- angled tria/ngle heimg Tinown, to solve the triangle. We then have the sutu of the other angles given, or the suTTi of the angles opposite the gi/oen sides (Euc. 32, 1. Ch. 18, 1.). By the theorem of Art. 49 we find half the difference of these same angles. Half the differ- ence added to half the sum of the angles will give the greater, and half the difference subtracted from half the sum will give the less an- gle. The remaining side may then be found by the theorem of Art. 48. Thus in triangle AB 0,\i we have the sides, a and J, and the included angle C given, then by Art. 49, a + h tan. \ {A + B) tan. ^ (180° - (J ) a-l~ ian. ^ (A— B)~i2Ln. I {A — B) _ cotj C ~tan. ^(J.-^)' therefore tan. \{A — B) = ^^ tan \{A-\-B) a-\-i i{A-B) added to J ( J. + B) or to i (180° - C) will equal A, and ^{A — B) subtracted from i{A-\-B) or from i (180°- C) will equal J?. We shall then have the two sides, a and h, and all the angles given to find the side c. SOLUTION OF OBLIQUE-ANGLED TELiNGLES. 55 Then by tlie theorem of Art. 48, sin. A a a sin. G sm. (J sm. ^ Solve the triangle when there are given : Example 1. c—lS,b = 1, and angle A = 62°. Am. a = 10.293, C= 96° 85' 43", B = 32° 24' 11". 2. c = 12, 6 = 8, and angle A = 42'. Ans. a = 8.0818, B = 41° 28' 47", C = 96° 31' 13". 3. 6 = 4, « = 16. A — 56°. Ana. B = 13° 26' 43"', C= 111° 33' 17", a = 14.092. 4. 6 = 9, c = 19, ^ = 50°. Ans. a = 14.906, B = 27° 33' 6", C = 102° 26' 64". 6. a=ll,c=: 17, -B=70°. uljM. 6 = 16.796, A = 37° 59' 3", C = 72° 0' 57". 6. a = 10,6 = 18, C=30°. 7. 6 = 543, c = 721, A =65°. 8. 6 = 543, c = 721, A = 75°. 54. Two sic?^, = 1 sin. A. Sm. A a ^ ' a We thus find i>. ABD = 180° - {A-\- B). (2) Then ^^ ^^""^^^ AD. c sin. D G Bin. ABD sia. D Thus the triangle ABD 18 solved. To solve the triangle AB 0,we shall have, as in (1) for sin. D, sin. (7= — sin. A ; but as A OB is the sup- plement of ADB, after finding D, we subtract D from 180° to find A OB. Then ABO= 180° - {A OB + A), ox - DOB - A. ^ , AO sin. ABO Lastly, AO B O sin. A a sin. ABO sin. ^ Solve an oblique-angled triangle when there are given : ■ Example 1. A — 14° 45', a = 415, b = 432. Ans. B = 61° 20' 10", - 43° 54' 50",- c = 341.47. % B- 67° 30', 6 = 310, c — 292. 3. (7= 41° 15' S", c = 891, a = 311. 4. S = 49° 30', 6 = 6, c = 6. Am. Two solutions, ^ = 64° 38' 55", C— 65° 51' 6", a = 5.9422. or 16° 21' 5", or 114° 8' 55", or 1.8512. 60 THE ELEMENTS OF PLANE TRIGONOMETRY. 6. B=S5°, i = lll, c=123. Am. Two solutions, A - 105° 32' 12", 0= 89° 27' 48", a = 186.46. or 4° 27' 48", or 140° 82' 12", or 18.06. 6. C=26°, c=llS, i = 191. Am. Two solutions, A = 1W 26' 10", B = 44° 84' 60", o = 266.016. or 19° 34' 50", or 135° 25' 10", or 91.194. 7. C =r 81° 30', c = 116, 6 = 191. S. A= 40°, a = 129, c = 165. 57. In order to solve an oblique-angled triangle ■when its three sides are given, we establish three prin- ciples : 1. That the tangent of an angle is equal to the sine of the angle divided ly its cosine / 2. That the sine of half an angle is equal to the squa/pe root of half the difference helmeen 1 a/nd the cosine of the angle', and, 3. That the cosine of half an angle is equal to the square root of half the sum of 1 a/nd the cosine of the angle / sum and difference being used in the algebraic sense. 58. The tangent of an angle is equal to its sine di- vided by its cosine. Suppose A to be a given angle, acute in the right- hand figure, obtuse in the left-hand figure. A Then tan. A = sm, cos. A Make the triangle ABO, right-angled at £, the tri- angle of reference. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 61 CB OB A C Tan. A = jj^ ((2) Art. 4 and (2) Art. ^) = j^ sin. ^ AC ((1) and (4) Articles 4 and 44). COS. J. 69. The sine of half an a/ngle is eqmH to the squa/re root of half the difference ietween 1 a/ad the cosine of the whole angle. 1. Let MA L be an acute angle ; then . , J /l — COS. A wi.\A = ^ -^ . With A as centre and any radius A O, describe a semicircle, BGD, meeting J.Z in Z>, and J.Z pro- duced in B, and cutting AM'\a.C. Draw the straight lines B C and CD, and from O draw CE perpendicu- lar to A D, meeting AD &t E. The angle B = ^A (Euc. 5 and 32,, I. Oh. 25, and Cor. 1. 18, 1.) ; dn.^A = sin. ^ OE CB' . ,„, CE^ BExED (sin. Bf (written sm. 'B) = -^^ = jgj^y^^j) (Euc. 8, VI. Ch.l3,III.)=^; 63 THE ELEMENTS OF PLANE TRIGONOMETRY. Ajex *(-ff) therefore ^m^B^^j^= ^^^ = — i{l— COS. A) ; therefore sin. ^A^ sin. JB =/ 1-cos-A 2. Let us take the obtuse angle B AO; then, also, sin. \ A = From A as centre, with radius A C, describe the semicircle B G D, and complete the figure, as in the previous article. The angle BDC=^BAC; therefore sin. J J. = sin. B D C Sin. ^B D C =^ CE CD' CE^_ DExBE BE BExB B ~ BD CD-" BA + AE 2BA =*^+^.) AE Now ^ .S'is a minus quantity (Art. 43), and ^—^ = — AE — — - = — cos. A ((4) Art, 44). SOLUTION OF OBLIQUE-ANGLED TRLA.NGLES. 63 Therefore sin. iA = sin. JBD C = _ /l — COS. A V 60. TTie cosine of half cm angle is equal to the squa/re root of one-half the sum of 1 and the cosine of the whole angle. First, let the angle be an acute angle, as the angle MAL. Then cos. f A ■■ / V' 1 + cos. A Construct the figure as for the first part of Art. 59. Cos. iA^ COS. -5 = Cos.^^ = jbje:^ BC BE-" BE BA-^AE BG^~~BExBD~BD~ 2 AC AE\ =*0+fl) AE Now 7-^= COS. A ((4) Art. 4) ; A ' therefore cos. iA=^ cos. B ■■ /: / 1 -f- COS. A Next let A be an obtuse angle, as the angle BAO. Then also cos. i ^ = V ^l±_52!i^. 64 THE ELEMENTS OF PLANE TRIGONOMETRY. Q Construct the figure as for the second part of Art. 59. Cos. i ^ = cos. BDC--=^^ Cos. ^BD C . DE" CD D^ CD^~ BEx B B~ BB AB-AE / AE \ 2AB -^v~Aor AE But cos. A = -7-7,, and is itself a minus quantity A (j (Art. 43, (4) Art. 45), and therefore - ^^is + cos. A. A Therefore cos. i A— cos. B B C- /■ y 1 + cos. A ~2 61. The three sides of a tricmgle being hnown, to sohe the triangle. Let AB Che an oblique-angled triangle, of which the sides a, b, and are given. It is required to find the angles. From B draw a perpendicular BB to AC, or AC produced. (1) AB = AB cos. A = o cos. A (Art. 30). (2) Also, a» + 2 5 X ^ 2> = J* + c^* (Euc. 18, II. Ch. 15, III.). SOLUTION OF OBLIQUE-ANGLED TRIANGLES. B In (2) substitut& the value of A J) already found in (1) and «* -f- 2 5 c cos. A = b''-\-o\ (3) Therefore cos. J. =- 25c (4) Subtracting each member of (3) from 1, and we , ^ , 2&C — J'-c^ + a* a' — ih-c)'^ have 1 — COS. A = :; ' — r; ~ 25c 2&C _ {a-{-o — b) {a-\-i — c) _ ~ 25e Now let — -~ — = s ; a-\-c — h - a + 5 — c tnen z = s — o ; z = s — c ; and 2 J-f-c — a 2 8 — a. Dividing (4) by 2, extracting the square root of both members, and substituting the values given for a-\-o — h , a-{-h — e z , and 2 ^ But COS. A /, 2 = / (,_5) (s-o) ho /■ i/ 1 — COS. A ~2 = sin. i J. (Art. 59). 66 THE ELEMENTS OF PLANE TRIGONOMETRY. (5) Therefore -sin. IA=J (*-^) (^-^) . Again, adding 1 to both members of (3), (6) i + eos.^ = ^^^+|i^^^ = (^+f^^ ^^ ^ 25c 25e {a^1) + e){h + o-a) ~ 2bo Dividing (6) by 2, extracting the square root, and , . . , , a-\-h4-c ,i-\-o — a substituting values for and , 2 2 /l -f- COS. A __ ! s {s—a) V~~2 ~V~b^' /■ But 1 + cos. A (7) Therefore cos.i^= ./ ^^^~^^ - V he Dividing (5) by (7) we have (8>) ^^"-^^ = tan.i^= / (^-^) {B^c) "^ ' Cos.iJ. 4/ s (s—a) We can prove in a. similar manner : l(s, — ,\ Co^n\ l~, • 1 D /\B—o) (s—a) , „ / « (s—i) em. ^ B= ./ ^ '-^ ^'; COS. ^B= ./ —^ '-; y ca y ca and tan. ^ B = AllASfz:^. ^ V s (s-i) Also : /(s-a) (s- i)_ p„s 1 /Y_ /iJlzA. sm, and tan. ^ O = ^ /(s-a) (s-i) 8 (S—C) SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 67 62. If the angle, as A, B^ in tlie figure, be an obtuse angle, the expressions in (5), (7), and (8), of the pre- ceding article will still be of the same form. For we shall have : (l)a^-2ixAB = l^-\-o^ (Euc. 12, II. Ch. 16, III.). Now AI> = oX COB. BAD — c X — cos. A (Art. 46), and substituting this value in (1) we shall have cos. A = J^ + c' 25o -, the same expression as in equa- tion (3) of the preceding article. Therefore we shall have the same expressions for sin. ^ A, cos. ^ A, tan. ^ A, as in the preceding article. Find the angles of a triangle when — ExAMPM 1. a = 6, i = 6, c = 4. Ans. A = 82° 49' 10", C= 41° 24' 34". 5 = 66° 46' 16". 2. a =11, 6= 13,c = 16. Am. ^=43° 2' 66", C == 83° 10' 22". 5 =63° 46' 42"- 3. « = 25, 6 = 26, e = 21. Am. ^ = 66° 15' 4", C = 63° 53' 46". .B = 69°51'10". 4. a = 222, b = 318, e = 406. Am. A — 32° 6'?' 1", C = 95° 61' 66". B - 81° 10' 68". 5. a = 400, 6 = 340, e = 250. 6. a = 1410, b — 2160, c = 3142. 1. a = 620, b = 348, c = 192. 8. a = 412, 6 = 662, c = 330. 63. Second method of solving a triangle when the three sides a/re known. Let AB Che any triangle of which the three sides 68 THE ELEMENTS OF PLANE TRIGONOMETBY. are known. From B draw a perpendicular -BjP upon A C, the longest side, dividing the triangle into two right-angled triangles, A B Z>-and B D O. {!) BC^=OD''^BD^ (Euc. 47, 1. Ch. 14, III). (2) J.^* = ^i?^+^z>^ Subtracting equation (2) from equation (1), BC^-AB^= CD^-AD\ Factoring, {BC-^AB){BC-AB) = AC{OD-DA). A _{a-\-c) {a —c) ~ l ■ Equation' (3) will give the difference of the seg- ments of the base. Half the hose (or half the sum of the segments) added to half the diff'erenoe of the seg- ments of the hase will give the greater segment, CD ; and half the difference of the segments of the base svb- tracted from half the 'base will give the smaller seg- ment, AD. Then, in each of the two right-angled triangles ABD and B OD, we have the hypotenuse and a side given, to solve the triangle (Art. 26). Solve the triangle when the sides are given : Example 1. a = 8, i = 6, c = 4. Am. Segments of a — 5.2S, A = 104° 28' 39", 5 = 46° 84' 3", 2.7B, 0=28° 57' 18". 2. a = 219, J = 91, c = 245. Ans. Segments of « = 203.4'79; A = 62° 61' 11", 0= 96° 26' 45". 41.521, B = 21° 42' 4". S. a- 1140, 5 = 718.9, c - 627. CHAPTER VII. RELATIONS OF THE TEIGONOMETEIOAL RATIOS TO EACH OTHER. ^TRIGONOMETRICAL RATIOS OF TWO ANGLES. Art. 64. The t/rigonomebrical ratios of cm angle in- teroTumgedhle. Let B A C 'hQ the triangle of reference for the angle A. AB ' AG Sm. A = — — ; COS. A AG' Squaring and adding, r^ y§\ / ^' ■\S / / i JB H A. B C B D he two acnte angles, whose sum \& AB D. It is required to find the sine of ^ ^ Z> in terms of the sines and cosines oi AB C, and of O B D. Denote the angle AB Ohj x, and GB D by y. In A B, take any point A, and from A draw A P perpendicular to B C, and A H perpendicular to B D. From P, the foot of the perpendicular from A upon B O, draw PK perpendicular to B D, and PL per- pendicular io AJE[. The angle LAP equals C B D, that is, y. ^g ^Z-fP ^ AL , PX J.^ ~ AB AP 8m.ABP ^ X ^Z , PK BP AB AB -^ AP^ BP AB' Now, sin. ^ ^Z> = sin {x -\-y); 4 74 THE ELEMENTS OF PLANE TRIGONOMETRY. Also, ■■ sm. X AL AP = COS. y ; PK BP' sin. y ; and -r-v, = cos. a; ; A H Therefore, sin. (a? -|- S') = sin. a; cos. y + sin y cos. ». Again, cos. (« + y) = cos. x cos. y — sin aj sin y. For, the same construction being made — ^ ^^^ BE BK-LP BK LP Cos. ABB = AB AB AB AB BP BK AP LP ~AB^BP AB^AP' But cos. ABB = cos. {x-\-y); BP BK AP Also, AB cos. X ; BP COS. y ; AB :sin.a;: and sm. y ; ^P Therefore cos. (a? -f- y) = cos. a? cos. y — sin. a; sin. y. C Next, let one of the angles, AB C,he an acute an- gle, and tlie other, OB D, be an obtuse angle. Denote A B Cby X, and CBD by y. BELATIONS OF THE TKIGONOMETEICAL RATIOS. 75 From A (any point in A B\ draw A P perpendicu- lar to B G, A JS perpendicular to B J) produced. From P, where A P meets B C, draw PK perpendicular to BD produced, and draw PL perpendicular to ^^ produced. The salient angle, AB J), is the sum oi AB G and GBJ), and equals {x-\-y). AB R is the triangle of reference for ^ ^Z> (Art. 66). Also (Euc. 4, 6. Ch. 7, III.), LAP, which is similar to P KB,{^e triangle of reference for GB 2?, or y\ may be used for the tri- angle of reference for y. The angle Z^ P is equal to the angle P^-E By Art. 41, the lines AH, AL, and B K, B H, are negative. We shall put the minus sign before them to indicate the fact. Now, -AL PK AP -AL PK PB ~ AB '^ AB~ AB^ AP '^PB^AB' But sin. ABL> = sin. (a? + y) ; ^^^ AB "^ ™" "'' "Xp" "^ ^gP~ ^^^^ ^^'^' **^ PK . ;,PS = cos. y ; -p-^ = sm. y ; and -j-^ = cos. x. Therefore sin. (as + y) = sin- as cos. y -\- sin. y cos. x. . 7?n ZlIM -BK-PL Agam, cos. ABD— ^^ = j^ BP -BK AP PL ~AB^ BP ~AB^AP' But COS. ABL> = COS. (» + y) ; 76 THE ELEMENTS OF PLANE TRIGONOMETRY. Also, BP AB =. COS. X ; AP . ^PL sin, 33, and -;-^ AP -BK BP PK cos. y ((4) Art. 44) ; = Bm. y; AB ""'""' ^P PB Therefore, cos. (a; + y) = cos. as cos. y — sin. a; sin. y. If both 83 and y are obtuse, it may be proved in a similar manner that — Sin. (cB -|- y) = sin. x cos. y + sin. y cos. x. Cos. (a; + y) = cos. a; cos. y — sin. a; sin. y. 68. 7b eaipress the sine and the eosme of the differ- ence of two cmgles in terms of the si/nes wnd cosines of the gi/uen angles. Denote one of the angles by x, and the other by y; then, (a) Sin. (x — y) = sin. x cos. y — sin. y cos. x. (b) Cos. {x — y) = cos. x cos. y -\- sin. x sin. y. M B K TT First, let both angles be acute. In the figure let the angle MB J) he denoted by x, and the angle MB A by y; then AB D will be (a; — y). From A, any point in ^ B, draw ^ P perpendicu- lar to B M, and A H perpendicular to B D. From P, the foot of the perpendicular A P, draw PK perpen- dicular to B D. From A, draw A L parallel to B D. A L will be perpendicular to P K. RELATIONS OF THE TRIGONOMETRICAL RATIOS. 77 The angle LP A is equal to MBD, that is, to w. „. ,-,„ AH PK-PL PE PL Sm. AB D = AB PK X AB PB AP PB'^AB AB Now, sin. A BD = sin. (x — y); PK . PB AP X AB PL AB AP' also, -^-= = sin. x P B AB COS. y; AB = sm. y; and PL COS. K ; AP Therefore, sin. {x — y)^ sin. x cos. y — sin. 3/ cos. x. BE BK-irAL BK , AL Cos. ABB AB BE X AB BP . AL also. ~ BP^ BA^ AP But cos. ABD =■ cos. {x — y); BE BP AL X ^5 ^.^ = cos. X ; = cos. y ; ^P = sm. X : and sm. j^; AP AB Therefore, cos. (x — y) = cos. x cos, y + sin, a? sin. y. Next, let one angle he obtuse and the other acute. Lr yuA K B M O In the figure, let MBP, the obtuse angle, be de- 78 THE ELEMENTS OF PLANE TKIGONOMETEY. noted by x, and let ABMhe the acute angle, denoted by y. Then ABDis (x — y). The construction is similar to the construction of the preceding figure. From A, any point in A B, draw A H perpendicu- lar to B D, and A P perpendicular to B M. From P, the foot of the perpendicular A P, draw PK perpen- dicular to BD produced. Also, from A draw A L parallel to -5Z>, and meeting P K produced through P, at L. The angle LP A'\^ equal Xo P B K, and the triangle Z P ^ is similar to the triangle P B K, the triangle of reference for MB D, that is, x. AH PK+PL ^m.ABD 'AB~ AB PK PB , AP PL y^-T-r.-^-ir^y^ ~ PB'^ AB^ AB'^ AP Now, sin. ABD = w[i. (x — y); , PK . PB ^AP . also, -p-g = sm, X ; — — = cos. y ; and — — = sm. y. -A-gain, - 2p = ~j>£ = ^^- * ^^*) ^^- **) 5 AP Therefore, sin. (x — y) = sin. x cos. y — sin. y cos. x. _ .„„ ^5^ -XB-{-AL Cos. ABD = j^ = -^ -KB PB . AL AP x-r^ + -niX PB "^ AB^ AP AB' Now, cos. ABD = cos. {x — y); RELATIONS OF THE TRIGONOMETRICAL RATIOS. 79 - -KB PB AL PE also, -y^ = COS. ^; -- = cos.y, Jp = p^g A. P = sin. X ; and -p^- = sin. y. AB ^ Therefore, cos. (as — y) = cos. a; cos. y + sin. x sin. y. If both X and y are obtuse, it may be proved, in a similar manner, sin. (x — y) = sin. x cos. y — sin. y cos. a?. cos. (» — y) = cos. 33 cos. y -\- sin. ce sin. y. 69. By addition of equation (a) of Art. 6T to equa- tion {a) of Art. 68, we have (a) sin. (x-\-y) -\- sin. {x — y) = 2 sin. a; cos. y. By subtraction, we have (5) sin. (» + y) — sin. (x — y) = 2 sin. j^ cos. x. By addition of equation (5) of Art. 6Y to equation (5) of Art. 68, we have (c) COS. (as + y) + cos. {x — y) ^ 2 cos. « cos. y. By subtraction, we have (d) COS. (« — y) — COS. {x-\-y) = 2 sin. a; sin. y. 10. In equations of preceding article, let a; + y = a and x — y^i. Then, finding values of x and y in terms of a and J, a+ 5 « — 5 Substituting these values in the preceding equations, in order : , . . . . T « . a + 5 a — h (a) sm. a + sm- o = 2 sm. — -— cos. — -— . ,,, . ,, „.« — 5 a+5 (5) Bin. a — sm. o = 2 sm. — — cos. — r— . 80 l^HE ELEMENTS OF PLAJSTE TRIGONOMETBY. 7„ a + 5 a—h (o) COS. a -\- COS. = 2 cos. — - — cos. — - — - ■r ^ . a-\-l> . a — i (a) COS. — COS. a = 2 sm. — - — sm. , — - — - 2 2i The above formulae are useful iii loga/ritJvmic calcu- lations. 71. Dividing (») of the preceding article by (5), we have , , sin. a + sin. h a -\-h a — h (a) — : : — ;- = tan, — ~ — cot. — - — am. a — sm. o 2 2 a + i , tan. — - — a — i tan. 2 Dividing (c) of the preceding article by {d), we have ,,, cos. a + cos. J> a -\- h a — h (o) ; — ■ = cot. — T — cot. — - — COS. — COS. a 2 2 a-\-h cot. — ^ a tan. 2 . 72. Tan.(. + y)= ^^°-f + ^) cos. [x + y) _ sin, a; cos, y -\- sin, y cos, x COS. 03 cos. y — sin. x sin. y Divide numerator and denominator of the last frac- tion by COS. X COS. y : , , , tan. X 4- tan. y (a) tan. (aj -4- w) = — -— — ^• ^ ^ V I ^z i_ tan. a? tan. y RELATIONS OF THE TRIGONOMETBICAL RATIOS. 81 In a similar manner it can be shown — ,,, , , , tan. X — tan. y (h) tan. (x — y) = — —. ^^ ^ ^' 1 + tan. X tan. y Again, as cot. {x-^y)= ^^^_^^_^^^ ((/) Art. 64), , , . , , , 1 — tan. X tan. y (c) cot. (a? + y) = V I y-' tan.iB + tan.y cot. X cot. y cot. X cot. y — 1 1 1 cot. y -|- cot. X cot. aj cot. y cot. a? cot. y + 1 {d) cot. (a; - y) = — ,^ ^-f-- ' cot. y — cot. 03 (o) may also be proved directly, thus : cos. {x ■\- y) cos. X cos. y — sin. x sin. y cot. lx-\-y) = — — - — ; — - = — ; — -, . sm. (aj + y) sm. aj cos. y -}" sin. 2^ COS. a; Divide numerator and denominator of last fraction by cos. X COS. y. 1 — tan. X tan. y cot. (» + v) = — : :— — ■ — . tan.a; + tan.y Divide numerator and denominator of same frac- tion, by sin. X sin. y. , ^ cot. X cot. y — 1 cot. (» + «) = : ; — . ^ ' "' cot. y + cot. a; {d) may be proved in a similar manner. 73. Sine amd cosme of an angle m terms of the sine and cosme of half the am,gle, amd in terms of tangent of half the angle. THE ELEMENTS OF PLANE TBIGONOMETBT. In equations (a) and (5) of Art. 6t let 2 sin. X y = x, then (a) sin.2a5 = „ . COS. X ■■ 2 sin, X COS. m = — = 2 tan, sec* ,x X 2 tan. GOS? X X 1 -|- tan.* X ' (5) COS. 2 a: = cos.* a; — sin.* a? = 1 — 2 sin.*® = 2cos.*a; — 1 sin.* X 1 COS.* X 1 — tan.* x 1 1 -(- tan.* X Ti. Tangent and cotangent of an angle m terms of the tangent and cotam,gent of half the angle. In equations {a) and (c) of Art. 72 let y = x, then 2 , , ^ „ 2 tan. cc cot. x 2 cot. a; (a) tan. 2 a; = (5) cot. 2 a; = 1 — tan.*a7 1 cot.*a) — 1* cot.*aj 1 — tan. *« cot.* a! — 1 2 tan. a? 2 cot. x 75. /Sme, cosine, OAid tangent of half am angle m terms of the cosme of the a/ngle. From (5) of Art. 73 we have, by transposition, 2sin.*a! = 1 — cos.2a!; and 2cos.*aj = 1 + cos.2a!. r (a) Therefore, sin. x= ./ — COS. 2 X . 2 (5)cos.a; = yiil£?^il^; RELATIONS OF THE TRIGONOMETRICAL RATIOS. 83 equations which have already been derived by geomet- ric methods (Articles 59 and 60). Dividing (a) by (b), we have r 1 1 (o) tan. a? = / — COS. 2 X -(- COS. 2 X Example 1. Find sin. 3 a; = sin. (2 k + ic) in terms of sin. x. 2. Find cos. 8a; in terms of cos. x. 3. Find cos. 4x in terms of sin, x. 4. Find cos. 6x in terms of sin. x. 6. Find sin. 6z in terms of sin. x and cos. x. 6. Prove sin. (30° + x) + sin. (60° + x) = J (1 + v s) (sin. x + cos. x). 1. Prove COS. (30°+ a;) + cos. (60°+ a;) = ^ (1 + 4/ 3) (cos. x - sin. a;). 8. Prove sin. (46° + a;) + cos. (46° + x) = V"2 cos. x. 9. Prove sin. (45° + x) + cos. (46° + x) = sin. (46° — x) + cos. (45° — x). , „ . , „ , 2 (1 + tan." x) 10. Prove tan. (46° + x) + tan. (46° - x) = ^^ '-. (1 — tan.' x) 4 tan. X 11. Prove tan. (x + 46°) + tan. (x - 45°) = . ' ' 1 - tan.2 * 12. Show that sin. 75° + sin. 15° =V\ 13. Show that sin. 75° - sin. 16° = cos. 45°. 14. Show that cos. 22^° - cos. 6U-y - ^ ^ 2 16. Show that cos. 67i° + cos. 22^ = 4/ t_Ll. ' 2 16. Show that sin. (90° + x) = cos. x. 17. Show that cos. (90° + x) = —sin. x. 18. Show that tan. (90° + x) = —cot. x. 19. Derive the sine, cosine, and tangent of an angle of 18°. Suggestion. Let x = 18° ; 8x = 90° ; 2 x = 36° ; 3 x = 64° j there- fore, sin. 2 X = cos. 3 x. Expand both terms and solve the equation. 20. Derive the sine, cosine, and tangent of an angle of 16°. Suggestion. Let 2x= 30°; then x = 16°. Use Art. 14 and equa- tions of Art. 76. 84 THE ELEMENTS OF PLANE TRIGONOMETRY. 21. Derive the sine and cosine of an angle of 7° 80'. 22. Derive the sine, cosine, and tangent of an angle of 22° 80'. 23. Derive the sine and cosine of an angle of 168° 46'. 24. Derive the sine and cosine of an angle of 176° 16'. 26. Derive the sine and tangent of an angle of 9°. 26. Derive the sine and tangent of an angle of 24°. 27. Derive the cosine and secant of an angle of 6°. 28. Derive the sine and tangent of an angle of 168°. 29. Prove tan.'' x sec' x — sec' a; + 1 = tan.* x. sin. X 1 30. Prove sec' x + — — = . COS.' X 1 — sin. JO 31. Prove sec. x cosec. x tan. x — sec' x. a b sec' fl ab s' — 6' tan.' e a' cos.' 9 — 6' sin.= 9 n sec' fl n 82. Prove- 83. Prove ■ _ 1 + m' tan.' cos.' 6 + n' sin.' fl 34. If tan. 3 x cos.' 3 a; = tan. x cos.' x, find x. Ans. x — 22° 80'. 35. If a — 6 tan. x = (a — b) Vl + tan.' x, find tan. x. ■ ab ±(a- Ana. - a' — 2a6 36. If cos.'^ = sin. 2^ sin. , find ^. 37. If 1 — o' cos.'a: = 2 a' sin.'a;, find sin. x and cos. Xt 38. If a' cos. 'a; sin.'a; + sin.'a; + a' cos.^a; = cos.'a;, find sin. x. Ans. Sin. x , /I - «» 39. Given sin. x + cos. 2x — i; find x. 40. Given sec. x + tan. a; = f ; find a;. 41. Given tan. x + cot. a: = 10 ; find x. 42. Given sin. x + cos. x = i; find x. CHAPTEK YIII. INTEESE TEIGONOMETEIO FUNCTIONS. — CIECULAE JttEABUEE OF AN ANGLE. Aet. Y6. If two quantities are so connected, that if one of them changes the other changes also, the second is said to be 2t,fimction of the first. Thus the sine, cosine, and tangent of any angle are functions of the angle, for when we change the angle the sine, cosine, tangent, etc., also change. In this sense, the trigonometrio ratios of an angle are called its trigonometric functions. If we consider the angle and any one of its func- tions as two variables, the angle wiU be the imde- pendent variable, and the function the d&pendent va- riable. Thus in the equation y = sin. x, when x changes y also changes, but the change of y depends upon the change of x. Now, if we wish to consider the angle as a function of one of its trigonometric ratios — that is, if we wish to represent the angle as the dependent, and the ratio as the independent variable, we use a system of in- verse notation, in which the negative exponent — 1 is ' employed. In the equation y — sin. x, y is a, function of x ; but 86 THE ELEMENTS OP PLANE TKIGONOMETBY. if we wish to express a? as a function of y, we indicate ^e relation between y and a? in the form : X = sin.""^y, and read the equation, x is an angle whose sine is y. The relation between the angle and its other trigo- nometric ratios may be expressed in the same way. Thus X = cos.~*2!, X = tanr^w, s = sec."' t> may be read ; x equals an angle whose cosine is s / x equals an angle whose tangent is w / s equals an angle whose secant is -. The expressions sin."^^, cos.""^s, ta,TX.~^w, sec.~^7-) etc., may stand alone, and are called inverse trigono- metric f unctions. They may be read: an angle whose sine is y, an angle whose cosine is s, etc. 77. In a ci/rcle whose radius is unity, to find the length of an are of a/ny number of degrees, mim/utes, and seconds. Denote the circumference of a circle by C, its radius by R ; then 67= 27r^ (Ch. Art. 40, Y.). Making radius equal to unity, we have C = 2 tt. That is, 2 TT is the length of the circumference of a circle whose radius is unity. As there are 360° in the 1 IT whole circumference, 1°= — of 27r, or . 360 ' 180 l-^of^of27r, or 60 360 ' 60X180" CIRCULAR MEASURE OF AN ANGLE. 87 1 = — of — — -, or 60 60 X 180' 60 X 60 X 180 Denote the arc by a ; then, generally, the length of an arc of n degrees in a circle whose radius is unity {^ being any number, whole or fractional), is expressed by the equation, nir ■'^^iTo- 78. In the ordinary measure of an angle and its in- tercepted arc by degrees, minutes, and seconds, the unit of an angle is the -^th part of a right angle, and the unit of arc is the -^th part of a quadrant (Ch. Art. 54, IL). In the circular measwre of an angle and its inter- cepted arc, the wnit of cmgle is an angle at the centre of a ctrcle, subtended by an arc equal to the radius of the circle / and this are is the tmit of arc. 79. The angle at the centre of any circle subtended by an arc equal to the ra- dius of the circle is a con- stant quantity. In the figure, suppose AD to be an arc equal •^J""^-- ^ in length to the radius, A B, of the semicircle ADO. Denote the radius by H ; H will also represent the arc A D. The length of the semicircle, ADO, equals ir R (Ch. Art. 40, Y.). A Ti Tt AT) . , , = -^^ (Euc. 33, YI. Oh. 19, II.) 2 right angles ADO^ ' ' ' R 1 ^ 88 THE ELEMENTS OF PLANE TRIGONOMETRT. ., , . ^ -r, 2 right angles therefore ABD= — ° ?^. „ , 2 right angles , . , , , , xsut — — IS independent of the radius of the circle, and, therefore, a constant quantity. Therefore the angle AB D,ox an angle at the cen- tre of any circle subtended by an arc equal to the radius of the circle, is a constant quantity. 80. Let BAG be any angle at the centre of a cir- cle, and be denoted by A. Denote the intercepted arc, B C, by a. Let B E, an arc equal to the radius of the cir- cle, be denoted by r, and let B AE, the subtended unit of angle, be denoted by u. A a Then — = - (Euc. 33, VI. Ch. 19, H.) ; but, as u is the unit of angle, we have ^=^(Ch.20,IL). That is, the cvroulwr measv/re of am, cmgle is ex- pressed ly the ratio of the subtendvng arc to the radrnts in a circle described from the vertex of the angle as a centre. 81. Four right angles at the centre of a circle are subtended by the whole circumference, or by 'iiirr (where r denotes the radius). Therefore, by the pre- ceding article, CIRCULAR MEASURE OF AN ANGLE. 89 4 right angles = — - = 2 tt ; that is, the circular measm-e of 4 right angles is 2 tt. TTV The circular measure of 2 right angles is — = t ; TTT IT of 1 right angle is — = — ; etc. 82. If, in the equation ^ = -, we let r = 1 ; that is, if we find the circular measure of an angle at the centre of a circle whose radius is unity, we derive the equation A:= a; that is, the circular measure of any angle, at the centre of a circle whose rad/ms is unity, is equal to the length of the a/rc subtending that a/ngle. 83. Now, let A be the circular measure of any angle, and n the number of degrees, minutes, or sec- onds in the angle, n being integral or fractional ; then, by Arts. 82 and T7, nir A = 180' which is the general expression for the circular meas- ure of any angle, when the degrees, minutes, or sec- onds which it contains are given. In this system of measuring angles, an angle of : 360 4 right angles or an angle of 360 = tt or 2 ir ; or 4 right angles = 2 ir = 6.2831854 ; 180 2 right angles = tt = jr = 3.1415927 ; ^ ^ 180 90 ir 1 right angle = jr = - = 1.5707963. ^ ^ 180 2 90 THE ELEMENTS OP PLANE TBIGONOMETKY. 46 IT an angle of 45 or ■*■ right angle = v = -= .7853982 ; & ^ ° ^ 180 4 60 v an angle of 60° =: — ^ = - = 1.0471976 : " 180 3 30 V an angle of 30° = ir = - = .5235988 ; ^ 180 6 an angle of 1° = -^ = .01745329 : ^ 180 an angle of 1' = — .000290888 ; ^ 60 X 180 ' an angle of 1" — = .00000485. ^ 60 X 60 X 180 84. To Jmd the number of degrees, mw/utes, and seconds in the unit angle of cireular measure. In Art. T9 the unit angle is shown to be equal to 2 right angles TT 2 right angles are measured by 180° : Therefore the unit angle = 180° 180° -K 3.1415927 57°.296779 = 57°17'44.8." 85. To change the measv/re of a/n a/ngle i/n degrees, minutes, or seconds, to the eqvmalent circular measure/ aTid con/oersely, to cha/nge the ci/rcula/r measure of an a/ngle to its measure in degrees, minutes, or seconds. Let A be the circular measure of any angle, n the number of degrees, minutes, or seconds which it con- tains, n being any number, integral or fractional. Ac- cording to Art. 83, CIRCULAR MEASURE OF AN ANGLE. 91 (1) A = — -; therefore, 180° (2) w = A — = A x57°.295779. To change degree measure into circular measure we use equation (1) ; to change circular measure into degree measure, we use equation (2). 1. Show that tan.-' 1 + tan.-' (— 1) = ir. 1 VF ir 2. Show that ain.— » — + sin.— » = _. 2 2 2 3. If sin.—' X + sin.—' — x = — , find x. Arts, x = ;^. 8 8 2VI9 ,33 ,4 6 4. Prove sin. — = sm.— — — sin.—' — . 65 5 IB 2 1 sin.— 1 — h sm.— t — . 8 8 sm.- 1 — . 8 6. Prove sin.— ' I ■ 1 V 24 / 6. Prove sin.-' (^ 3 Vs + 2 i^ j ^ ^^_, i + 29 6 1 '!. Prove tan.—' — = tan.—' — + tan.—' — . 14 . 6 4 16 ,13 1 8. Prove tan.—' — =; tan.— — — tan.—' — . 17 7 3 9. rind the number of degrees, minutes, and seconds, in an angle whose circular measure is — . Ans. 12° 61' 28.7". 14 10. Find the number of degrees, minutes, and seconds, in an angle . , 27r whose circular measure is — . 18 11. Find the number of degrees, minutes, and seconds, in an angle 6 6 whose circular measure is — , that is, — of the unit angle. 6 6 Ana. 68° 45' 17.77". 92 THE ELEMENTS OF PLANE TRIGONOMETRY. 12. Find the number of degrees, minutes, and seconds, in an angle of which the circular measure is .IGS. IT 13. What is the circular measure of an angle of 11° 15' ? Am. —. ^ 16 14. What is the circular measure of an angle of 1° 30' ? 15. What is the circular measure of an angle of 7° 30' ? 16. What is the number of degrees, minutes, and seconds, in an arc 40 feet long, in a circle whose radius is 35 feet ? arc 180° Solution. Degrees in arc = degrees in angle =: unit x = radius ir 40 8 X —- 57°.2957'79 X - = 65° 28' 61.2". 36 7 17. What is the number of degrees, minutes, and seconds in an angle subtended by an arc of 25 feet in a circle whose radius is 15 feet ? CHAPTEE IX. AITGLES HAVING COEEESPONDING TEIGONOMETEIC FTmcnONS OF THE SAME NTJMEEICAL VAIUE. TABLE OF IMPOBT- AUT TEIGONOMETEIC FUNCTIONS. ANGLES GEEATEE THAN FOUR BIGHT ANGLES. POSITIVE AND NEGATIVE ANGLES. Aet. 86. Angles having correspondmg trigonome- tric f mictions of the scmve numerical vahie. In the figure, let XZ' and Y Y', two lines at right angles at O, be the mitial lines (Ai-t. 41). Beginning with XOY,&adi going from right to left, call XO Y the first quadrant, Y X' the second quadrant, X' O Y' the third quadrant, and Y' O X ^efovHh quadrant. Let two lines, AH and EK, be drawn making equal angles with XX' and intersecting XX' and one 94 THE ELEMENTS OF PLANE TRIGONOMETRY. anotlier at the same point, O. Then will the angles XOA, EOX\ X' OH, and KOX be equal to one another. If, now, we take the points A, E, H, and K, equally- distant from O, and join the points A and K by the straight line A K, and the points JS and H by the straight line EH, we shall have four equal right-angled triangles, BOA,EOG,GOE, and KOB(^ViQ. 4,1, Ch. 20, 1). Besides these four right-angled triangles, there can be no others equal to them, having their bases and per- pendiculars in the same relative position, with respect to the initial lines and the point 0. Eeckoning the angles from X (from right to left) through the four quadrants, there will be, therefore, only four angles, having equal tricmgles of reference (Art. 3). The triangle of reference for an acute angle will be in lAie first quadromt ; for an dblmse angle, in the second qitadrcmt ; for an angle greater tha/nh two right angles but less than three right angles, in the thi/rd quadrant ; for an amgle greater thorn three right amgles but less than four right angles, in the fourth quadromt. The nimwrioal values of the six trigonometric func- tions, sine, cosine, tangent, cotangent, secant, and co- secant, of these four angles will he the same (Euc. 4, VL Ch. 4, III). Thus the angles B A, B E, and the salient angles BOS and B OK, will have equal triangles of reference, and, consequently, equal numerical values for the corresponding trigonometric functions, sine, co- sine, tangent, cotangent, secant, and cosecant. ANGLES OP EQUAL TRIGONOMETKIO FUNCTIONS. 95 87. To find repressions for the four cmgles whose irigonoTnetric functions home the same numerical vahie. In the figure, denote the angle BOA by a, then each of the equal angles, EOG,OOH, and KOB, will also be denoted by a. BOA,BOE,^% salient angles BOHa^diBOK, are the angles which have the same numerical values for their corresponding trigonometric functions, BOA^a. ^ 6> ^= 2 right angles — i? O G^ = 180° - a = TT - a. ^ <9^=2right anglesH- G OH=1^0°-\-a = tr-{-a. B0K=4: right angles - KOB = 360° - a = 27r — a. Denoting then any acute angle by a, the four angles, which have the same numerical values for their cor- responding trigonometric functions, are : a, 180°— a, 180°-fa, 360°— a; or, a, TT — a, ir-\-a, 27r — a. 88. To find the signs of the trigonometric fvmctions of angles ha/oing equal tria/ngles of reference. In the figure, let the construction be the same as in Art. 86, In the first quadrant, the perpendiculwr and "base of the triangle of reference, AO B, are positive (Art. 41) ; 96 THE ELEMENTS OF PLANE TKIGONOMETRT. therefore, as the hypotenuse is positive, all the trigono- metric fwfietions of an angle in the first quad/rant a/re positi/ve, all the ratios having like signs* in both their terms. In the second quadrant, Y OX', \hQ perpendicular is always positi/ue, while the iase is alwajB negative / therefore, where the triangle of reference is in the sec- ond quadrant, the trigonometric functions of the angle in which the perpendicular occurs with the hypotenuse are positi/oe, both terms of the ratios having like signs ; and the trigonometric functions in which the J>ase occurs are negati/oe, as the two terms of these ratios have oppo- site signs. That is, of an obtuse angle the sins and coseca/nt are positive, but the cosine, tangent, cotangent, and secant are negative. In the third quadrant, X'O Y', the perpendicula/r and hase are both negative,- therefore, where the tri- angle of reference is in the third quadrant, the trigono- metric functions of the angle in which the perpendicu- lar or hase occur with the hypotenuse are negatme y but those in which the iase and perpendicula/r occur to- gether 2i.re positi/oe. That is, of an angle greater than two right angles and less than three right angles, or greater than tt and less than — , the sine, cosine, secant, and. cosecant are negative, but the tangent and cotangent are positive. In the fourth quadrant, Y'O X, the perpendicula/r is always negative, while the hase is positi/oe; therefore, when the triangle of reference is in the fov/rth quad- rant, the trigonometric functions of the angle in which ANGLES OF EQUAL TRIGONOMETRIC FUNCTIONS. 97 fae pefy)en(M and <— 2 + + + + + + + > — and < TT 2 ■ + - - - + + , Stt > IT and < -=-^ 2 - + - - + - + > — and < 27r 2 ' - - + + - - + 89. Particular angles having equal 1/riangles of ref- erence. (1) In the figure of Art. 86, let AO B or a, equal ; then the expressions, a,^- — a, ir -\- a,2'rr — a (Art. 87), reduce to 0, tt, and 2 tt ; Therefore the corresponding trigonometric functions of 0, TT, and ^ir, are equal in numerical value (Art. 39). As the perpendicular of the triangle of reference 5 98 THE ELEMENTS OF PLANE TRIGONOMETBT. disappears at 0, at ir, and at 2 tt, the signs of the trigo- nometric ratios, in whicli the 'perpendAoular occurs, "will be undetermined; but, as the base remains, the signs of the ratios in which the base occurs with the hypote- nuse will be determined by the sign of the base, and will therefore \i%posiime at and 2 w, but negative at ir. TT (2) Let AO B ox a equal - ; then the expressions 2 a, IT — a, 7r-\-a, 27r — • a, reduce to - and — - ; there- 2 2 TT fore the corresponding trigonometric functions of - and 2 — - are the same in numerical value (Arts. 35, 36, 37, 2 38). As the base of the triangle of reference disappears at - and at — — , the signs of the ratios in which the lase 2 2 occurs will be undetermi/ned ; but as the perpendicular remains, the signs of the ratios in which the jperpen- dicuT/r occurs with the hypotenuse, will be determined by the sign of the perpendicular, and will therefore be posvtive at -, but negaime at -— • TT Let a = - ; then, by Art. 87, (3) — , — -, — -, , will_ have their corresponding trigonometric functions the same in numerical value (Art. 14). FUNCTIONS OF PARTICULAR ANGLES. 99 Let a = — ; then, (4) -, — -, — -, — — , will have their corresponding i 4: 4 tt trigonometric functions the same in numerical value (Art. 15). Let a = —; then, o (5) -, -— -, -— , -— , will have their corresponding 000 o trigonometric functions the same in numerical value (Art. 16). The signs of the functions (3), (4), and (5) will be determined by Art. 88. (90) The following table will show the results of the preceding article : 100 THE ELEMENTS OF PLANE TRI60N0MEXEY. & s ^ o o 1-1 - 8 8 1 J5 <» iH |C<) 1 1 ?. « CO ^h 1 1 1 £h Y 1 t ^h iH 1 01 > 1 & s J|« ^1« 1 1 ©» t-t|c» 1 « k Jl|« 7 8 S o o 1 1 ^|« 1 ^ 1 IH |C(| 1 ^l» ^ « 0= 1 ^h 1 " 1 1 .H 1 o c3 £h 1 ^1" 1 1 ^ T 1 t a o 1 1 8 8 1 Jh fh|« 1 1 ^ N CO J§h ^i« 1 1 I 1 ^ 1H ^l« -^H 1 1 r-t |« 1 ^|« CO 1 t!|si - 8 8 O o rH S fc|=0 ^|« ^ Ol »H [m ^l« CO S l=H ^1" -- ^ .%\« '^ ^ o CO t|» rH|« %\« % M CO ^h ^ ca o o O o - 8 8 1 i j J OQ 1 ■g 1 ANGLES GREATER THAN FOUR RIGHT ANGLES. 101 (91). Angles greater them four right angles. C If the line A G \)Q conceived as first coinciding with A B, and then to revolve about A, from right to left, till it again coincide ~B with A B, it will pass through four right angles. If, after coinciding with A B, it still continues to revolve, from right to left, till it takes the position of ^ C (in the figure), it is said to make with A B am, angle greater than fov/r right a/ngles. Let C A B be denoted by a. Then the angle, greater than four right angles, which C A makes with A B, will be expressed by 27r -\- a. If C A be revolved twice through four right an- gles, and then continue to revolve till it takes the position A C; the angle, greater than eight right an- gles, which A G makes with A B, will be expressed by 47r -|- a. Generally an angle greater than four right angles may be expressed by 'inir + a, where n is any inte- gral, and a is the excess of the angle over ^nir. 92. Trigonomet/rio functions of angles greater than fov/r right angles. It is evident, from the figure, that the trigonometric functions of the angle 'inir + a will be the same as those of a, as they will have the same triangle of ref- erence. Therefore, to jmd the trigonometric ratios of am, angle greater than four right a/rhgles, we subtract from the a/ngle four right angles, or some multiple of fov/r right amgles, amd take the trigonometiric ratios of the remaining amgle. 103 THE ELEMENTS OF PLANE TRIGONOMETRY. Thus Bin. (2?nr + a) = sin. a. tan. (2«7r + o) = tan. u. sec. (2?iir + a) = see. a. COS. (2nir + o) = cos. u. cot. (2nir + o) = cot. a. cosec. (2»i7r + o) =: cosec. a. versin. (2«Mr + a) = versin. a. 93. It has been shown that an acute angle, an obtuse angle, an angle greater than two right angles, an angle greater than three right angles, an angle greater than four right angles, all have equal tri- angles of reference (Arts. 86, 91). Therefore the relations (established for acute and obtuse angles, in Chapter VII.), of trigonometrical func- tions of the same angle to each other, and the relations of the trigonometrical functions of the sum or differ- ance of two angles to the functions of the angles, and the relations following these, may also be considered as established for any angle whatever, as all these rela- tions depend upon the triangles of reference. 94. Positive and negative angles. If a horizontal line, ^£, be taken as an initial line, and a straight line, as A C, be made to coin- cide with A B, and then to revolve about A from right to left, assuming various posi- >E tions, the angles which it then makes with A B are positive angles. If the line AC he revolved about J. in a contrary direction (i. e., as the hand of a watch), the angles which it then makes with A B are negative angles. POSITIVE AND NEGATIVE ANGLES. 103 Thus, in the figure, B A C is a positive angle ; £ A£Jie a negative angle. 95. Belations iePween the trigonome^io fimctions of positive angles, and of equal negative angles. In the figure, let K B be the initial line; BAG be a positive angle ; and 5^^ an equal nega- ^^ tive angle. I)enote the angle BAG by a; '^ ^3^ 1^ — B then ^ J. ^ will be de- noted by — a. From G, any point in A G, draw a perpendicular to A B, and produce it to meet ^ ^ in ^. The two triangles of reference, CJ.^and HAE, are equal in all respects, except that the sign of EH is negative, while the sigh of GHiB positive ; therefore the trigonometric functions of — a will be numerically equal to those of a, but the signs of the functions of the two angles' in which the perpendicular occurs will be unlike, while the signs of the functions in which the base occurs with the hypotenuse will be the same (Art. 41). . . ^ ES -GH Thus, sm. ( - a)=^^= ^^ = -sm- «; , , EH -GH tan.(-a)=^^=-^-y=-tan.a; , , AE AG sec.(-a)=^^=;j^=sec.a; . , , AH AH ' cos.(-a) = j^ = -^ = cos.a; , ^ AH AH 104 THE ELEMENTS OF PLANE TEIGONOMETKY. Ai: AC Cosec. (- «) =;^^= _ (j^ = — cosec. a ; ^^ . , . ^ AH AH V ersin. (— o) = 1 — -7-^ — 1 — 7~77 = versm. a. 9ir 15jr Example 1. Find the sine and cosine of — , and of 8 8 2. Find the Bine and cosine of a, ir — a,ir + a, and 2 ir — o, where o = 16°. 3. What angles areexpressed in degrees by ir(re ± i), n being or any integral number ? Write the sines of these angles. 4. Write the cosines for the series of angles expressed by ir(«±J), n being 0, or any integral. 5. Write the tangents of the series of angles 7r(n±i), n being 0, or any integral. (2» + l)ir, 6. Write the trigonometric ratios for the angles n being 0, or any integral. Bec.°(^ + |) ^ 1. Prove """""^^^^^ = • COS. z Stan. sin, 8. Prove 1 + (■^ \ 1 + tan. X „ , .. sm.« cos.a; 9. Prove that a; + -J y = ir, if 1 = 0. 1 — C08.y sin. a; 10. Find two circular measures of fl which will satisfy the equation IT COS. 9 + cos.'e = sin." fl. ■ Am. 6 = — and fl = ir. • 3 11. Find two circular measures of 9 which will satisfy the equation 3 IT ir 2 cos. sm. 29 — 2 cos. 29=2 cos. 2 9 sm. 6. Am. 9 = and — . 2 6 12. Find one circular measure of

J. at right angles to A B. From C draw C^at right angles to 106 THE ELEMENTS OF PLANE TRIGONOMETRY. AB or AB produced. From B draw the tangent B fl" meeting A G produced at B.. Also, from G draw GF perpendicular to AD and from I) draw the tan- gent D G. Let A G, produced through the centre, meet the cir- cumference at M, and the tangent B H 2it H. From M draw MIST perpendicular to A B, or A B produced. The sine of an angle (or its subtending arc), is the perpendicular drawn from one extremity of the arc to the radius, or radius produced, passing through the other extremity of the arc. Thus, sin. ^ ^ C or sin. B G= C E. The trigonometric tangent of an angle (or its sub- tending arc), is the part of the geometric tangent at one extremity of the arc, terminated by the point of con- tact and the produced radius passing through the other extremity of the arc. Thus, tan. BAG or tan. BG=BII. The secant of an angle (or its subtending arc), is the part of the produced radius between the centre and the* tangent. Thus, sec. BAG or sec. BG=AE. The cosine, cotcmgent, and cosecant of an angle, (or its subtending arc), are the sine, tangent, and secant, respectively, of the complement of the angle or arc. Thus : cos. BAG, or cos. B G= sin. GA D = e,m..GI>=.GF; cotan. B AG or cot. B G= tan. GA D = tan. GJD = DG', oa^QcBA G or cosec. B C=sec. GA D = sec. GJD = AG. TRIGONOMETRIC FUNCTIONS AS LINES. 107 The versed sine of an angle (or arc) is the distance between the foot of the sine (drawn from one extemity of the arc) and the other extremity of the arc, measured on the radius, or radius produced. Thus, versin. B A O, or versin. B C= B E. It is evident from the figure that the sines and co- sines of all angles not 0, or not multiples of a right angle, are less than the radius, that is, less than 1 (1 standing for the unit of length) (Euc. 16, III. Ch. 3, II.). By bisecting the arc of the quadrant B D, and draw- ing tangents, it can be shown that a tangent may be less than radius, equal to radius, or greater than radius ; that is, less than 1, equal to 1, or greater than 1 (Euc. 6 and 19, 1. Oh. 27 and 28, I.). It is also evident from the figure, that the secants of all angles not 0, or not equal to two or to four right angles, are greater than radius, that is, greater iihan 1. At 0° the sine is 0, the tangent is 0, the secant 1, and the cosine 1. At 90° the sine is 1, the tangent co, the secant oo, and the cosine 0. 97. Lines from above, perpendicular upon A B, or A B produced are positive ; and lines from lelow, per- pendicular to AB are negatWe. Perpendicular lines proceeding from A B, or A D produced, to the right are positive, while those proceed- ing to the left are negative. Also, the radius and lines measured on the radius or radius produced, measv/red from the centre to or through one esctremity of the arc are positive ; but lines measured on the radius produced through the centre in 108 THE ELEMENTS OF PLANE TRIGONOMETRY. the opposite direction from the extremity of the a/ro are negative. Thus, all the lines representing the trigonometric functions of an acute angle are positive. Of the lines representing the trigonometric func- tions of an obtuse angle, the sine and cosecant are posi- tive ; but the cosine, tangent, cotangent, and secant are negative. That is, (in the right-hand figure, page 105), OS and A O are positive ', A E, B H, D G, and A H are negar tive. If (in the left-hand figure) we have an angle, BAM (salient), greater than two right angles : its sine MN, cosine A N, secant A H, and cosecant A G are negative ; but the tangent B H and cotangent D G are positive. If (in the right-hand figure) we have a salient angle, BAM, greater than three right angles : its sine M N, tangent B H, cotangent D G, and cosecant A G are negative ; but the cosine A N and secant A Haxe positive. The versed sine in all cases is positive. The results of thia article will, on comparison, be seen to agree with the results of Art. 88. 98. From the similar triangles BAH and EA O {see figure on page 105), BE'. OEv.BA : AE{oxCF); BAY. OE therefore, BH= -^ ; or, denoting the angle BAG'\i-^ A, and the radius by B, tan. A = '—- • TRIGONOMETBIO FUNCTIONS AS LINES. 109 sin. A but, as i? = 1, tan. A = In a similar manner therefore AM = sec. A COS. A' In a similar manner AH : AC :: AB : A£!; B^ 1 COS. A cos. A' From similar triangles, A CF and AOD, AG : AO w AD : AF{= GE); B^ 1 therefore, AG =^ cosec. J. = — = -. sin. A sm. A In a similar manner : — DG : FO :: AD : ^i!"(= <7^); ,, „ r. /7 i. ^ ^ ''OS- ^ cos. A tnereiore, D G = cot. ^ = — : -^ = -; -. sm. A sm. A Again, from similar triangles, AD G and AB H, DG : AB :: DA : BH; B^ 1 therefore, D G = cot. ^ = = . tan. A tan. A These results have already been obtained by the method of ratios (Art. 64). 99. In Arts. 14 and 15 we have obtained the tri- gonometric functions of an angle of 30° and of 45° respectively. By means of the solution of Example 19 at the end of Chapter YII., we can find the trigono- metric functions of an angle of 18°. By Art. 75 we can then obtain, from these functions, the functions of an angle of 15°, 22^°, and 9° ; and then the functions of an angle of 7° 30', 11° 15', 4° 30' ; and then the func- tions of the halves of these angles, and so on. Also by means of Arts. 67, 68, and 72, we can obtain the func- tions of the sums and differences of the angles thus found. 110 THE ELEMENTS OF PLANE TEIGONOMETKT. By these methods, however, we can ascertain the trigonometric functions of only a part of the angles of the first quadrant. By the method given in outline in the following article, we can ascertain the trigonometric functions of any angle of the quadrant. 100. Outline of a method of const/ructvng tables of t/rigon- ometrio f mictions of angles of the first quadrant. Let A he & small acute angle ; CA D its triangle of reference. From ^ as a centre with a radius A O describe the arc C E, intersecting the sides of the angle in 6" and E. li D A C were one of a number of equal angles at the centre of a circle, CD would be half the side of a regular polygon inscribed in the circle, By increasing the number of sides of the polygon, that is, by decreas- ing the size of the angle at the centre, CD would very nearly equal the arc CE {Gh. 14, Y.). Consequently, CD if the angle A were small enough, ——would approxi- C A G E CD mately equal 77-7. But -—7 = sin. A ((1) Art. 4), and L/ A A C E — — = the circular measure of A (Art. 80) ; therefore, of a very small angle the sine approximately equals the circular measure. In the construction of trigonometric tables, the angle A is taken as very small, and the sine of the angle is taken as equal to its circular measure. The angle is sometimes taken as 1' and sometimes as 10". CONSTRUCTION OP TRIGONOMETRIC TABLES. HI Suppose, then, A equal to 1', and that its sine is taken as equal to its circular measure; then: Sin 1' = .00029089 (Art. 83). Cos. 1' = Vi - sin." 1' = 4/(l + sin.l')(l-sin.l') = |/1.00029089 X .99970911 = .999999957. Now by equations {a) and (c) of Art. 69, Sin. (« + y) = 2 sin.iB cos.y — sin.(a; — y). ~ Cos. {x + y) = 2 COS.* cos.y — cos.(a! — y). In these equations let y = 1', and let x be in succes- sion 1', 2', 3', etc., and we shall have for the sines (1) Sin.2' = 2 sin.l' cos.l' ; (2) Sin.8' — 2 sin. 2'cos. 1' — sin.l' ; (3) Sin.4' — 2 sin.3' cos.l' — sin.2', etc. ; and for the cosines (4:),Cos.2' = 2cos.'l'-l; (5) Cos.3' = 2 cos.2' cos.l' — cos.l' ; (6) Cos.4:' = 2 cos.3' cos.l' — cos.2' ; etc. By substituting in equation (1) for sin. 1', its value, and for cos. 1' its value, we obtain sin. 2' ; then, by sub- stitution of values in (2) we obtain sin. 3' ; and so on. By substituting in equation (4) for cos. 1' its value, we obtain cos. 2' ; and then by substitution of values in (5) we obtain cos. 3' ; and so on. Thus, starting with the sine and the cosine of 1', we find the sine and cosine of 2', of 3', of 4', etc. The tangents, cotangents, secants, and cosecants are derived from the sines and cosines by means of the equations of Art. 64. , sin. 2' „, 1 , Thus, tan. 2' = 5, ; sec. 2' = -—— ; etc. ' cos. 2 COS. 2 It is not necessary to obtain the functions of all the 112 THE ELEMENTS OF PLANE TRIGONOMETRY. angles of the quadrant by the above method. For in- stance, suppose we haVe obtained the functions of an- gles up to 45°, at intervals of 1'. Then, denoting by x the excess of any angle ov6r 45°, we have — sm. (45° + x) = cos. (45° - x) ; cos. (45° + ») = sin. (45° - x) ; tan. (45° -\-x) = cot. (45° — x) (Art.-5) ; since 45° -|- a; is the complement of 45° — x. Let the angle, whose functions are required, be 47° 35'. Then 1 = 2° 86', and sin. 47° 35' — cos. 42° 25' ; COS. 47° 35' = sin. 42° 25' ; tan. 47° 35' = cot. 42° 26;' etc. The right-hand members of these equations are supposed to be al- ready obtained. Therefore we shall know their equals, sin. 47° 36', etc. The results obtained by the methods of Art. 99 may also be used to test the accuracy of the work by the method of this article. CHAPTEE XI. DEFIOTTIONS. — THBOEEMS OF EIGHT-ANGLED TEIANGLES. Aet. 101. Spherical Teigonometet teaches the metli<3ds of finding the unknown parts of triedral an- gles from certain known parts, by means of the solution of spherical triangles. 102. A spherical fnam.gle is formed by the inter- section of ^Q planes of a triedral angle with the surface of the sphere, the vertex of the triedral angle being at the centre of the sphere. The sides of the spherical triangle are arcs of great circles (Oh. Art. 26, YHI.), which measure the face angles of the triedral angle (Ch. Art. 53, II.). The angles of the spherical triangle are equal to the diedral angles of the triedral angle (Ch. 16, YIII.). 103. The face angles of a triedral angle are assumed to be each less than two right angles. Therefore each side of a spherical triangle will always be assumed as less than 180°. Each angle of a spherical triangle is less than 180°. (a) The trigonometric functions of a side of a tri- angle are the trigonometric functions of the angle measured by the side. 104. A right triedral angle is a triedral angle one of whose diedral angles is a right diedral angle (Ch. Art. 43, VI.). 114 THE ELEMENTS OF SPHERICAL TEIGONOMETEY. A EIGHT-ANGLED Spherical triangle is a spherical tri- angle one of whose angles is a right angle. It is formed on the surface of a sphere by the planes of a right trie- dral angle, whose vertex is at the centre of the sphere. The hypotervuse is the side opposite the right angle. 105. A guadrcmtal triangle is a spherical triangle one of whose sides is a quadrant. It is formed on the surface of a sphere by the planes of a triedral angle (whose vertex is at the centre of the sphere), one of whose face angles is a right angle. 106. In a right- wngled spherical tricmgle, not a quadrcmtal tricmgle, the three sides a/re each less than 90° ; or, of the three sides, one is less than 90°, and the other two are each greater thorn, 90°. Let OABC\)Q a right triedral angle, of which each of the face angles is less than 90°. The planes of these face angles will form on the surface of the sphere a triangle ABC, each of whose sides is less than 90' (Art. 102). Let the triangle be represented on the sur- p THEOREMS OF RIGHT-ANGLED TRIANGLES. 115 face of a hemisphere di which the base is the circle BGB'C. Let the planes of the face angles, AOO and AOB, produced intersect the surface of this hemi- sphere in the arcs CAC, BAB' (which are semicircles (Oh. Art. 32, YIII.), and let the face BOG coincide with the base of the hemisphere. By hypothesis the triangle ABO has' its three sides each less than 90°. In the triangle BAO', BA is less than '90° ; BO' and A 0' are each the supplement of an arc less than 90°, and are, therefore, each greater than 90°. In the triangle B'AO, AO is less than 90° ; B'O and B'A are each greater than 90°, being each the supple- ment of an arc less than 90°. B'OO' = BOG; there- fore the arc B'G' = the arc BO, and is less than 90°. Consequently in the triangle B'AO', one side, B'O', is less than 90°, while the other two sides, B'A and O'A, are each greater than 90°, being each the supplement of an arc which is less than 90°. 116 THE ELEMENTS OF SPHEEICAL TRIGONOMETRY. The triangles ABO, BAG', CAB', and BAG' are the only, four kinds of triangles, not quadrantal, iato which the surface of the hemisphere can be divided. The four triangles, on the surface of the other hemi- sphere, formed by producing the planes AOG and AOB to meet the surface of that hemisphere, will be symmetrical to these (Oh. Art. 63, YIII.), and wiU therefore have their sides either each less than 90°, or one side less than 90°, and the other two sides greater than 90°. The principle stated therefore holds true for the whole surface of the sphere. lOY. Two angles or two arcs, or an angle and an arc, are said to be of the sa/me species when both are less than 90°, or both are greater than 90°. 108. In a right- cmgled spJierical tricmgle, a side about the right angle cmd its opposite angle are always of the same species. Let a triangle^^C be constructed on the surface of a hemisphere having its sides each less than 90°, and let the figure be completed as in Art. 106. Then the hypotenuse wiU be greater than either of the other sides. For from G draw, ia the plane BOC,& straight line, CD, perpendicular to the radius OB. CD will be perpendicular to the plane AOB (Ch. 18, VI.), and therefore perpendicular to the straight line AD, in that plane, drawn from D io A (Ch. Art. 6, VI.). Now OD + DA >0A (Euc. 20, I. Ch. 17, 1); therefore OD + DA > OB, and taking away OD DA >DB; therefore the chord CA is greater than the chord CB (Euc. 47, 1. Ch. 4, VI.), and the arc CA is greater than THEOREMS OP RIGHT-ANGLED TRIANGLES. H7 the arc CB. Since the arc CA is greater than the arc GB, the angle GBA is greater than the angle CAB (Ch. 26, Vm.) ; but CBA is an angle of 90° ; there- fore the angle CAB is less than 90° ; but the side CB is by hypothesis less than 90° ; consequently the side CB and its opposite angle are of the soume species. In a similar manner it can be proved that AB and the angle ACB are of the same species. In the triangle BAO, BA and BCA are both less than 90° ; BC is the supplement of BC, which is less than 90°, and is therefore greater than 90° ; but its op- posite angle BA C is also greater than 90° being the supplement of BAC, which has just been proved to be less than 90°. In the triangle BAC, the two sides BA, B'C, and their opposite angles B'C A, B' AC are aU greater than 90°, being the supplements of parts of the triangle BA C, which are aU less than 90°. In the triangle B'AC, B'C and its opposite angle 118 THE ELEMENTS OF SPHERICAL TEIGONOMETBY. A are both less tlian 90°, while S'A and the opposite angle B'O'A are both greater than 90°. The principle of the article holds true for the trian- gles on the surface of the hemisphere represented by the figure ; it also holds true for the triangles symmetri- cal to these on the surface of the other hemisphere, and therefore holds true for the surface of the whole sphere. If the triangle had two right angles, the sides oppo- site these would be quadrants (Oh. Art. 89, YIH.), but the remaining side and its opposite angle would be of the same species (Ch. 16, YIII.). 109. In a right-cmgled spherical tricmgle, the sine of either of the sides about the rigM angle is' equal to the prodMct of the sine of the hi/pot&nuse ly the sine of the a/ngle opposite the side.* ^^ Let ABG be a right-angled ^ ~^ — ~-— ^P triangle on the surface of a sphere \y^m /\ '^^ose centre is the vertex, 0, of ■^\ "•■-. / \ the triedral angle, which forms \ ^y<. I the triangle by the intersection Jii"6) and ^i^(9 are right-an- ''^^^^~~^::^ C gled triangles by construction, V/^^ ^~~7l DFE is the flome wngle of ^^\ ""• .. / I the diedral angle whose edge is \ '"/{_ r OA, and therefore is equal to V<:£___li^ the angle A of the spherical tri- '' angle ABC (Ch. Arts. 39 and 45, YI. Prop. 16, Yin.). Denote the sides opposite the angles of the spherical triangle by small letters of the same name. sin. a = sin. BOG {(a) Art. 103) = ^((1) Art. 4) ^DF DE, ~ BO DF' T)F but — = sin. 00 A = sin. h : BO T)W and — - = sin. DFE= sin. A ; BF . ' . (1) ( sin. fls = sin. 5; sin. A. By drawing FD, and ED perpendicular to 00, it can be proved that (2) sin. c = sin. h sin. G. 110. In the preceding article the demonstration has been applied to a figure drawn to represent a triangle 130 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. of whicli the sides are each less than 90°, but the theo- rem is true for all right-angled triangles. Let ABC represent a triangle on the surface of a hemisphere. Let the three sides, a, b, and c, be each less than 90°. Then the angles, A and C, are each less than 90° (Art. 108). Complete the figure as in Art. 106. Not considering the right angle, of the five parts (three sides and two angles) of each of the triangles BAG', B'AC, and BAG, one part, as in, the triangle BAG, is the same as a part of ABC', or two parts, as ia the triangle B'AC', are equal to parts in the triangle ABC, and the other parts are the supplements of parts of the triangle ABC. Therefore the theorem is true for these triangles (Art. 46, {a) Art. 103). Thus in the triangle BAG, sin. B'G= sin. a; sin. AG= sin. 5; sin. CAB' = sin. A ; but, according to previous article, sin. a = sin. 5 sin. A ; substituting, sin. B'C= Bm.ACx sin. GAB'. THEOREMS OP EIGHT-ANGLED TRIANGLES. 121 In a similar manner the theorem may be proved true for the triangles BAG' and B'AC. If A and £ were both right angles the theorem would still be true, for in that case AO axiA BG would be quadrants (Oh. Art. 89, VIII.), and sin. a, sin. 5, and sin. A would each be 1 ; also c would equal C (Ch. 16, Yin.). 111. In a right-angled spherical triamgle, the cosine of the hypotenuse is equal to the product of the cosi/nes of the other two sides. Ox ,Let ABG be a triangle right- angled at B, and on the surface of a sphere whose centre is 0, the vertex of the right triedral angle OABG. Then we are to prove that cos. h = cos. a cos. c. Construct the triangle DEF as in the figure of Art. 109. 6 123 THE ELEMENTS OP SPHERICAL TRIGONOMETRY. COS. b = COS. A00=^ ((4) Art. 4) OB OF X OD OE' but —— • — COS. BOG— COS. a ; and 77^= cos. AOB OD , OS — COS. c ; therefore cos. h = cos. a cos. c. This proposition may be proved to be true for all right-angled trian- gles by applying the principle of Art. 46, with regard to the cosine, to the figure of the preceding article. 112. In a right-angled spherical l/ria/ngle the ta/nn gent of either side about the right angle is equal to the jprod/uot of the ta/ngent of the opposite angle by the sine of the other side.* Let ABO be a spherical tri- nnT ~~;f ^c- angle, right-angled at B, formed \^^m /I ^y ^ right triedra;! angle whose -^V ' .. y/ j vertex is at the centre, O, of the \ ^^ / sphere. ^-<:^___-^ Then we are to prove tan. a = tan. A sin. c. Construct the triangle DEFas in Art. 109. Then DEO, and EEO, are triangles right-angled at E; and DEE, and DEO, are triangles right-angled at E. (See Art. 109.) tan. a = tan. BOC= ^ ((2) Art. 4) ; EO ^=^X ^=tan. DEEX sin. AOB EO EF EO = tan. A sin. c ; therefore * This theorem may be remembered by its resemblance to (2) Art. 32. THEOEEMS OF RIGHT-ANGLED TEUNGLES. 123 (1) tan. a = tan. A sin. e. By drawing ^D and BB at right angles to OC it may be proved (2) tan. G = tan. O sin. a. 113. In a right-angled spherical triangle, the tan- gent of either side about the right angle is equal to the product of the cosine of the adjacent dbUque a/ngle iy the tangent of the hypotenuse* Construct the figure as in Art. 109. Then we are to prove tan c = cos. A tan. 5. .^o FF EF FD tan.. = tan.^O^ = — =— X— ; but ^ = COS. DFE= cos. A ; FD WT) and = tan. AOC = tan. h ; FO , • . (1) tan. c = Qos. A tan, b. In a similar manner it may be proved (2) tan. a = cos. C tan. h. The theorem of this article and the theorem of the preceding article can both be shown to be true for all right-angled triangles, by applying Art. 46 to the figure of Art. 110. 114. The Circular Parts of a right-angled spherical triangle are the sides about the right angle, the comple- ments of the hypotenuse and the obUgue angles. The right angle is not considered one of the circu- lar parts. Of the five circular parts any one may be taken as the middle pa/rt, and then the two next to this, one on the one hand and the other on the other (not counting * This theorem may be remembered by its resemblance to (2) Art. 29. 134 THE ELEMKNTS OF SPHERICAL TRIGONOMETRY. the right angle as a part) are called adjacent jpoHs, and the remaining two, each separated from the middle part by an adjacent part, are called opposite parts. Thus u, c, 90°— b, 90°— A, and 90°— C, are the circular parts of the triangle ABC right-angled at B. If a is the middle part, adjacent parts are c, 90°— O; opposite parts are 90°-^, 90°-5. If c is the middle part, adjaeent parts are a, 90°— A ; opposite parts are 90°— 6, 90°— C If 90°— 6 is the middle part, adjacent parts are 90°— -4, 90°— C; opposite parts are a, c. If 90° - ^ is the middle part, adjacent parts are 90°— S, c; opposite parts are o, 90°— O. If 90° — C is the middle part, adjacent parts are a, 90°— 6 ; opposite parts are c, 90°—^. 115. Narpv&fis rule of the Cirffidao' PaHs. The sine of the mMdle pa/rt is equal to the prockiot of the tamgents of the adjacent pa/rts ; and the sine of the middle part is equal to the product of the cosines of the opposite paHs. Let ABO he a spherical triangle right-angled at B. We shall take each circular part in succession as a middle part ; be- ginning with 90°— 6, and next taking 90°- C, and so on. Taking 90°— J as the middle part we are to prove sin. (90°-5) = tan. (90°-^) tan. (90°- C7); or (Art. 5), (1) cos. J = cot. A cot. C. Now, according to Art. 112, (a) tan. a = tan. A sin. c, and (5) tan. c = tan. O sin. a. THEOREMS OF RIGHT-ANGLED TRIANGLES. 135 Divide equation (5) by equation («), putting the second member of (J) over the first member of (a). tan. G sin. a _ tan. c tan. a tan. A sin. o tan. C COS. a = ((5) Art. 64), or tan. A COS. c (c) COS. a COS. G = -. = cot. A cot. O ' ' tan, A tan. G ((/) Art. 64) ; but, according to Art. Ill, COS. 5 = COS. a cos. c ; therefore from (c) COS. J = cot. A cot. (7. Again, sin. (90°— 5) = COS. a cos. c, or (2) cos. J = COS. « COS. c (proved under Art. 111). Next take 90°— (7 «s middle pa/rt, and we are to prove sin. (90°— C) = tan. a tan. (90°— I), or (3) COS. C= tan. a cot. 5, and sin, (90° -(7) = COS. g cos. (90°—^), or (4) COS. G = COS. c sin. ^. To prove (3) we have from Art. 113 cos, G tan, h = tan, a ; therefore COS, G— — '-— = tan, a cot. J ((/) Art, 64). tan. I \\J I J To prove (4) /j\ n 1. I. -L sin. a ^ , cos. 5 (a) cos, 6' = tan, a cot, o = X COS. a sin, h ((5), (^, Art. 64); sin, A = ^^ (Art. 109). sin. 126 THE ELEMENTS OF SPHEBICAL TEIGONOMETBY. Also COS. h = COS. a cos. o (2) ; substituting these values for their equivalents in, second mem- a ber of (d) cos. O = eos. sin. A. Let a be the middle part, then we are to prove (5) sin. a = tan. c tan. (90°— (7) = tan. c cot. C; (6) sin. a = cos. (90° -5) cos. (90°—^) = siu. h sin. A. To prove (5) sin. «=-^5Hif (Art. 112) tan. C/ = tan. a cot. C ; (6) is proved under Art.' 109. Let be the middle part, then (7) sin. e = tan. a cot. ^ ; proved in the same manner as (5). (8) sin. = sin. 5 sin. C (Art. 109). Lastly, let 90° — A he the middle part, then (9) cos. A — cot. 5 tan. c ; proved by Art. 113 ; see proof of (3). (10) COS. A = cos. a sin. C ; cos. I ^ an^ ^^g^^ ^^^ ^j^ ^^^ ^^^ (e) cos. A of Art. 64), sin. O sin. S cos. c sm. sin. J tuting these values in (e) cos. .4. = cos. a sin. C. cos. h = cos. a cos. c ; substi- CHAPTEE XII. SOLOTTON OF EIGHT-ANGLED TEIANGLE8. Aet. 116. Any two parts (in addition to the right angle) of a right-angled spherical triangle being known, the triangle can be solved, and the v/iiknown parts can be obtained in terms of the knawn parts, by applying Napier's rule. 117. To find any unknown part from two known parts, care should be taken in applying Napier's rule to make such a selection of one of the three parts for a middle part that the other two should- be either both adjacent parts or both opposite parts. 118. Care should also be taken to observe the signs of the trigonometric functions, as the sign of the result wUl determine generally whether the arc or angle is < 90° or > 90° (Art. 46). 119. The two parts, which may be given (in ad- dition to the right angle) to solve a right-angled spher- ical triangle may be : 1. the hypotenuse and a side ; 2. the hypotenuse and an angle; 3. the two sides about the right a/ngle; 4. a side and an adjoicent angle; 5. a side about the right angle and the opposite amgle; 6. the two angles. 120.. Case I. — The hypot&n/use and a side, of a 128 THE ELEMENTS Of SPHERICAL TRIGONOMETRY. right-angled spherical triangle, being known, to solve the triangle. In the triangle ABC, right-angled at B, suppose h and a to be known. It is required to find c and the angles A and' O in terms of a and i. To find G take 90°— & as the middle part, and a and o as the opposite parts ; then COS. h = COS. a COS. c (Art. 115) ; , , COS. i (ffl) or COS. c = . COS. a To find A take a as a middle part, 90°— 4 and 90°— 5 as op- posite parts ; then sin. a = sin. A sin. h ; whence sin. a (5) sin. A = sin. h To find O take 90°— (7 as a middle part, a and 90^ 5^s«(^Voew#-pa¥ts;-.th^ (f^ (c) COS. <7 = tan^ a cot. h. --^ ^"^ArifroEaTequation {a) c is obtained by means of its cosine, the sign of — '— will determine whether c is cos. a < 90° or > 90° (Art. 46). If a and i are both < 90° -0 COS. h or both > 90° COS. a wiU be positive, and cos. c will also be positive, and therefore c will be < 90°. If a and b are not of the same species (Art. 107), COS. b will COS. a be negative, and consequently cos, c will be negative, and, therefore, o will be > 90°. SOLUTION OF EIGHT-ANGLED TRLINGLES. 129 Though from (S) A is obtained by means of its siae, A can have only one value, since A and a are of the same species (Art. 108). Also as O is found from its cosine, the sign of tan. a cot. 5 will determine whether <7 is < 90° or > 90°. Again G and o must be .of the same species (Art. 108). As -check on work, form an equation in which only the three required parts occur. In the pres- ent case after c. A, and G are found, if the results are correct, the equation made by applying Napier's rule to 90°— C as a mid- -^ die part, g and 90°—^ as opposite parts, that is COS. <7= cos. c sin. A, should be a true equation. If it prove not to be a true equation, there must be some error in the previous calculations. 121. Case II. — The hypotenuse cmd an angle of a right-angled spherical triangle lehig hnown, to sdhe the t/riam^le. B being the right angle, suppose h and A are given to find G, a, and c. Take 90°— 5 as a middle part, 90°— J. and 90°— C as adjacent parts ; then cos. l = cot. A cot. G (Art. 116) ; ^ cos. J (a) cot. G = 7. ^ ^ cot. A To find a, use it as a middle part, and 90°— J., 90°— h as opposite parts. (5) sin. a — sin. I sin. A. 130 THE ELEMENTS OF SPHERICAL TBIGONOMETBY. To find c, take 90°— J. as middle part, and o, 90°— h as opiposite parts. COS. A = tan. c cot. 5, whence c^ cos.^ -3"'^^Sa' (c) tan. JLi,„'*x^ - .i.-^ cot. & As G is found by means of its cotangent, and as c is found by means of its tangent, the sign of the equivalent frac- tion in each case will determine whether the quantity is < 90° or > 90° (Art. 46). Though a is found by means of its sine, it can have but one value, being < 90° or > 90°, ac- cording as ^ is < 90° or > 90° (Art. 108). As check on work, use the equation sin. a = cot.. O tan. c. 122. Case III. — The two sides about the right angle of a right-angled spherical triangle heim^ hnown, to solve the triangle. Suppose a and c, about the right angle B, to be known, it is required, to find A, i, and O. To find A take c as a middle part, and 90°—^ and a as adjacent parts ; then, sin. = cot. A tan. a (Art. 115) ; / \ J. A sin. c (a) cot. A = . tan. a To find h, take it as middle part, (5) cos. b — COS. a COS. o. To find C take a as a middle part, and e, 90°— C as adjacent parts. SOLUTION OF EIGHT-ANGLED TRLiNGLES. 131 sin. a — cot. C tan. c, or sin. a (c) cot. C= tan. G As A and (7 are each found by means of its cotan- gent, and as h is found by means of its cosine, eacb of these quantities can have only one value, which will be determined by the sign of its trigonometric function (Art. 46). As check on work, use the equation cos. b = cot. A cot. C. 123. Case IV. — A side about the right angle, of a right-angled spherical triangle, and the adjacent oblique angle being known, to solve the tri- angle. B being the right angle, suppose c and A are known ; it is required to find b, O, and a. To find b, take 90°—^ as the middle partj and 90°— 5, G, as adjacent parts. cos. A = cot. b tan. c (Art. 115), or , cos. A {a) cot. ■C as middle part, and c, tan. e To find O, take 90' 90°— A as opposite parts. (5) COS. C — cos. c sin. A. To find a, take c as middle part, and », 90°—^ as adjacent parts. sin, c = tan. a cot. J., or sin. c (g) tan. a cot. A 132 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. As the required parts are found by means of the cotangent, cosine, and tangent respectively, each can have but one value, which -will be determined by the sign of its trigonometric function (Art. 46). As check on work, use the equation cos. C = cot. h tan. a. 124. Case Y. — A side about the right angle, of a right-angled spherical triangle, cmd the oppo- site cmgle hei/ng Tmovm, to sol/oe B being the tight angle, a and A are given ; it is reqiured to find 5, C, and G. To find 5, take o as a middle part, and 90°— J., 90°— 5 as opposite parts. sin. a = sin. J sin. A (Art. 115) ; ... , sin. a (ob) sm. = -. sin. A To find C, take 90°— J. as middle part, and a, 90°— C as opposite parts. cos. A — COS. a sin. C; whence ,,, . „ COS. A (p) sin. (7= . COS. a To find 0, make it the middle part, and a, 90°— -d a^aoent parts. (c) sin. c = tan. a cot. A. As, in this case, each of the required parts is ob- tained by means of its sine, each part will have two values (Art. 46), and there will be solutions answering to two triangles, each of which will contain the given parts. SOLUTION OF EIGHT-ANGLED TRIANGLES. 133 The required parts of one will be supplements of the required parts of the other. As check on work, use the equation sin. c = sin. i sin, O. 125. That there will be two right-angled spherical triangles containing the same given parts, when those parts are a side and the opposite, angle, will be evident from the accompanying figure. Let BA'OAC be an ungula in a hemisphere of which the base is the circle, APA', and the centre 0. The lune ABA'C will be the base of the wedge (Ch. Art. 90, VIII.). Let a plane BCPO be passed through the pole, P, of the circle ABA', through the centre, 0, and any point, B, of the arc ABA' (Ch. Art. 33, VIIL). The wedge will be divided into two right triedral angles, OB AG, OBA'C, having the face angle, BOO, in com- mon, and having the diedral angles, A and OA', the 134 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. same, since these are parts of the diedral angle AA'. The remaining parts of the one are the supplements of the corresponding parts of the other. The lune will be divided into two right-angled tri- angles, corresponding to the right triedral angles, hav- ing the side a in common, and having the angle A equal to- the angle A' (Oh. 16, VIII.), and having- the remaining parts of the one the supplements of the cor- responding parts of the other. This case of the solution of right-angled spherical triangles is called the CMnbiguous case. 126. Case YI.—The imoo an- gles of a right-angled spherical triangle leing known, to solve the t/ricmgle. B being the right-angle, sup- SOLUTION OF BIGHT-ANGLED TRIANGLES. 136 pose A and C to be known ; it is required to find h, a, and c. To find h, make 90°— 5 the middle part ; then c («) COS. h = cot. A cot. C (Art. 115). To find a, make 90°— J. the a iniddle part. COS. A = sin. C cos. a, or ,Tx cos. ^ (o) cos. a = — — -. sin. O To find c, make 90°— C the middle part. cos. O — COS. c sin. A, or , , COS. G (c) COS. e = -. sin. A As check on work, use the equation COS. h = COS. a COS. c. Suppose A, B, C to be a right-angled spherical triangle of which the hypotenuse, 6, is 100° and the side, a, Is 60°. Bequired to find the other parts. COS. b ■- : COS. a COS. e. COS. h COS. a COS. 100° A sin. a - . • . sin. A — sin. b sin A. sin. a sin. 60° sin. b = — quantity. COB. 60° . • . e> 90°. Log. COS. 100° = 9.2396T0 Log. COS. 60° = 9.698970 Log. COS. 69° 40' 40.7" 9.640700 .-. c=:110° 19' 19.3° sin. 100° A of same species as a. .-. ^<90°. Log. sin. 60°= 9.937B31 Log. sin. 100°= 9.9933B1 Log. sin. 61° 34' 61" 9.944180 136 THE ELEMENTS OF SPHERICAL TKIGONOMETRY. COS. C = tan. a cot. 6. = tan. 60° cot. 100° = — quantity. .-. 0> 90°. Log. tan. 60°= 10.238561 Log. cot. 100°= 9.246319 Log. COS. 72° 13' 2.27° 9.484880 .-. C= lOr 46' 87.73". (Chech) COS. C= COS. c sin. A. Log. COS. c = 9.640700 Log. COS. A = 9.944180 9.484880 Log. COS. (7=9.484880 (Examples worked by six-place logarithm tables, and answers given to nearest tenth of second.) Solve a right-angled triangle when there are given Hypotenuse and a side. Example 1. Hypotenuse, 72° ; side, 13°. Am. Side, 71° 30' 35" ; angles, 13° 40' 54" ; 85° 41' 53". 2. Hypotenuse, 64° 40' ; side, 137° 60'. Hypoiemise and an angle. 3. Hypotenuse, 73° ; angle, 16°. Ans. Angle, 85° 31' 14" ; sides, 14° 19' 48.7" ; 72° 26' 11.8". 4. Hypotenuse, 92° ; angle, 95°. . Ans. Angle, 68° 15' 10" ; sides, 96° 23' 5" ; 68° 9' 65.6". Two sides about the right aaigle. 6. 63° and 42°. Ans. Hypotenuse, 70° 16' 67*" ; angles, 71° 10' 25^" ; 45° 18' 2". 6. 16° and 116°. Ans. Hypotenuse, 114° 55' 20.5" ; angles, 17° 41' 40" ; 97° 39' 24.4". 7. 100° and 135°. A side about the right angle and an adjacent aiigle. 8. Side, 39° ; angle, 61°. Am. Hypotenuse, 69° 5' 29.4" ; side, 48° 37' 34.3" ; angle, 47° 10' 45.6". SOLUTION OF RIGHT-ANGLED TRIANGLES. 137 9. Side, 100° 85' ; angle, 50° 2'. Ans. Hypotenuse, 96° 50' 37.6" ; side, 49° 32' 65.3" ; angle, 98° 5' 31.6". 10. Side, 112° 4' ; angle, 100° 6'. A side about the right angle and an opposite angle. 11. Side, 25° 16' ; angle, 36° 13'. „ , 46° 18' 15.5" ; ^«s. Hypotenuse, jg3„^^, 4^ g„. . 40° 7' 40" ; 63° 8' 36.4". " ®' 139° 52' 20" ; °'°S"^' 116° 51' 26.6". 12. Side, 114° 2' ; angle, 102° 16'. , „ ^ 69° 9' 42.5"; ^ns. Hypotenuse, jjQ„g(j,j^_g„. . 29° 8' 13.2" ; 31° 23' 62.6". ^ ®' 150° 51' 46.8" ; ^^ ' 148° 36' 7.4". 13. Side, 62° 10' ; angle, 74° 1'. Two angles. 14. 56° and 40°. Atis. Hypotenuse, 36° 30' 34" ; sides, 29° 32' 49.2" ; 22° 28' 45.6". 16. 131° 20' and 110° 15'. Ans. Hypotenuse, '71° 3' 57.6" ; sides, 134° 44' 41" ; 117° 26' 63.9". 16. 60° and 120°. CHAPTEE XIII. QUADBANTAL TBLANGLES. Aet. 127. A quadrantal triangle is formed by a trie- dral angle (at the centre of a sphere), one of whose face angles is a right angle (Art. 106). Thus suppose OABOi& a triedral angle (at the cen- tre, 0, of a sphere), having one of its face angles, AOC, a right angle. Then the planes of the faces of the trie- dral angle will form, by their intersection with the sur- face of the sphere, a quadrantal triangle, ABO, of which the side, AC, which measures the right angle, AOC,^& a quadrant (Oh. Art. 53, II.). QUADRANTAL TRIANGLES. 139 128. Thepola/r tvicmgle of a quadromtal tricmgle is a right-angled tria/ngle. Let ABO be a quadrantal triangle, having its side, AC ovb,2L quadrant. IMA'B'C be i^iQjpolar triangle of ABC (Ch. Art. 6Y, VIII.). Denote the sides of the triangle ABC by a, b, 0, and the sides of A'B'C' by a', V, and c', respectively. Then A'B'C is a right-angled triangle, right-an- gled at B'. ^' + 5 = 180° (Oh. Arts. 69 and 70, VIII.) ; there- fore B' = 180°- h = 180°- 90° = 90°. 129. Any two pwrts of a quadrantal triam,gle, in addition to the side .which is a quadrant, being known, we can solve the polar right-angled triangle. For the two given parts of the quadrantal triangle are the supplements of two parts of the polar triangle (Ch. 18, VIII.). Two parts of the polar triangle, in addition to the right angle (Art. 128), will there- fore be known, and the polar triangle can be solved (Art. 116). Thus let ASO be a quadrantal triangle, having the side, b, a quad- rant. Let A' SO' be the polar triangle of ABC. Denote the sides of the triangle ABC by a, h, e, and the sides of A'ff C by a', V, and c', respectively. 140 THE ELEMENTS OF SPHEKIOAL TRIGONOMETRY. (1) a' - 180°- A ; (2) V = 180°— B ; (3) c' - 180°^ C ; also (4) A = 180° — a; (5) B = 180° — 6 = 90° ; (6) C" = 180°— c (Ch. Art. 70, VIII.). From equations (1) to (6) inclusive, it is evident that if we know any two parts oiABO, besides 6, we know two parts of A'BC besides the right angle B. 130. The polar right-angled triangle being solved, and all its parts being known, all the parts of the quadrantal triangle will be known, as the sides of the one are the supplements of the angles of the other (Oh. 18, YIII.). 131. To solve a quad/ramtal triangle we derive from Arts. 128, 129, and 130, the following rule : Take the supplements of the gi/ven pa/rts of the quad- rantal triamgle for the gi/ven parts of the polar right- angled triangle / solve the pola/r triamgle, and take the supplements of the pa/rts fovmd as the regwi/red pa/rts of the quadrantal triangle. If in the quadrantal triangle we have given, in addition to the quad- rant, the other two sides, in the polar right-angled triangle we have given, besides the right angle, the two angles ((4) and (6), Art. 129). If in the quadrantal triangle we have given any two angles, in the polar right-amgled triangle we have given two corresponding sides ((1), (2) and (3), Art. 129). If in the quadrantal triangle we have given, in addition to the quad- rant, a side and an angle, in the polar right-angled triangle we have given, QUADRANTAL TRIANGLES. 141 besides the right angle, an angle and a side ((4) and (1) or (4) and (2) etc., Art. 129). In any example. all these relations will be best seen by drawing the quadrantal triangle and its polar triangle. Example. Solve a quadrantal triangle of which there are given, in addition to the side which is a quadrant, a side equal to 64°, and an angle between the side and quadrant equal to 120°. Let 6 = 90° ; a =; 64° ; C= 120°. .-.^ = 90°; ^' = 116°; d — 60°. Solution of A!B'Q' faUs un- der Case IV. (Art. 123). COS. A = cot. b' tan. c'. COS. A' cos. 116 cot. 6' = ■ — — quantity . ■ . 6' > 90°. tan. c' tan. 60 Log. COS. 116° = 9.641842 Log. tan. 60° = 10.238561 Log. cot. 15° 47' 49.43" = 9.403281 b' = 104° 12' 10.67'. sin. c' = cot. A' tan. a' COS. C" = sin. A' cos. c' : sin. 116° COS. 60° = + quantity. .-. C"<90°. Log. sin. 116°= 9.963660 Log. cos. 60°= 9.698970 . tan. o'=- sin. c' sin. 60° cot. A cot. 116° = — quantity. . •.a'>90°. Log. sin. 60° = 9.937531 Log. cot. 116° = 9.688182 Log. cos. 63° 17' 41.96" = 9.652630 Log. tan.60° 86' 44.8" = 10.249349 C"= 63° 17' 41.96'. a'= 119° 23' 16.2'. (Cheek.) cos. C— cot. b' tan. a'. b'= 104° 12' 10.67" Log. cot. 6'= 9.403281 Log. tan. a'= 10.249349 0"= 63° 17' 41.95 Log. cos. 0"= 9.662630 "'= ^^^° ^3' 16-2" •.jB= 76° 47' 49.43". . e = 116° 42' 18.05." .A= 60° 36' 44.8". 142 THE ELEMENTS OP SPHERICAL TEIGONOMETRY. Solve a quadrantal triangle when there are given, in addition to the side which is a quadrant : ExAUFLE 1. The two angles adjacent to the quadrant, 76° and 104°. Ans. Angle = 86° 24' 36.6" ; sides = 16° 26' 84.8" and 103° 82' 2V". 2. The two other sides, 70° and 106°. Am. Angles, 84° 24' 11.9'; 69° 15' 45.4"; 106° 69' 15.3". 3. A side 60°, angle between the side and quadrant 126°. CHAPTEE XIV. THEOREMS OF OBLIQUE-ANGLED TELiHGLES. Art. 132. If from apomt m the smfaceof a hemi- sphere, which is not the pole of its hase, a/rcs of great circles cure d/ra/um to the circumference of the great cir- cle, which is its base, of all these arcs the geeatest is that which passes thbough the pole, the least is that which when produced passes through the pole, and of J 3 ^ ^ A j^ ^.j:?^ h>\ y\--^^'.'- -■■''■/ \&\ -^ ^■■■•■■.■■•'•■■'!!.- '■' // \ \'a\ /^■""/ y-^- 1 > ...V a L/- t^^ 1\\ "i / 3^ ic the others that which is nearee to the greatest is geeatee tha^ the mobe eemote / amd from this point eqttal AEcs of great circles can he drawn only est pairs. Thus, let the figure PBEGG represent a hemi- sphere having for its base the great circle BEGG. 144 .THE ELEMENTS OF SPHERICAL TRIGONOMETRY. Let P be the pole of the base (Ch. Art. 27, VIII.), .Suppose A to be any other point on the surface of the hemisphere. From A let the arcs APB, A C, AD, AE, AF, and AK, be drawn to the circumference, BECQ. P Then, of these arcs APB is the greatest, AC is the least, and AD, which is nearer to APB, is greater than AE, which is more remote; ^£'than AE, etc. Let M be the centre of the sphere of which the fig- ure represents the hemisphere. Join PM; then PM is perpendicular to the plane of the circle, BECG {Ch. Art. 27, VIII.), and the planes, BPCM and BFCM, are perpendicular to each other (Ch. 17, VI.) ; also the arcs BA G and BEG are perpendicular to each other (Ch. Art. 68, VIII.). BC is the diameter of the sphere common to the two semicircles BPCM and BFCM (Ch. Art. 32,VIII.). Now, let a perpendicular be let fall from A upon the plane of the circle BECG. It will lie in the plane BACMifih. Art. 51, VI.), and at L will meet at right angles the line of intersection, BC, of the two planes BAGM and BFCM {Gh. Art. 6, VI.). THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 145 Draw the straight lines LD, LE, LF, and LK, in the plane BFCO; also draw the straight lines J. 5, AD, AE, AE, AK, and AC. Now, M is the centre of the circle BFCO (Ch. Art. 26, YIII.), and Z is a point in the diameter which P is not the centre; therefore LB is the greatest line which can be drawn from L to the circumference BFCG, ZC is the least, and LD, which is nearer to LB, is greater than LE, ZZ'than LF, etc. (Euc. 7, III. Ch. 17 and 24, 1.). Therefore of the straight lines, drawn from the point A to the points B, D, E, F, K, and C, AB is the greatest, AC \.% the least, and AD is greater than AE, :4^than ^i^(Euc. 47, 1. Ch. 4, VI.). Consequently, since in equal circles, or in the same circle, greater chords cut off greater arcs, the arcs being less than semicircumferences (Ch. 6, II.) of all the great circle arcs, drawn from A to points on the circum- ference BFCG, APB is the greatest, AC \s, the least, AD is greater than AE, AE than AF, and AE than AK, these arcs being arcs of equal circles 146 THE ELEMENTS OP SPHERICAL TKIGONOMETKT. and all less than semicircumferences (Ch. Art. 29 and 32, VIII.). Again, from the point L equal straight lines can be drawn to the circumference BFCG only in pairs (Euc. 7, III. Ch. 20, I.). Consequently, equal straight Hues, and equal arcs of great circleSj can be drawn from A to the circumference, BFGG, only in pairs (Euc. 47, I. and 28, III. Ch. 4, VI. and 5, II.). 133. Let every point on the surface of the hemi- sphere be projected upon the plane of its base (Ch. Art. 66, VI.) ; that is, let the hemisphere itself be pro- jected upon the plane of its base. , Let the figure BEOQ represent the hemisphere thus projecjied, with the arcs also projected in the lines BPd,i>AH,AK, and AE. The arc BPC of the preceding figure will be pro- jected in the straight line BC. Therefore the straight line j5P<7will represent an arc on the surface of the hemisphere perpendicular, to the arc BKCQ. THEOREMS OF OBLIQUE-ANGLED TRIANGLES. I47 By this method of projection A OJT, A OB, A 02), and ABD, ABE, etc., are made to represent triangles right-angled at O and at B, respectively. 134. If arcs of great circles are drawn to the cir- cumference of the base of a hemisphere, from a point on the sv/rface not the poh of the hase, of the two adja- cent omgles made hy these arcs with the droiimference of the hose, the one on the side toward the longer per- pendicular is an obtuse angle, and the other angle on the side toward the shorter perpendieula/r is acute. Let the hemisphere be represented as projected on the plane of its base, and let AB, AD, AE, AK, be arcs drawn from A on the surface of the hemisphere (not the pole of the base) to the circumference BECG. Of these arcs let BAO be the one passing through the pole of the base. BAG will be perpendicular to the arc BEGG (Oh. Art. 58, YIII.). In the right-angled triangle ADB, AB i& > AD (Art. 132) ; 148 THE ELEMENTS 01" SPHERICAL TRIGONOMETRY. . ■ . ADB is > ABJ) (Ch. 26, VIII.), i. e., is > 90°, or is an obtuse angle. Again, in the right-angled triangle ADO, AC ia <^i?(Art. 132); . • . the angle ABC is < the angle ACD (Oh. 26, VIII.), i. e., < 90°, or is an acute angle. In a similar manner it may be proved that the an- gles AEB, AKB, are obtuse, and that the angles AEO and AKO are acute. 135. A. jperjpendioula/i' wpon cm a/rc, which coincides with a side of an oblique-angled spherical triangle, drawn yy-om dm cmgle opposite this side, falls withm the triangle, if the other two cmgles are both acute, or are both obtuse / but falls without if one of these angles is acute while the other is obtuse. Let HAK, HAE, etc., be triangles represented on the surface of a hemisphere which is projected on the plane of its base (Art. 133), Let these triangles have the common vertex A, not the pole of the base. Let THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 149 P be the pole of- the base. Let the great-circle arc, BAG, be drawn through P, and let it meet the circum- ference of the base in two points, B and O. The angles at B and (7 will be right angles (Oh. Art. ,58, VIII.). Suppose the triangle HAK to have the angles H and K acute ; also suppose the angles AEK, ABK, and A GH are acute. As the angle (7 is a right angle, A CM is greater than H; therefore AC i^ less than AH {Ch. 26, VIII.), and must be the shorter perpendicular (Art. 132), for in the same manner it may be proved less than AK or AE, or less than AB or AG; therefore it is the least arc from A, and it is perpendicular by hypothesis. Now, sls AC IB the shorter perpendicular, the acute an- gle H lies toward AC; for the same reason, the acute angle K lies toward AC (Art. 134) ; that is, Clies be- tween H and K, or the shorter perpendicular, A C, falls within the triangle AKH. In the same manner it may be shown to fall within the triangle HAE. 150 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. ^C having been shown to bathe shorter perpen- dicular, AB must be the longer perpendicular. As by hypothesis ADE is acute, ADB must be obtuse, and must lie toward the longer perpendicular (Art. 134) ; and, since AGH is acute, AGB must be obtuse, and must lie toward the longer perpendicular; therefore AB lies within the triangle -4Z>6'^, of which the angles AGB and ABG are obtuse. In the same manner it may be proved that AB lies within the triangle AEG, the angles ^6^^ and AEG being obtuse. Next let AEK^ie a triangle having the angle AEK acute, but the angle AKE obtuse ; then both perpen- diculars, ^(7 and AB, lie without the triangle. Since ^^^is acute, AC lies to the right of AE (Art. 134). Since AKE is obtuse, AKC must be acute, and AC lies to the right of AK; therefore AC, the shorter perpendicular, lies without the triangle. Again, since AKE is obtuse, AB lies to the left of AK. Since AEK is acute, AEB must be obtuse, and THEOEEMS OF OBLIQUE-ANGLED TRIANGLES. 151 A£ lies to the left oi A^; therefore the longer per- pendicular also lies without the triangle. 136. In a spherical triangle — 1. the sum of two an- gles is 180°, if the sum of the opposite shjes is 180° ; 2. is less than 180°, if the stm of the opposite sides is less than 180° ; 3. is greater than 180°, if the sum of the opposite sides is gkeatee than 180°. Suppose ABO to be a spherical triangle projected on the plane of the base of the hemisphere ABDB'. Let D'D be the projection of the arc drawn through C perpendicular to the arc AB produced. 1. Suppose (a + 5) = 180° ; then {A^B) = 180°. If a=.T), then a and h each = 90° ; and A and B each = 90° (Ch. Arts. 38 and 58, YIIL). .-. ^-1-^ = 180°. Let a and 5 be unequal. Then a and h must lie on the same side of DD'. For if J were on one side of DD' and a on the other side, occupying the positions CA and CB' respectively, 153 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. they would together form an arc, AGB', equal to a semicircle, and the figure ABB'G would be a lune and not a i/riangle (Ch. Arts. 32 and 90, VIII.). Therefore a and 5 both lie on the same side of J)D'. Produce ^ C to meet the circumference of the base of the hemisphere in B'. Then A CB' is a semicircumfer- ence, and is equal to 180° (Ch. Art. 32, VIII.). Then J + (75' = 180° = 5 + a ; .-. CB' = a=CB; and the angle OB'B = GBB' (Ch. 23, VIII.) = CAB (Ch. 16, VIII.); ■■ - OBB'= CAB ; add to each CBA ; CBB'+ CBA = CAB + CBA = A + B; but CBB'+ CBA = 180° ; therefore A-\-B = 180°. 2. Suppose a-^h < 180°; then A-^B< 180° . Then C cannot be the pole of the arc ABE\ for if THEOREMS OF OBLIQUE-ANGLED TlllANGLES. 153 Cwere the pole, a and 5 would eacli be 90° (Oh. Art 37, VIII.) ; that is, a-\-h would equal 180°, which is con- trary to the hypothesis. If a = 5, CD being the shorter perpendicular from G upon BB\ a and S are both greater than CD (Art. 132) ; therefore B and B< are each <; 90° (Art. 108), if CB' and CB are a and h ; therefore their sum is less than 180°. Suppose now a and 5 unequal, and a < 5. Pro- duce ^Cto meet the circumference of the base of the hemisphere at E. From C draw the arc CE'— CE. Then h+CE^ 180° (Oh. Art. 32, VIII.). .'.{a-\-l) 180° ; then also A-\-B> 180°. Now, if O were the pole of AB, a and h would each equal 90° (Oh. Art, 37, VIII.), and a-\-l would equal 180°, which is contrary to the hypothesis ; therefore is not the pole of AB. li a = h, a would lie on one side of 01)' and b on the other side (Art. 132). Now, if we take and O' on the circumference of the base, each at the distance of a quadrant from <7, since a and h (ii a — h) are each 156 THE ELEMENTS OF SPHERICAL TRIGONOMETBY. greater than 90°, a and 5 must eacli meet the circum- ference OD'O' to the left of and 0' (Art. 132) ; there- fore in this case the angles A and B are each obtuse (Art. 134), and their sum is greater than 180°, since OD' is the longer perpendicular from (7.* If a and l are unequal, then one of them, as a, might be below CD', and the other, as h, might occupy one of two positions (Art. 132), as CA or CA', one below CD' and the other above CD'. If a be below CD', and h above CD', then A'CB wiU be the triangle to be considered. CA'B and CBA' are each obtuse in this case (Art. 184), and there- fore CA'B-\- CBA' > 180°. * This may be provfid by a method similar to that employed in the note for 2 ; or ji and B may be proved obtuse directly, thus : 0D'> CS.-.B>D' (Ch. 26, VIII.) ; i. e., S>90°; in the same way the equal angle is > 90° ; .-. A + B> 180°. THEOREMS OF OBLIQUE-ANGLED TRIANGLES. I57 If a and 5 are both below CD\ tben CAB is tbe triangle to be considered. Produce ^Cand BO to meet the circumference of the base of the hemisphere again at -S"and K. Then A(JR and BGK are semicircumferences (Oh. Art. 32, VIII.), and each equals 180°. . • . «+ CK= 180° and 5 -{-CH= 180°, or ffl + J + CB:-\- CK= 360°, but a + J > 180°. .-. CH^CK<\m°; therefore by the 2d case (1) CHK^CKH 180°. 137. The converse of the preceding article is true, that is, the sttm of two sides of a spherical triangle is 158 THE ELEMENTS OF SPHERICAL TRIGONOMETET. 180°, if the SUM of the opposite angles is 180° ; is less than 180°, if the sum of the opposite angles is less than 180° ; is GEEATEK than 180°, if the sum of the opposite ANGLES is greater than 180°. Let two angles of a spherical triangle be denoted by A and B, and the sides opposite them by a and S respectively. 1. If ^ + 5 = 180°, a + J = 180°. Let A! and B' be the corresponding angles of the polar triangle, and al and V the sides opposite these. 4' Then A = 180°-a', B = 180°-5', A' = 180°-a, B' = 180°-5 (Ch. 18, YIII.) ; - • . substituting for A and B their equivalents in 1, 360° -(a' + JO = 180° ; or a' + &' = 180° ; consequently A' -\- B' = 180° (Art. 136). Substituting values for A' and B', 360°-(a + h) = 180°, or a + 5 = 180°. %liA-\-B< 180°, a + 5 < 180° ; ^ for, passing to the polar triangle, 360°-(a' + 5')<180°, or a' + J' > 180° ; consequently A'-\-B'> 180° (Art. 136) ; that is, 360°-(a + J) > 180°, or a + 5 < 180°. THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 159 3. liA-{-B> 180°, a + i > 180° ; for, passing to the polar triangle, 360°-(ffl' + 5') > 180°, or a' + 5' < 180° ; consequently A'^B' < 180° (Art. 136) ; that is, 360°-(a-|- 5) < 180°, or a + 5 > 180°. 1 88. The sines of the sides of a spherical tria/ngle m-e proportional to the sines of the opposite angles. Thus let ABC be kny spherical triangle, of which the angles are A, B, and C, and the sides opposite them are a, I, and c, respectively. Then sin. a _ sin, A sin. i sin. B From C let the arc CD be drawn perpendicular to the side c, or (in the right-hand figure) to o pro- duced, meeting it in the point I). Then ABGia equal to the sum or to the diiference of two right-angled tri- angles. In the triangle A CD, taking CD as a middle part and 90°— J, 90°— CAB as opposite parts, sin. C2> = sin. i sin. CAB (Art. 115) = sin. h sin. A (Art. 46). , In the triangle BCB, also taking CB as a middle part and 90°— a, 90°— -S as opposite parts, 160 THE ELEMENTS OP SPHERICAL TKIGONOMETKT. sin. CD = sin. a sin. B\ . • . sin. a sin. ^ = sin. 5 sin. A, or , , sin. a sin. A {a) In a similar manner it can be proved ,,, sin. a sin. A (*) -• = -• — n- sm. c sm. 6 From (a) by alternation we have sin. A _ sin. B sin. a sin. 5 From (J) by alternation sin. ^ sin. C Bin. a , , sin. A (c) — — sm. a sm. c sin. B sin. J ; therefore sin. G sm. c If the perpendicular coincided with the eide a, B would be- a right angle, and we should have from {a) sin. a = sin. b sin. A (Art. 36) an equation already established (Art. 109). 139. Suppose a to be > h, then A\6> B (Oh. 26, VIII.). TUEOKEMS OP OBLIQUE-ANGLED TRIANGLES, sin. a sin. J. 161 sin. 5 sin. B sin, a -\- sin, h _ sin. A -\- sin. B sin. A (Art. 138); ■ J. ■ A ■ „(Cli. Art. 10,111); sm. a— sm. 6 sin. J.— sm. B ' /' tan.i(^ + 5) _ t an.i(^+^) tan. i («-5) - tan. J (^-^) ^^''^ Art. 71). 140. If an arc of a great circle be drawn from one of the angles of a spherical triangle perpendicular to the opposite side, or to the opposite side, produced, then (1) The smES of the segments of the sdde, on which the perpendicular falls, will he proportional to the nu- MEKICAl cotangents of the ADJACENT ANGLES. Thus, in the triangle ABG, if the perpendicular CD is drawn from G to c, or o produced, sin. AD _ c ot. A sin. BD ~ cot. B' For in the triangle A CD, taking AD as a middle part and 90°— CAD, CD, as adjacent parts, (a) sin. AD ^ cot. CAD tan. CD (Art. 115) ; and in triangle BCD, taking BD as a middle part and CD, 90°— B, as adjacent parts, (5) sin. BD = cot. 5 tan. CD; dividing equation (a) by equation (J), 162 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. («) sin. A£> cot. OAB cot. A (Art. 46), sixx. BD cot. B cot. B cot. A being the numerical value of cot. A. If the sign as well as the numerical value of cot. A be taken into account, equation (c) for the left-hand figure, or in the case in which the perpendicular falls within the triangle, would be unaltered ; but for the right-hand figure, or in the case in which the perpen- dicular falls without, (c) would become , 7. sin. AD —cot. A ^ ' sin. BD ~ cot. B ' (2) The COSINES o/" the segments of the side will ie proportional to the cosines of the sides adjacent to them. In triangle ACB, taking 90°— & as a middle part, and CD, AD as opposite parts, cos. AD COS. CD = cos. b (Art. 115) ; and in triangle BOD, taking 90°— a as a middle part, and BD, CD as opposite parts, cos. BD cos. CD = cos. a ; dividing the first of these equations by the second, / s COS. AD cos. i (e) = . COS. BD COS. a THEOREMS Or OBLIQUE-ANGLED TRIANGLES. 163 141. Let ABC be a spherical triangle, and from C let an arc of a great circle, GD, be drawn perpen- dicular to the opposite side. Let the side a be >5, then is A >B (Ch. 26, YIIL). De- note the segments BD and AD by « and y, respectively. 1. Let the perpendicular fall within the triangle. Then, tan. \{x-y) = '!°-^f~-f j tan. i c. For ^ = £2^ ((1) Art. 140) sm. X cot. B = *;^((/)Art.64); tan. A ^l^-'^ + ^^^-y = tan. ^ + tan, .g ^^^ sm. 33— sin. y tan. J.— tan. 5 sin. A |_ sin. ^ cos. J. cos. B ..,. . , ^,, ■((5) Art. 64); sin. A sin. B cos. ^ cos. jB therefore t^P-iC'^ + y) ^ Bin.(^+^) ^^^_ tan. I (cc-y) sin. (^-^) (a) Art. 67, (a) Art. 68) ; or tan. 4- (x—y) = — ^-^^ '- tan. i c,* since » -j- V = c. ^^ ^'^ sin.(^+^) ^ ' ^^ *If a is >S and a and S are each >90°, A and jB are also >90° (Art. 136). Then y is >a; (Art. 132), and the equation above will be sin. (B-A) tan. i (y-x) - - — tan. i c. sm. (5 +.4) 164 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. The last equation may be written {a) tan. i {x-y) = «ii^- H^-g)eoB.i(^-^) ,^^^ , ^ ^^ ^^ ^^ sin. J(^ + ^)cos4(^+^) ^ ((a) Art. 73). 2. Let the perpendicular fall without the triangle. BD and AD, denoted by x and y, are external segments. A 'c—^£ Then tan. i(x4-y) — — ^-r J- tan. A c. ^^ ^^^ sin. (^ + ^) ^ For $^ = =^ ((^) Art. 140) sin. X cot. ^ cot. CAD , . . ,o\ • tan. ^ ,, _^, . , ... (Art. 46) = ((/) Art. 64) ; cot. ^ tan. <7J.X> sin, a; + sin, y ^ tan. gAZ> + tan. ^ ™ ^ ^^^^ ^^ jjj, sin. a?— sin. y tan. CAD— tan. B sin. 'CL4Z> . sin. B cos. (7J.-0 ' COS. B {{h) Art. 64) ; sin. CAD _ sin. B COS. C4Z) COS. B But .B ia <^ (Ch. 26, VIII.), therefore B—A is a negative quantity, and sin. (£-^)=-sin^ (A-B) (Art. 95). Also tan. ^ (y — x) — — tan. \ (x—y) (Art. 96). Substituting these values our new equation becomes like the equar tion above — ^ sin. (A—B) . tan. i {x—y) = - — tan. i c. sin. (A+B) THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 165 therefore ^"°- * (^+y) ^ ?El(^4^+^((«) Art. 11), tan. J (a!-y) sin. {OAD-B) ^^ ' " (a) Art. 67, (a) Art. 68). Now, CAD = 180°— J., and Bubstituting the value of CAD in the right-hand member of the last equation, tan. ^ (x+y) _ sin. {180°-(J.-:^)} tan, I (x-y) ~ sin. {180° -{A+B)} t&n.^{x + y) ^ sin. (A-B) ~sin. {A + Bi sin. (A-B) , or smee x—y=G, tan. c. tan. A c sin. (J-+^) The last equation may be written (5) tan. i fe+2/) = BJ^.^ (^-^) cos, j (^-^) ((a) Art. Y3). + ■ 142. From the angle, 6', of a spherical triangle, let an arc of a great circle, CD, be drawn perpendicular to the opposite side or opposite side produced. Suppose a>l, then is J. > ^ (Oh. 26, YIII.). Denote the segments of the base by x and y, re- spectively. 166 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. Then tan. ^{a + l) tan. J (a— 5) = tan. ^ {x+y) tan. I {x—y). COS. i COS. a H^^ ((2) Art. 140), COS. X COS. 5— COS. : COB- y-COB. X ^^^ ^^ ^0^ jjj^^ ^^ COS. t + COS. fl! COS. y + cos. i tan. ^ (a + J) tan. -^ (a— 5) = tan. ^{x + y) tan. J {x—y) ((5) Art. n, (/) Art. ©4). In the left-hand figure x-\-y = c; in the right-hand figure x—y = c ; therefore (a) when the perpendicular falls within the triangle tan. ^{a + i) tan. i (a—h) = tan. ^ c tan. J (x—y) ; * (5) when the perpendicular falls without the triangle, tan. ^{a + i) tan. ^ {a—h) = tan. J (a? -f y) tan. ^ c. * When a is >b and a and b are each >90°, y is >a; (Art. 132), and equation (a) above would become tan. i{a + b) tan. ^ (i— a) = tan. -J c tan. -J (y— k). But -J (6 — 0) is a negative quantity, tan. i (J— a) = — tan. i (a— 6) (Art. 96) ; also tan. i (y — i) = — tan. -J (x—y). Substituting these values our new equation becomes like the equa- tion above — tan. ifa + b) tan. i (a—b) = tan. i c tan. jg (x—y). THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 167 143. In a spherical triangle, the tangent of half the SUM of TWO SIDES equals the eatio of the cosine of HALF the DIFFEEBNCE tO the COSINE of HALF the SUM of the OPPOSITE ANGLES, multiplied hy the tangent of half the THBBD SIDE ; (md the tangent of half the diffee- ENCE of TWO SIDES equols the EATIO of the sine of half the diffeeence to the sine of half the sum of the op- posite angles, multvpUed by the tangent of half the THIED SIDE. Thus, a, h, and c denoting the sides of a spherical triangle, and A, B, and C denoting the angles respect- ively opposite these, a also being greater than 5, cos. \ (A-B) (a) tan. ^{a-\-h) (i) tan. ^ (a—h) ■ cos. ^{A-\-B) sin. i (A—B) tan. A c. s' tan. sin. J(^+^) 1. When a perpendicular, drawn to the third side from the opposite angle, falls within the triangle, i. e., when A and B are both acute or both obtuse (Art. 135). Denote the segments BD and AD by x and y, re- spectively. 168 THE ELEMENTS OP SPHERICAL TRIGONOMETRY. , tan, i (ffl 4- 5) ^ tan. ^ jA+B) ,j^^ ^g^, ^ ^ tan. J (a-h) tan. ^ {A-£) (2) ta,n.ir(x—v)— — ^ ^-7-; =rtan. + c {{a) Art. 141). t an. ^ {a+h) tan. ^ (a— 5) tan. ^ {x-y) Multiply equations (1), (2) and (3) together, mem- ber by member," cancel like terms in numerator and (3) : tan. I- c ((a) Art. 142). denominator of the resulting fractions, and extract the square root of each member of the final equation, and (4) tan.i(a+5): cos.^(.A-^) tan. cos,^(J.-|-i?) Now equation (1) may be written tan. ^ {a—h) _ tan. \ {A—B) tan. ^ {a+l) ~ tan. J {A-\-BJ Multiply the equation in this form by equation (4) and (5) tan. i ia-h) = 'l^'t^^T^^ tan. J o. sm. ^ (-4+^) 2. When the perpendicular falls without the trian- gle, i. e., when A is obtuse and B acute. THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 169 The proof is the same except that in equations (2) and (3) we have tan. ^ (aj+y) in place of tan. J {x—y) ((5) Art. 141, and (5) Art. 142). 144. In a spherical triangle, tJie tangent of half the SUM of two ANGLES equoLs the eatio of the cosine of HALF the DIFFEEENCE tO the COSINE of HALF the SUM of the OPPOSITE SIDES, multvplied hy the cotangent of HALF the THiED ANGLE; and the tangent of HALF the DIFFEEENCE of TWO ANGLES eqUOls the EATIO of the SINE of HALF the DIFFEEENCE tO the SINE of HALF the SUM of the OPPOSITE SIDES, multiplied Iry the cotangent of half the THIED ANGLE. Thus, in the triangle ABC, a being >h, (a) tan.i(^+^) = ''''"■ \^^~^l cot. i 0. COS. f \a-\-o) (b) tan. H^-B) = ?i^iifc|.) cot. * O. ^^ ^^ ' sin.|(a+5) Let A'B'C be the polar triangle of ABG. Since by hypothesis » is > ^, 180°— a is <180°-5; but J.'=:180°-ffl, and ^'=180°-5 (Ch. 18, VIII.); 8 IT'O THE ELEMENTS OF SPHERICAL TRIGONOMETRY. therefore -4' is ((5) Art. 6T) sm. sm. G o . (ffl+J+c) . (5+0 -a) 2 sm. i — ! — —^ sm. ^^-! ' 2 2 = r-7^ {{d) Art. 70) ; sm. sm. c ' substituting values of and , ^/.^ 1 . . 2 sin. s sin. (s— ffl) (./) 1+cos. A = ^ — —r^ '-. sm. sm. c Divide equation (e) by equation (/) and extract the square root of both members. / 1 — cos. A _ /sin. (s— 5) sin. (s— c). r 1+cos. J. r sin. s sin. (s— a) / or, since tan. \ A=^ ./ -^ — '- — ((c) Art. 75), r 1+cos. J. tan. iA= A°-(^-^)Bi"- 5IZ). T sin. » sin. («— ffl) In a similar manner it can be proved /: in /sin. (s— c) sin. (s— a) tan. \ B= ./ ^^ ^^ ; r sin. s sin. (s—b\ 176 THE ELEMENTS OF SPHERICAL TRIGONOMETRY tan, J p__ /sin. {s—a) sin. (s— J) sin. 5 sin. {s — c) 147. To find an expression for the cosine of a sroe m TEKMS of the angles of a spherical triangle. Let ABC be a spherical triangle, A, B, O being the angles, a, i, c being the sides opposite these. Let A'B' C be the polar triangle oiABC, A', B', C being its angles and a', V, & being the sides opposite these. /i\ M COS. a'— cos. V cos. o' ,, . . , ^ .-. (1) COS. A = ; ((a) Art. 145). sin. h' sin. o' A'=im°-a; a'=nO°-A', h'=180°-B; c'=180°-(7(Ch. Arts. 69 and 70,Yin.); substituting these values in (1) -COS. ^ — COS. B COS. O , . . .„. (Art. 46), or — COS. a : sin. B sin. O , V COS. A-\-Q.o%. B COS. O (a) cos. a = ! . sin. B sin. C. In the same manner it may be proved ,-,. V cos. ^+cos. G cos. A {h) cos. J = P—^- — ; sm. 6 sm. A THEOREMS OF OBLIQUE-ANGLED TBIANGLES. 177 COS. 0+ COS. A cos.B (c) COB. C ■■ sin. A sin. £ 148. To Jmd an egression for th£ tangent of one HALF a SroE, IN TERMS of THE ANGLES of ffl Sphsricol trimigle. Let A, B, C denote the angles of tlie spherical trian- gle ; let a, h, o denote the sides opposite these respect- ively, and let S denote — — — ^^ ; then / x . 1 / — cos. /S'cos. (/iS'— J.) (a) tan. * « = . / ^^ H- ' V (cos. S- B) COS. {8- C) r (5) tan. i 5 = V -cos- >y cos. (^-^) ^' V COS. {S-C) COS. (S- A) (c) tan. i c = / -cos.^cos.(^-^)_ r cos. (aS'-^) cos. (-S*-^) , ,x T^ cos. J.+COS. 5 cos. C {a) Jb or COS. a = ! — — . sin. B sin. C Subtract each member of this equation from 1. 1 — {cos. J.-fcos. (^-|-(7)J /^Ts . , „-s 1— cos. a = ■ — ' ^^ — ! — '- ((b) Art. 67) sin. ^ sin. C ^^ ^ ' —2 COS. — ■ ■ — cos. — ' 2 ! ((c) Art. TO). sin. B sin. C. Now, let,(l),AS'= ^^^ ; (2), -^^ = S-A; 178 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. (3), 4±^b5 = S-B ; (4), 4±^IZ^ = s- a Substituting values, from (1) and (2), in the last equation .V . — 2cos. xS'cos. (/S'— ^) (e) 1-cos. a = . J. . r> ■ sin. Js sm. 6 Again, adding 1 to both members of equation (d) c c~ 1+cos. a — . ' / ^ — I (b) Art. 68) s]n. B sm. C „ A+B-C A+O-B 2 cos. —z COS. — ■ 2 sin. B sin. C 2 COS. (S-B) COS. (S-C) or. sin. B sin. C (/) 1+cos. a (see (3) and (4)). Divide (e) by (/") and extract the square root. A -cos. a /■ 1+cos. a V' —COS. S cos.{/S—A) COS. {S-B) COS. {S-C)'^'^' r tan. i a = V ^^^f -^^i^^. ((«) Art. 75). y COS. ip—B) COS. (xs— 17) In a similar manner it can be proved THEOREMS OF OBLIQUE-ANGLED TBL^^GLES. 179 r tan. i I = / -COS. ^ COS. (^-^) r ca%A8-C\co%.{8-A COS. {S- C) COS. (/S-^) ,1 / — COS. /S'cos. (>S'— C") tan. * c = . / '^ '-- V coB.{S-A)cos.{S-£) 149. In order that we may find real values for tan. i a, tan. J h, and tan. ^ c, the quantity under the radical must be positive. This quantity will be posi- A' tive if cos. /S" is negative (that is, if —cos. 8 is positive), and cos. (S—A), cos. (S—B), cos. (8— G) are all posi- tive. N'ow, 2 8 16 >180° and <540° (Ch.29, VIII.); therefore aS is >90° and <2Y0°; consequently cos. 8 is negative (Art. 88). Again, cos. (8— A), cos. {S—£), cos. {8— C) are all positive. For, let A'B'C be the polar triangle of ABC. Then a'= 180°-^; S'=180°-^; c'=180°-C (Ch. Arts. 69 and YO, VIII.). But a' <{h'+6') (Ch. 25, VIII.), or, (180°-^)< {360°-(^+^)K 180 THE ELEMENTS OF SPHERICAL TBI60N0METRT. ...^±^=^^ '-. \ sm. c sm. a , jf. , ri /sin. 5 sin. (s— c) (/) COS. \C=J —, ^-— ^ r sm. a sm. 6 IK-. < — 2cos.xS'cos.("^-(-^-^)"^^-('^TA); r sin. Csin.^ / (/) COS. i C = ,/ COS. (^-^) COS. (^ -^)^ • r sin. ^ sin. _2 162. Gauss's Equations. In a spherical triangle , , sin. \ {A.-\-B) COS. \ (a—h) \aj ! COS. ^ C COS. ^ COS. \ {A-\-B) _ COS. i (a+5) . (5) sin. I- G COS. i c , s sin. \{A—B) _ sin. |- (a— 5) . COS. f (7 sin. \ c ,^ cos.i(.A— ^) _ sin, i (a+h) sin. i G sin. -^ c. T^ sin. J. sin. ffl ,, x . . ^„o\ For --— - = -^— {{a) Art. 138), sm. B sm. o sin.J.+sin.^ sin. a+sin. 5 ,™ . ^ ^„ ttxv -. -. = : (Oh. Art. 10, III.). sm. A sm. a THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 183 2 Bin, i (A+JB) COB, j (A~B) sin. A BUT. a "^ By alternation, sin, i (A+B) COS. j (J.-^) _ ein.^ _ sin. (7 sin.^(a-|-J) COS. ^(a— J) ~~ sin. a ~ sin.c ((c) Art. 138) = ™-*<^coB.i(7 sin. ^ c COB. ^ c {{a) Art. 73) ; therefore, again by alternation, ,-^s sin, i (A+B) COS. j (^ -B) Bin. -J C COS. i C _ Bin, f (a+h) cos. ^ (a— &) sin. ^c COS. I^c COS. i (^+^) ^ tan.i. ^^_ ^ COS. i (A-B) tan. ^ (a+b) ^^ ' ' tan. i (^+^) ^ COS. i {a-l) ^ ' cot. i C COS. i (a+5) ^^ ' ' Multiply together equations (1), (2), and (3), member by member, cancel like terms in numerator and denomi- nator, and extract the square root of the result, and . ,, sin. * {A-\-B^ cos. \ (a—V) , . • / x\ (4) ^-i — '—-^ = -^ '- (equation (a)). COS. \ O cos. -I c Divide equation (1) by equation (4), and coB.H^-^) ^ 8inJ>+5) sin. ^ O sm. ^ c Multiply equation (2) by equation (5). 184 THE ELEMENTS OF SPHEBICAL TRIGONOMETBY. ^g^ C OS. i {A+£) ^ COS. i (a+h) ^^^^^^.^^ ^^^^ sin. ^ O COS. i G Again, tan.i(A-^) ^ smJ^«-5) ^^^^ ^^_ ^^^_ cot. ^ C sin. 1^ (b+J) Multiply this last equation by equation (5), and ^^^ sin.i(A-^) ^ Bin, i («-5) ^^^^^^.^^ ^^^^_ cos. ^ C sin. ^ c OHAPTEE XY. SOLTTTION OF OBLIQUE-AN&LED TKIANGLE8. Aet. 153. Of the six parts of a spherical triangle (the three sides and the three angles), it is necessary to know three in order to sol/oe the triangle. To solve a right-angled triangle three parts were given, viz., the right angle and two other parts. (Arts. 116 and 119.) To solve a quadrantal triangle three parts were given, viz., the quad- rant and two other parts. (Arts. 129 and 131.) 154. There are six cases of the solution of oblique- angled spherical triangles, which may be classified under four heads, as the known parts are, in (1) Case I. — The three sides ; (2) Case II. — The three angles ; (3) Two sides and an angle, under which head are Case III. — Two sides and an included amgle, and Case IV. — Two sides and an angle opposite one of these sides ; (4) Two angles and a side, under which head are Case Y. — Two ambles and an included side, and Case YI. — Two angles and a side opposite one of these angles. 155. Case I. — The three sides of an oblique-angled spherical triangle iei/ng hnown, to solve the tria/ngle. Use formulas {a), (h), and (c). Art. 146. For check on the work use (c) Art. 138. 186 THE ELEMENTS OF SPHERICAL TRIGONOMETEY. Example. Suppose ABG to be a sphericid triangle, of which a=60°, 6=40°, c=76°. a - 50° 6 = 40° e = 76° _ a + 6 + e _166° *~ 2 ■" 2 s—a— 33° s-b- 43° s—c = 1° :83° r sin sin. s sin. (s — a) ^l/| sin. 43° Bin. 7° sin. 83° sin. 33° Tan. i B -. Log. sin. 43° = 9.833783 Log. sin. 7° = 9.085894 Ar. CO. log. sin. 83° = 0.003249 Ar. CO. log. sin. 33° = 0.263891 2 I 19.186817 Log. tan. 21° 24' 38.47"= 9.6934086 . . ^ = 42° 49' 16.94". . /sin. (s—c) sin. (s—a) r sin. s sin. (s— J) =i/W r em sin. 7° sin. sin. 83° sin. 43° Log. sin. 7° = 9.085894 Log. sin. 33° = 9.736109 Ar. CO. log. sin. 83° = 0.003249 Ar. CO. log. sin. 43° = 0.166217 2 I 18.991469 Log. tan. 17° 23' 14.18"= 9.4967346 .-. .5= 34° 46' 28.36". Tan. iC = i/ "°- ('-") ^'°- ('-'>) V sin. s sin. («— c) ' sin. 33° sin. 43° sin. 83° sin. 7° Log. sin. 33° = 9.736109 Log. sin. 43° = 9.833783 Ar. CO. log. sin. 83° = 0.003249 Ar. CO. log, sin. 7° = 0.914106 2 I 20.487247 Log. tan. 60° 17' 18.23" = 10.2436286 .-. (7=120° 34' 36.46". SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 187 , _, . , sin. A sin. B sin. O (Check.) —. = -: — r = — : sin. a sin. o sm. c Log. sin. 42° 49' 16.94" = 9.83232'? Log. sin. 34° 46' 28.36" == 9.756140 Log. sin. 60° = 9.884254 Log. sin. 40° = 9.808067 1.948073 1.948073 Log. Bin. 120° 34' 36.46° = 9.934977 Log. sin. 76° = 9.986904 1.948073 Find the angles of a triangle when the sides are : Example 1. 75°, 100°, 65°. Ans. 68° 8' 64.4", 108° 61' 46.2", 60° 33' 33.3". 2. 67°, 83°, 114°. Ans. 49° 3' 20", 63° 22' 18.3", 124° 38' 8.6". 3. 70°, 140°, 80°. Ans. 41° 22' 18.7°, 163° 7' 14.7°, 43° 60' 32.1". 4. 120°, 80°, 60°. 6. 100°, 69°, 61°. 156. Case II. — The three angles of an oblique-an- gled spherical triangle hemg Tmmjon, to sol/oe the triangle. Use formulas (a), (5), and (c) of Art. 148. For check on work use (c) Art. 138. Example. Suppose the angles A, B, and C of a spherical triangle to be 130°, 60°, and 74° respectively, required the sides. Tan.ia=i/.-''°^-^'^''«-(^-^) r c A = 130° B = 60° = 74° s- _A + B + 2 264° _ 2 ^- -A = 2° S- -B = 72° S- -C = 68° =/ cos.(,S-£)cos.(;8'-(7) — COS. 132° cos. 2° cos. 72° COS. 58° Log. (— COS. 132°) = 9.826511 Log. COS. 2° = 9.999735 Ar. CO. log. COS. 72° = 0.510018 Ar. CO. log. COS. 68° = 0.275790 2 I 20.611064 Log. tan. 63° 40' 17.56"=10.305527 .-. a = 127° 20' 36.1". 188 THE ELEMENTS OP SPHERICAL TRIGONOMETKT. r cos.(5-C)cos.(,8'-^) y coa.{S-A)cos.{S—B) _ / — C03."l32° COS. 72^ _ / - COS. 132° coa. 68° y COS. 58° COS. 2° y COS. 2° COS. 72° Log. (- COS. 132°) = 9.825511 Log. (— cos. 132°) = 9.825511 Log. cos. 72° = 9.489982 Log. cos. 58° — 9.724210 Ar. CO. log. COS. 58° = 0.275790 Ar. co. log. cos. 2° = 0.000265 Ar. CO. log. COS. 2° = 0.000266 Ar. co. log. cos. 72° = 0.510018 2 I 19.591548 2 | 20.060004 Log. tan. 31° 69' 66.81"= 9.795774 Log. tan. 46° 58' 39.05''=10.030002 . • . 6 = 63° 59' 63.62". . • . c = 93° 57' 18.1". ^Ch.ck.) sm^^sin^^sin^ sm. a sm. o sin. c Log. sin. 130° — 9.884254 Log. sin. 60° = 9.937531 Log. sin. 127° 20' 38.1" = 9.900377 Log. sin. 63° 59' 53.6" = 9.963654 1.983877 1.983877 Log. sin. 74° — 9.982842 Log. sin. 93° 67' 18.1" = 9.998966 1.983877 Find tlie sides of a triangle when the angles are : Example 1. 64°, 75°, 117°. Ans. 68° 6' 36.4', 88° 43' 1.4", 113° 6' 38.7" 2. 136°, 72°, 102°. Ans. 147° 23' 9.2", 47° 33' 14.4", 130° 37' 46.8". S. 128°, 75°, 100°. 4. 143° 3', 119° 12', 110° 36". Ans. 140° 10' 43^", 111° 34' 17.4". 86° 48' 47.9". 5. 129' 8' 20", 106° 8', U2° 12' 40". 157. Case III. — Two sides and an included angle of a spherical triangle ieing known, to solve the triangle. To find the angles : first, find the half sum and the half difference of the angles, opposite the given sides, by Art. 144 ; then add these quantities for the greater angle, and subtract the half difference from the half sum for the less angle. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 189 To find the third side use {a) or (J) of Art. 143, or use Art. 138. If the third side is found by Art. 143, for check on work use (c) Art. 138, otherwise use a formula of Art. 143. Remabe. — When i (a— 6), and consequently -J (A—S), is a very small ' quantity, 1 c is better found by (a) of Art. 143 than by (6) of same article, since the cosines of small angles differ from one another less than the sines (as will be seen from an inspection of the tables), and the effect of a small error in .4 or ^ would be greater if sin. ^ {A—B) were used rather than cos. \ (A—S). On the other hand, if 1 (a+^p), and consequently \ (A + B), is near 90°, to find 1 e (6) Art. 143 should be preferred, since near 90° the sines differ less than the cosines. Example. Suppose ABC to be a spherical triangle, of which the side a=76°, 6=56° 10', and the angle C=126°. a= 1&° b= 65° 10' p \{a + b) = 131° 10' 2 z 65° 36' ; :^ ^^ *(«-*) = 20° 60' 10° 2S'; ^^ 2 e 10 = : 62° 30'. tan .i{A + B) COS. COS. cot, l(« + i) ■ iC; cos. 10° 26' cot. 62° 30' COS. 66° 35' Log. COS. 10° 26' =9.992783 Log. cot. 62° 30'=9.716477 Ar. CO. log. COS. 66° 36'=0.383662 Log. tan. 51° 4' 59.77''=10.092922 sin. i (a—b) tan. i {A-B) = ^ ' cot. 1 C; sin. l(a+6) sin. 10° 25' = cot. 62° 30' ; sin. 65° 36' 190 THE ELEMENTS OF SPHERICAL TRIGONOMETRT, Log. sin. 10° SB' = 9.25Y211 Log. cot. 62° 30' = S.IUill Ai. CO. log. sin. 65° SB' = 0.040690 * Log. tan. 5° 54' 5.34" = 9.014378 i (A+B) = 61° 4' 69.7'7" i {A-B) = 6° 54' 5.34" A = 56° 59' 5.11" 5=45° 10' 64.43" COS. i< A + B) tan. ic = tan. i (a+b); COB. i{A-B) cos. 51° 4' 59.7'7" = tan. 65° 35'. cos. 5° 54' 5.34" Log. cos. 51° 4' Bg.'?'?" = 9.798092 Log. tan. 66° 35' = 10.342972 Ar. CO. log. cos. 5° 54' 6.34" = 0.002308 (Cheek) Log. tan. 54° 17' 24J" 10.143372 . • . c = 108° 34' 49^' sin. A sin. B sin. sin. a sin. b sin c Log. sin. 56° 59' 6.11" = 9.923616 Log. sin. 76° = 9.986904 1.936612 Log. sin. 45° 10' 54.43" = 9.860868 Log. sin. 125° = 9.913366 Log. sin. 66° 10' = 9.914246 Log. sin. 108° 34' 49j" = 9.976763 1.936612 1.986612 Solve a spherical triangle when there are given the two sides and the included angle : Example 1. Sides 112° 30', 60° 16' ; angle 36° 40'. Am. Angles 142° 36' 17.7", 84° 47' 58.1" ; side 62° 29' 67.4". 2. Sides 77° 41', 64° 16' ; angle 122° 13'. Ans. Angles 69° 69' 36.7", 46° 0' 49.2" ; side 107° 21' 2.7", 3. Sides 140°, 113° 22' ; angle 110° 16'. Ans. Angles 142° 41' 58.2", 120° 3' 69.9" ; side 84° 17' 40.3" 4. Sides 106°, 66° ; angle 40°. SOLUTION OF OBLIQUE-ANGLED TEUNGLES. 191 158. Case IV. — Two sides and cm angle (Opposite one of these sides being hnown, to solve the triangle. The angle opposite the other given side can be found by means of Art. 138. The third side may be found by either of the for- mulas of Art. 143. Formula {a) should be preferred if I" (ffl— 6) is a small quantity, but formula (b) if ^ (a+5) is near 90° (see remark under preceding Art.). The third angle may be found by either of the for- mulas of Art. 144. Formula (a) is to be preferred if ^ (a—h) is a small quantity, but formula (5) if ^ {a-\-h) is near 90° (on the principle given in remark of preceding Art.). As check use Art. 138. As the first required part is found from its sine, and as the sine of an acute angle and the sine of its supple- ment are the same, both in value and ia sign, the part found may be either less than 90° or greater than 90° (Art. 46). In some examples, therefore, there may be two tri- angles having the given parts, and in such examples we shall obtain two solutions, each of which will be correct. This will be evident from the following figure. Suppose the given parts to be a, h, and the angle A. Let the triangle be projected upon the plane of the base of the hemisphere, of which the circumference contains the side <; as a part of it. Let CD be the pro- jection of the perpendicular from O (to the arc ABD). From Ctwo arcs can be drawn to the circuriiference ABB', each equal to a, x)ne on each side of CD (Art. 132). Thus the triangles ABO and AB'G each con- tain the angle A, the side b, and the side a. 192 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. There will not always be two solutions under this case. For instance, if a +5= 180°, and A is given, B will be known, since ^+^=180° (Art. 136). 159. Two sides cmd cm angle opposite one of these heing given, to deternmoe whether the pb; then there will be one solution. For suppose A to be >90°, then £ must be <90° (Art. 136) ; but suppose J. to be < 90°, then B must be <90°(Ch. 26, VIIL). (h) Let a be < 5, and -d < 90 ; there will be two solutions, since £ might have two vali;es, one < 90 and greater than A, or another >90°, and therefore >A (Oh. 26, VIIL), and A+B would still be <180° (Art. 136). Thus, suppose a+6<180°, and .4=36°, £ might be 60° or 130°, as with either value A is <,B, and ^ + .B<180°. When a + 6<180°, and a is <5, and A is >90°, no triangle can be formed, for £ would also be >90 (Ch. 26, VIIL), and A+£ would be >180°, which is impossible (Art. 136). The first and second classes of examples may be illustrated by the accompanying figure, which represents the surfae* of a hemisphere pro- jected upon the plane of its base (Art. 138). CAB represents a spherical triangle. CD represents the perpendicular from C to the circumference 9 194 THE ELEMENTS OP SPHERICAL TRIGONOMETRY. of the base of the hemisphere. AGE represents a semicircumference (Oh. Art. 82, VIII.). CE and CE' represent two equal arcs. CB is an arc of a great circle, drawn from (7, equal to CB (Art. 132). 1. a + ft=180°=^C+OA 2. (a)a+6<180°; (a> S, ^ (==£■)> 90°)=^ C+ OS (a> h,A(=E) <90°)=^C+ CM'. (i) + 6 <180°, a<6 and A<^V=EG+ GA, otE'G+GA. 3. Let the sum of the given sides be greater than 180°. (a) Suppose a+i > 180° and a is 90°, B must be >90°, and so there will be but one solution ; for If A is <90°, A+B >180° (Art. 136), and there- fore J3 must be >90°, and can have 'but one value, and If A is >90°, ^ is >^ (Ch. 26, YIII.), and there- fore jS must be > 90°, and can have but one value. (5) If, now, a is >J and A >90°,B may be either < 90° or >90°, for, with either value of B, A+B might be >180°, and ^>^ (Oh. 26, VIII.), so that in this case there would be two solutions. Thus, suppose a+i>180°, a>6, and A— 110° ; B might be 75° or 106°. With either value of B, A is >B,A + B> 180°. When a + 6>180° and a is >6, and jl<90°, no triangle is formed; for^mustbe <,A,i.e., <90°, and ^ + 5 would be < 180°, which would be impossible. The last case, (3), when a+ 6> 180°, may also be illustrated by a figure. Let the hemisphere be projected upon the plane of its base (Art. 133). GA, GA', GE, and CB are all arcs of great circles drawn from G. GA and GA' are equal. GD is the longer perpendicular from C upon the circumference of the base of the hemisphere. (a) o + J>180°(a90°)=GB+C4' r '- (5) a + 6>186° (a>6, ^>90°)=a£'+ GA or GE+ GA'. Therefore, when two sides of a spherical triangle SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 195 and an angle opposite one of them are given (consider- ing only possible cases), If the sum of the given sides is less them 180°, and if the side opposite the gi/o&n, cmgle is less than the other given side ((5) 2), or, j^ the Slim of the given sides is geeatee thmi 180°, and if the side opposite the given wrigU is geeatee tham, the other given side {{b) 3) : There will be two solutions ; in all othee cases there will be one solution. Example. Suppose a=62° 40', 6=73° 13', and ^=44° 18'. As (1+6=135° 55'<180°, and as A is < 90°, there are two solu- tions. (See rule.) sin. 6 sin. A sin. T3° IS' sin. 44° 18' Sin.J5 = sin. 62° 42' 196 THE ELEMENTS OF SPHEBICAL TRIGONOMETRY. Log. sin. 73° 13' =9.981095 Log. sin. 44° 18' =9.844114 i (6 + a)=67° 51' 30" Ar. CO. log, sin. 62° 42' =0.061286 i (6— o)= 6° 15' 30" Log. sin. 48° 48' 20"=9.876494 S- 48° 48' 20" i {-B'+^)=46° 33' 10" i {B'-A)= 2° 15' 10" B =131° 11' 40" i (B +.4)=8V° 44' 50" i {B -.4)=43° 26' 50" Solving triangle AB'C, Tan. i .= '=°\^;;+j) tan.i(a+ J)=55!d^^; tan. 67° B7' 30" cos. I {S-A) ' COS. 2° 15' 10" Log. COS. 46° 83' 10" = 9!837390 Log. tan. 67° 67' 30" =10.392682 Ar. CO. log. COS. 2° 15' 10" = 0.000336 Log. tan. 59° 31' 65.62" =10.230408 c=^5'=119° 3' 61.26" Cot. i C= °°°-^f+") tan. i (A+S)= ''''-'V'.^'' \ tan. 46° S3' 10" COS. i (6-a) ^ ^ ' COS. 6° 16' 30" Log. COS. 67° 67' 30° = 9.674356 Log. tan. 46° 33' 10" =10.023561 Ar. CO. log. COS. 5° 16' 30" = 0.001832 Log. cot. 68° 18' 14.26"= 9.699739 0=^05' =136° 36' 28.82" Solving triangle ABC, Tan. i ,=^--nB+A) ^^ (i_,)^ Bin. 87° 44| 60" ^^^_ ^^ ^^, ^^, sin. i (B-A) sin. 43"^ 26' 50" Log. Bin. 87° 44' 60" =9.999664 Log. tan. 6° 15' 30' =8.963947 Ar. CO. log, sin. 43° 26' 50" =0.162610 Log. tan. 7° 37' 0.6"=9.126221 e=AB=16° 14' 1.26" Cot. i c=^JBJf±£) tan. i (^-^)= "''-^V^,'^": tan.43° 26' 60" sin. i (b-a) ^ ' sin. 6° 15' 30" Log. sin. 67° 67' 30" = 9.967038 Log. tan. 43° 26' 50" = 9.976449 Ar. CO. log. sin. 5° 16' 30" = 1.037884 Log. cot. 5° 67' 32.79"=10.981371 C=iACB=\r 56' 5.58" SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 197 ,„ , sin. 5 sin. ^ OB' sin. ^CB (Ciieck) ~ — ; = — : j^rr = — : -r sin. sin. Am^ Bin. A£ Log. sin. 48° 48' 20''=:9.8'76494 Log. sin. 136° 36' 28.52"=9.836948 Log. sin. 73° 13' —9.981095 Log. sin. 119° 3' 51.26"= 9.941S49 1.895399 1.896399 Log. sin. 11° 55' 5.58"=9.314952 Log. Bin. 15° 14' 1.26''= 9.419553 1.895399 Solye a triangle when there are given : Example 1. Sides, 62° 14', 50° 3' ; angle opposite latter, 35° 33'. Ans. Angles, 131° 25' 9.6", 42° 9' 5.8" ; side, 98° 36' 12.1" ; or 11° 45' 13.8', 137° 50' 64.2" ; or 15° 34' 49.7". 2. Sides, 135° 10', 115°; angle opposite first, 143°. Ans. Angles, 50° 40' 44|", 23° 13' 13.6" ; side, 27° 30' 35.2" ; or ■ 129° 19' 15|", 121° 27' or 91° 65' 60.5°- 3. Sides, 137° 2', 145°; angle opposite first, 151°. Ans. Angles, 166° 55' 16.8", 137° 28' 16.4"; side, 71° 61' 40°. 4. Sides, 132° 10', 63° 20' ; angle opposite latter, 73° 20'. 160. Case Y. — Two angles and an included side of a spherical triangle ieifng known, to sol/oe the triangle. To find the half sum and half difference of the sides, use formulas of Art. 143. The sum of these differences will be the greater side ; the difference of these differences wiU be the less side. The third angle may be found by Art. 144. Use formula (a) if ^ (a—h) is very email; but for- mula (b) if f . (a+5) is near 90°. As check on work use (c) Art. 138. Thus, suppose the parts given are A, B, and c. „ , , ,. cos. i (A — B) Tan. * (a+b) = ^^ ^ tan. i e ,. sin. * (A — B) tan. i (a-b) = . l),^' tan. i c sin. i [A + B) will give a and b. 198 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. eot. i C = ?2?4M*) tan. i (A+£), or COS. i (a—o) cot. i C= ""• 1 1"'^^^ tan. i (A-£), will give C. sm. i (a — o) Solve a triangle when there are given : Example 1. Two angles, 33° 16', 12° 14'; the included side, 49° 14'. Ans. Sides, 38° 14' 16.8", 13° 49' 67.2°; angle, 137° 50' 14.8". 2. Two angles, 140° 10', 110° 2' ; the included side, 125° 4'. Ans. Sides, 138° 37' 58.8", 76° 45' 68.8"; angle, 127° 30' 12.8". 3. Two angles, 78°, 56° ; the included side, 59°. Ans. Sides, 61° 33' 39.8", 48° 10' 69.3" ; angle, 72° 27' 31.8". 4. Two angles, 160°, 110°; the included side, 130°. 6. Two angles, 40°, 82 ° ; the included side, 75°. 161. Case VI. — Two a/ngles and a side opposite one of the given angles of an oblique-angled spherical tri- angle })evng hnown, to sol/oe the t^iamgle. The side opposite the other given angle may be found by Art. 138. The thi/rd side may then be found by Art. 143. Use formula {a) if one half the difference of the given angles is a very small quantity, but use (5) if one half the sum of the given angles is near 90°. (See re- mark under Art. 15Y.) The thi^d angle may be found by Art. 144. Use formula {a) if one half the difference of the sides opposite the given angles is a very small quantity, but use (J) if one half the sum of the same sides is near 90°. As check on work use Art. 138. As the first side is found by means of its sine, and as the sine of an angle and the sine of its supplement are the same (Art. 46), there may be two values given SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 199 to the side, either of which will satisfy the conditions of the problem. Consequently there may be two triangles with the given parts, and in such a case two solutions are possible, both of which will be correct. This will be evident from the accompanying figure, in which the surface of a hemisphere is represented as projected upon the plane of its base (Art. 133). Suppose the angles CAB and CBA and the side GB are the given parts. Let the arc CA be produced to meet the plane of the base of the hemisphere again at A' , so that AC A' is a semicircumference (Ch. Art. 32, Vni.). Suppose the given side CB is not equal to 90° ; then C is not the pole of the base (Oh. Art. 37, VIII.). Since C is not the pole of the base, from G two arcs CB and CB' can be drawn to the base equal to one another (Art. 132). The angles GBB' and GB'B are equal (Ch. 23, VIII.); therefore their sup- plements CBA and GB'A' are equal. Also A is equal to A' (Oh. 16, VIII.). 200 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. Thus we have two triangles GAB and OA'B', which have two angles in one equal respectively to two angles in the other, and the side opposite one of the equal angles the same in both. The side opposite the other given angle of one is the supplement of the side opposite the equal given angle of the other. Thus, CA opposite OBA is the supplement of CA\ which is opposite OB' A! (the equal of CBA). 162. Two angles and a side opposite one of these heing given, to determine whether these pa/rts helong to one or to two triangles / that is, whether we shall have one solution or two solutions from the gimen parts. From the given parts we may find the tuoo sides and an angle opposite one of these sides in the polar tri- angle (Oh. Arts. 69 and YO, VIII.). There will be as many solutions for the triangle whose parts are given as for the polar triangle. But we have already consid- ered the case of the triangle of which two sides and an angle opposite one of these sides are given (Art. 158). By passing to tlie polar triangle we can then find (under Case IV. and Art. 159) how many solutions it has, and thus determine how many solutions the triangle under this case has. We can also ascertain this independently by refer- ring to Ch. 26, VIII., and to Art. 137 of this book. Thus, suppose the given parts are the angles A and B, and the side a. The sum of the angles maybe, 1. 180°; 2. <180° ; or, 3. >180°. Under these three heads, if A=^B, a^=:h (Ch. 24, VIII.), and there is one solution. In the following cases we shall suppose A and B unequal. SOLUTION OF OBLIQUE-ANGLED TRLlbfGLES. 301 1. If A+B=1S0°, a+b also equals 180° (Art. 13Y), and there can be but one soluHon. 2. Let A+B be <180° ; and (a) Let A be >J3 ; there will be one solution ; for if a is <90°, b must be 90°, b must be <90°, since a+b <180° (Art. 137). (b) Let A he a; and the other >90°, such that a-1-5 <180°. Thus, suppose ^ + ^<180°, and^<5and o=40°; b might be 60° or 120°, for with either value a is 90°, is impos- sible, for if ^ is <^, a is < 6, . • . 6 is also > 90°, that is, a + 6> 180°, which is impossible (Art. 137). 3. Let A-\-B be >180°, and {a) Let J. be < ^, then there will be one solution ; for iiA90°, so that a-\-b should be >180° (Art. 137); or again, if «>90°, b must be >90°, because a is < 5 as by hypothesis A\& B, and a be >90° ; then there will be two solutions, for as J. is >B, ais >b (Oh. 26, VIIL) ; but b may have two values, one < 90°, and therefore < a, or one >90°, but still 180° (Art. 137). Thus, suppose A + B>180°, and a to be 120°, and A>S, tljen 5 might be 16° or 105° ; with either value a is >6, and a+6>180°. The case (under this head) in which A is >B, and a is <90°, is im- possible ; for as ^ is > 5, a must also be > J ; therefore 6 must be < 90°, that is, a+b is <180°, which is impossible (Art. 137). Therefore (considering only possible cases), when two 202 THE ELEMENTS OF SPHEKICAL TRIGONOMETRY. cmgles of a spherical triangle and a side opposite one of them are given, If the sum, of the gi/ven cmgles is less than i80°, cmd the wngle opposite the gimen side is less than the other gi/oen cmgle ((J) 2), or, if the swn of the gi/ven amgles is GEEATEE thcm 180°, and the angle opposite the gi/ven side is GEEATEE thoM the Other gi/oen, amgle {{i) 3), there will be TWO solutions ; in all othee cases orJ/y one solution. The two angles of a triangle are 126° and 110°, and the side opposite the first angle is 183°. It is required to solre the triangle. Let A = 125°, £= 110°, a = 133°. A + B>180°, and A is >B; therefore there are two solutions. „. , sin. 5 . sm. 110° . Sin. b = sin. a = sm. 133 sin. ^ sin. 125 Log. sin. 110° = 9.972986 Log. sin. 133° = 9.86412'/ Ar. CO. log. sin. 125° = 0.086635 Log. sin. 57° 1' 64.62" = 9.923748 b = CA= 67° 1' 54.62" 1 (A+B) = 117° 30' 6'= 0A'= 122° 58' 5.38" i l-A—B) = 7° 30' To solve ABO i(a + i) = 95°0' 67.31" I \a-b) - 37° 69' 2.69" SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 303 T»T, X ^ - «os-i(^ + -B ) ^ , , rx COS. 117° 30' '^'°- * " - ^^^:h:i=F) *'"'• *("+*)= COB. r 30- - ^^-^^ ^« "^ ^'-^^ Log. coa. 117° 30' = 9.664406 Log. tan. 95° 0' 57.31" =: 11.056661 Ar. CO. log. COS. 7° 30' = 0.003731 Log. tan. 79° 19' 39.48" = io.724798 e = ^5=158° 39' 18.96". r«t i/^-"°^- *(" + *)* ,/^.m COS. 95° 0' 67.31" ^„ , Cot.iC'= ) '-ta.n.i(A + B)= tan. 117 30 cos.i(a-6) ^^ ^ ' COS. 37° 59' 2.69" Log. COS. 95° 0' 57.31" = 8.941673 Log. tan. 117° 30' = 10.283523 Ar. CO. log. COS. 37° 59' 2.69" = 0.103374 Log. cot. 77° 58' 14.08" — 9.328570 A (7.8=155° 56' 28.16". To solve A' E O i(a + 6') = 127° 59' 2.69" i {a-b') = 5° 0' 57.31" cos 117° 30' Tan. ic' = — '- tan. 127° 59' 2.69" cos. 7 30 Log. cos. 117° 30' = 9.664406 Log. tan. 127° 59' 2.69" = 10.107439 Ar. CO. log. cos. 7° 30' = 0.003731 Log. tan. 30° 48' 60.62J" = 9.775576 c' = ^'.B' = 61° 37' 41.25". „ ^ , „, COS. 127° 69' 2.69" Cot. i C = -— tan. 117° 30' cos. 5° 0' 57.31" Log. cos. 127° 59' 2.69" = 9.789187 Log. tan. 117° 30' = 10.283523 Ar. CO. log. cos. 5° 0' 67.31" = 0.001667 Log. cot. 40° 7' 3.26" = 10.074377 A' a E = 80° 14' 6.52". (CTeci.) sin. B sin. C sin. 0" sin. h sin. c sin. c' Log. sin. 110° =9.972986 Log. sin. 165° 66' 28.16"= 9.610314 Log. sin. 67° 1' 64.62" = 9.923748 Log. sin. 158° 39' 18.96"= 9.561076 0.049238 0.049238 204 THE ELEMENTS OF SPHERICAL TRIGONOMETBY. Log. Bin. 80° 14' 6.52" =: 9.993662 Log. sin. 61° 3'?' 41.25" = 9.944424 0.049238 Solve a splierical triangle when there are given : Example 1. Two angles, 55° 2', 68° 10' ; the side opposite the first angle, 48° 42'. ^m. Sides, 68° 4' 58.6", 65° 26' 30.6" ; angle, 84° 5' 12.4"; or 121° 55' 1.4", 160° 12' 23.8" ; or 158° IS'Sl.l". 2. Two angles, 150° 22', 124° 12' ; the side opposite the first angle 149° 20'. Am. Sides, 68° 33' 28i-°, 143° 33' 22° ; angle, 144° 50' 21.8" ; or 121° 26' 31t", '73° 18' 35.6" ; or 111° 47' 4.9". 3. Two angles, 125°, 130° ; the side opposite the first angle, 114°. Am. Sides, 121° 18' 56.2", 98° 36' 17.2"; angle, 117° 33' 11.7". 4. Two angles, 12°, 25° ; the side opposite the second angle, 60°. SOLUTION OF OBLIQUE-AITGLED TEIANGLES BY MEAJS^S OF EIGHT-ANGLED TEIAifGLES. 163. The preceding cases can all be solved by divid- ing the given triangle into two right-angled triangles, by drawing a great circle arc from one of the angles to the opposite side, or to the opposite side produced, and by then solving the two right-angled triangles. Care should be taken, in drawing the perpendicular, to make one of the right-angled triangles formed con- tain two of the three given parts. (Art. 116.) Case I. — The three sides hemg gwen. From the angle opposite the longest side draw a per- pendicular to that side. The side will be thus divided internally into two segments. The half difference of the two segments can be found by formula {a) of Art. 142. This half difference added to the half side will give the greater segment ; and subtracted from the half side will give the less segment. We shall then have, SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 205 in each right-angled triangle, the hypotenuse and a side given to find the other parts. (Art. 120.) Case II. — The three angles ievng given. Find the sides of the polar triangle (Oh. Arts. 69 and 70, VIII.) ; solve that by the method of the pre- ceding case ; the supplements of the angles found will be the required sides of the given triangle. Case III. — Tioo sides cmd an i/ncl/wded a/ngle ieimg given. Suppose the given parts to be a, b, and C. Draw the perpen- dicular AD. Then in the tri- angle A OD we have the hypote- nuse and an angle, O, to find OD (Art. 121). BD^a~CD. Then c can be found by 2, Art. 140. A and B can be found by Art. 138. The three last parts can also be found by finding all the parts of the two triangles A CD, BOD. Case IY. — Two sides and an angle opposite one of these sides being given. Suppose the given parts to be a, I, and A. Draw the perpendicular from C upon c or c produced. C The figure represents one of the examples in which there are two solutions (see rule at the end of Art. 169). We shall only give one solu- tion, as the method is the same in both solutions. To solve the triangle A CD we have given the hy- 206 THE ELEMENTS OF SPHERICAL TRIGONOMETRY, potenuse h and the angle A (Art. 121). BD can be found by 2, Art. 140. Then to solve the triangle ^CZ? we have given the hypotenuse a and the side £D (Art. 120). Case V. — Two cmgles and an included side being given. Suppose the given parts to be A, B, and c. Draw an arc from B perpendicular to h. To solve the triangle ABB we have the hypotenuse c and the an- gle A (Art. 121). This will enable ^ us to find BD and the angle J[5Z>. If BJ) falls within the triangle, CBB—B-ABD; if without the triangle, CBD- ABD—B. Then to solve CBD we have given the side BD and the angle CBD (Art. 123). Case VI. — Two angles cmd a side opposite one of these being gimen. (See rule at end of Art. 162.) Suppose the given parts to be A, B, and a. Draw the perpendicular from G upon g (or upon o produced). To solve the triangle CBD we have given the hy- potenuse a and the adjacent angle (Art. 121). Having found BD, AD can be found by 1, Art. 140. A C can then be found by 2, 140. Or, by solving the triangle CBD we can find CD. Then to solve the triangle SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 307 A CD we have given a side, CD, and the opposite an- gle A (Art, 124). C will be the sum of A CD and BCD and G will be the sum of AD and BD^ if the perpen- dicular falls withim, the triangle ; but C will be the difference of A CD and B'CD, and c will be the differ- en-ce of JLi> and DB', if the perpendicular falls without the triangle. 164. It is well to solve oblique-angled triangles by right-angled triangles when there is any difficulty in arriving at a true solution by means of formulas ; for instance, when in Case I. s is very near 180°, or s—a, s— 5, or s— c is very near 0°; when in Case II. Sis, very near 90° or 270°, or 8- A, S-B, or 8-Ci& very near 90° ; when in Oases III. and IV. ^ (a+5) is very near 90°, or ^ (a—b) is very near 0° ; when in Cases V. and VI. i (A+B) is very near 90°, or ^ (A—B) is very near 0°. It is sometimes advantageous to combine the two methods so as to check the work. TH£ END. 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