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 UfliversitY and Scheel Extensien 
 
 SPHERICAL TRIGONOMETRY. 
 
 1889. 
 
 A. W. PHILLIPS, 
 
 Va/e University. 
 
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 Copyright, 
 
 1889, 
 
 By a. W. PHILLIPS. 
 
 Press ofJ.J. Little & Co., 
 Astor Place, New York. 
 
Spherical Trigonometry. 
 
 THE SPHERICAL BLACKBOARD. 
 
 § I. The student should construct his spherical triangles on a 
 globe. He will thus get a good idea of the meaning of each problem, 
 and the unknown parts of the triangle may be fairly well measured, 
 and a good check against large errors in his calculations will be 
 obtained. 
 
 A cylindrical cup, whose depth is half its diameter, or a hemi- 
 spherical cup, should be provided with a flat rim graduated to de- 
 grees. To this cup should be fitted a slated globe, so it will turn 
 easily in any direction. 
 
 The rim of the cup is the ruler used for drawing and measuring 
 the arcs of great circles. 
 
 Dividers may also be used for laying off the lengths of arcs, 
 their measures being taken from the globe or from the graduated 
 rim of the cup. 
 
 An angle is measured by laying off on each of its including 
 sides, or on those sides produced, 90°; the arc joining these two 
 points thus found will be the measure of the angle. 
 
 Two equal triangles may be made to coincide by direct super- 
 position, and two symmetrical triangles by turning one of them 
 
 inside out, and then superposing it on the other. 
 
 [A piece of tin-foil may be fitted to a sphere and figures cut out from it as 
 they are cut out of paper in plane geometry to make this superposition.] 
 
UNIVERSITY EXTENSION. 
 
 PROBLEMS OF CONSTRUCTION. 
 
 1. To draw one great circle perpendicular to another one. 
 
 2. To construct an angle at a given point equal to a given angle 
 
 already marked out on the sphere. 
 
 3. To construct an angle at a given point equal to an angle of a 
 
 given number of degrees. 
 
 4. Having a triangle given, to construct its polar triangle. 
 
 5. To construct a triangle having given in degrees : 
 
 (a.) the three sides. 
 
 {^.) two sides and the included angle. 
 
 (c.) two sides and an angle opposite one of them. 
 
 When are there two solutions ? When one? When none? 
 (d.) two angles and a side opposite one of them. 
 (e.) two angles and the included side. 
 (/.) the three angles. 
 
 6. To construct the triangles as above, having the given lines and 
 
 angles laid down on the sphere. 
 
 MODEL OF A PORTION OF THE SPHERE. 
 
 § 2. The relations between the sides and angles of a spherical 
 triangle may be best obtained from a pasteboard model of a portion 
 of the sphere. 
 
 The sides of a spherical triangle are the intersecting arcs which 
 
SPHERICAL TRIGONOMETRY. 
 
 planes passing through the centre of the sphere make with the sur- 
 face of the sphere. 
 
 The angles of a spherical triangle are the diedral angles formed 
 by the planes of its sides. 
 
 If a line be drawn in each of the two faces of a diedral angle 
 perpendicular to its edge at a given point, the plane angle formed by 
 these two lines is the measure of this diedral angle. 
 
 RIGHT-ANGLED SPHERICAL 
 TRIANGLES. 
 
 CONSTRUCTION OF MODEL. DERI- 
 VATION OF FUNDAMENTAL 
 FORMULAS. 
 
 § 3. A model for this purpose 
 may be made as follows : On a piece 
 of pasteboard lay off from a point O, 
 as a centre, the line O C equal to the 
 assumed radius of the sphere, and 
 describe the arc of a circle B A C B' 
 in pencil, O being the centre. Fig. 7. 
 
 Draw the several lines as in 
 this figure. The angles A O B, 
 A O C, COB' are to be of the 
 same size as in the figure ; but the 
 radius O C may be taken of any 
 
 Fig. 7. 
 
6 
 
 UNIVERSITY EXTENSION. 
 
 length, and the other lines, of course, must be taken proportional 
 toOC. 
 
 Cut half through the pasteboard on the reverse side along the 
 lines O C and O A, and cut out the entire figure O L K C L' O. Bend 
 the side O A H about O A till it is perpendicular to O A C, and bend 
 O C B' about O C till O B' coincides with O B. 
 
 Call the arcs B' C, A B, and A C respectively a, c, and b, and label 
 them accordingly. • These are the sides of the right spherical triangle ; 
 A is the right angle, C the angle at the base, and B the angle at the 
 vertex. See Fig. 8. 
 
 Fig. 8 
 
 Construct also three plane right triangles from separate pieces 
 of pasteboard. 
 
 I St. B D E, where the base and perpendicular are respectively 
 
 equal to D E and B D as found above. 
 2d. H A F, having FA and A H as base and perpendicular. 
 
SPHERICAL TRIGONOMETRY. 
 
 3d. K L C, having K C and K L as base and perpendicular. 
 
 Fasten these three triangles securely in their places to the sides 
 O B A and O A C of the model, leaving O C B' to swing on O B' as a 
 hinge. The several planes D B E, H A F, and L K C will be perpen- 
 dicular to the plane O A C and to the line O C, and the points and 
 lines of the triangles will meet the points and lines of the model, 
 which have corresponding letters. See Fig. 9. 
 
 The angles E, F, and C in these triangles are each equal to the 
 angle C of the spherical triangle. 
 
 DERIVATION OF FORMULAS. 
 
 § 4. In the plane right triangle D B E. Fig. 10. 
 
 D B = sin ^ , ,. ^ , 
 
 — the radius of the sphere, bemg unity. 
 
 The angle D E B = angle C of the spherical triangle ABC. 
 sin c 
 
 sin C = 
 
 sm a 
 
 (i) 
 
UNIVERSITY EXTENSION. 
 
 In the plane right triangle F AH. Fig. ii. 
 
 H A = tan c 
 
 A F = sin /^ 
 The angle H F A = angle C of the spherical triangle, 
 
 tan c 
 
 tan C = (2) 
 
 sin b 
 
 In the plane right triangle KCL. Fig. 12. 
 
 C L = tan a 
 
 C K = tan ^ 
 The angle K C L = angle C of the spherical triangle, 
 tan b 
 
 (3) 
 
 cos C = 
 
 tan a 
 
 In the plane triangles ODE and A O F. Fig. 10. 
 O F = cos ^ O D = cos r 
 
 O E = cos a O A = unity = radius of sphere, 
 
 cos a = cos b cos c (4) 
 
sin a 
 
 (s) 
 
 tan /^ 
 
 (6) 
 
 sin c 
 
 tan c 
 
 (n) 
 
 SPHERICAL TRIGONOMETRY. 9 
 
 If the model were made with c for the base and B the base 
 angle we could obtain the following relations : 
 
 sin d 
 sin B = 
 
 tan B = 
 
 cos B == 
 
 tan a 
 
 By combining the above formulas we may obtain 
 
 cos C 
 
 sin B = (8) 
 
 cos c 
 
 cos B 
 
 sin C = (9) 
 
 cos <^ 
 
 COS a = cot C cot B (10) 
 
 GROUPING OF FORMULAS. 
 
 § 5. It will be seen that the above are all the possible combina- 
 tions of the five parts of the right spherical triangle B, C, a, b, c, taken 
 in sets of three, and therefore we have all the cases that can arise in 
 their solution. For convenience of memory these ten formulas may 
 be arranged in two groups. 
 
 ist. Those which involve tangents and cotangents. 
 
 2d. Those which involve only sines and cosines. 
 
lO UNIVERSITY EXTENSION. 
 
 1ST Group. 
 cos ^ = cot B cot C or sin (co. a) — tan (co. B) tan (co. C). 
 cos B = cot ^ tan c or sin (co. B) = tan (co. a) tan c. 
 cos C = cot ^ tan l> or sin (co. C) ~ tan (co. a) tan d. 
 sin d = tan ^ cot C or sin d = tan ^ tan (co. C). 
 
 sin c = Un d cot B or sin c = tan ^ tan (co. B). 
 
 2D Group. 
 
 cos a = cos d cos c or sin (co. a) = cos ^ cos c. 
 
 cos B = sin C cos ^ or sin (co.B) = cos (co. C) cos d. 
 
 cos C = sin B cos c or sin (co. C) — cos (co. B) cos c. 
 
 sin 3 = sin a sin B or sin l^ = cos (co. a) cos (co. B). 
 
 sin <: = sin ^ sin C or sin c = cos (co. a) cos (co. C). 
 
 NAPIER'S RULES. 
 
 ^ § 6. If we draw a spherical triangle, making A the right angle, B 
 and C the remaining angles, and a, b, and c the three sides, and then 
 write comp. B, comp. a, comp. C for B, a, C, respectively, we shall 
 see that by excluding the right angle A we may state as a rule cov- 
 ering the formulas in the first group : 
 
 I. The sine of the middle part is equal to the product of the tangents 
 of the adjacent parts. 
 and for the second group — 
 
 II. The sine of the 7niddle part is equal to the product of the cosines 
 of the opposite parts. 
 
SPHERICAL TRIGONOMETRY. II 
 
 It will help the student to avoid the confusion of these rules if 
 he will observe that the words /^ngent and adjacent go together, and 
 also that the words c<?sine and opposite go together. 
 
 RULE FOR THE SOLUTION OF RIGHT SPHERICAL 
 TRIANGLES. 
 
 § 7. By I. and II. find directly from the tivo given parts each of the 
 
 re?nai?iing parts. 
 Check. — Substitute the proper values in a formula co7itaining the 
 
 three required parts. 
 
 EXAMPLES. 
 
 The student should select a sufficient number of examples from 
 any text-book for solution, and also plot these examples on the 
 globe, measuring the parts required in the solution. 
 
 PROBLEM. 
 
 Make the necessary computations for a sundial. 
 
 OBLIQUE-ANGLED SPHERICAL TRIANGLES. 
 
 EXERCISES ON THE GLOBE. 
 
 § 8. Construct the following triangles from the given parts, and 
 measure the remaining parts : 
 
 1. Given the three sides, 38°, 56°, 70°. 
 
 2. Given the three angles, 75°, 80°, 112°. 
 
 3. Given two sides and included angle, 80°, 74°, 60°. 
 
 4. Given two angles and a side opposite one of them, 60°, 80°, 94°. 
 
 5. Given two angles and the included side, 40°, 55°, 70°. 
 
12 UNIVERSITY EXTENSION. 
 
 6. Given two sides and an angle opposite one of them, 85°, 65°, 50°. 
 
 How many cases in this last ? 
 
 7. Draw the polar triangle for each case. 
 
 MODEL FOR OBLIQUE-ANGLED TRIANGLE. 
 
 A model may be made to derive the fundamental formula of 
 oblique-angled triangles as follows : 
 On a piece of pasteboard take O as a centre, with a radius O A, and 
 
 describe an arc A B C D. 
 Draw O B, and produce it to meet the perpendicular A E erected to 
 
 O A at A, and draw O C to meet the perpendicular D F erected 
 
 toODatD. 
 Cut out the piece of pasteboard O A E F D O. 
 Cut half through the pasteboard along O E and O F. 
 
 A 
 5" 
 
 c 
 
 Fig. 13. 
 
 Bend the two outside triangles so that O D and OA will coincide, 
 and fasten the pieces securely together, calling the radius OA 
 unity, and the sides of the spherical triangle respectively 
 opposite A, B, C, — a^ b^ and c. 
 
SPHERICAL TRIGONOMETRY. 13 
 
 F A E also measures the angle A of the spherical triangle. 
 A E = tan (T : A F = tan ^ 
 
 O E = sec c : O F = sec d. 
 
 In the triangle A E F 
 
 F E ^ = tan ^ d + tan ^ <: — 2 tan d tan c cos A. 
 
 In the triangle O E F 
 
 F E ^ = sec ^ (^ + sec ' c — 2 sec ^ sec ' c cos a. 
 
 Subtracting these two last equations and reducing, we have 
 cos a = cos d cos <: + sin ^ sin c cos A. (I.) 
 
 By § II Plane Trigonometry, advancing the letters, we may 
 get expressions for cos B and cos C. 
 
 § 9. By the principle of Polar Triangles, if we substitute in the 
 
 above formulas 
 
 A =180 — ^' ^=180° — A' 
 
 B= i8o — ^' ^ == 180° — B' 
 
 C = 180 — ^' c = 180° — C 
 
 we obtain, after reducing, and suppressing the accents : 
 
 — cos A = cos B cos C — sin B sin C cos a (II.) 
 
 etc., etc. 
 
 [The formula, before suppressing the accents, was true for all polar triangles, 
 and since every possible triangle may be included in these polar triangles, the 
 formula will hold generally.] 
 
 § 10. In a spherical triangle ABC, let B and C be the angles at 
 the base. The sides A B, B C, and C A are respectively c, a, b. 
 Drop a perpendicular A D from A on the side B C. 
 
14 UNIVERSITY EXTENSION. 
 
 By right triangle formulas we have 
 
 A D = sin ^ sin B == sin b sin C, 
 
 sin b sin c 
 or, ^ = — (III.) 
 
 sin B sin C 
 
 Note. — In this triangle the base a may be found by adding two segments BD 
 and C D together. These segments may be found by the rules of right triangles. 
 
 Summary of Formulas for solving Oblique Angled Triangles 
 without Logarithms. 
 
 § II. I. Three sides Formulas I. 
 
 2. Three angles Formulas II. 
 
 3. Two sides and included angle I. 
 
 4. Two angles and included side II. 
 
 5. Two sides and opposite angle III. III. note. 
 
 6. Two angles and opposite side III. III. note. 
 
 SOLUTION OF OBLIQUE-ANGLED TRIANGLES 
 
 BY THE USE OF LOGARITHMS. 
 
 COS a — COS b COS c 
 
 §12. From Formula (I.) cos A = 
 
 sm b sm c 
 Subtract each side of this ec^uation from unity. 
 By PI. Trig. 2 sin" >^ A = i — cos A. 
 
 |/^ 
 
 . , , , . ^ , sin (^ sin <r — cos a + cos b cos c 
 sm"" Yz K 
 
 2 sin b sin c 
 
SPHERICAL TRIGONOMETRY. 1 5 
 
 reducing 
 
 • x/ * i / COS ib — c) — cos ^ rp, ^u* 1 
 
 sm }4 A = i/ ^ [Prove this.J 
 
 2 Sin a sin c 
 
 By PI. Trig. Formula 13. 
 
 / a — l^ + c a ^ b — c 
 sin 5^ A = / ^^^ ^'^ 
 
 / 
 
 sm b sm c 
 
 a + b + c 
 Putting — s 
 
 sin J^ A = i/ sin {s-b) sin (s-c) (jv_ ^ 
 
 r sin /^ sin c 
 
 § 13. Add unity to each side of Formula (I.) and then prove in 
 a similar way. 
 
 cos >^ A = J sin . sin (.-<.) (^_) 
 
 r sin b sin <r 
 
 Dividing (IV.) by (V.) 
 
 -\/^ 
 
 tan '^ A = ^/ sin js-b) sin (.-.) (^I.) 
 
 sin s sin (^ — ^) 
 
 tan y2 B and tan yi C may be found by advancing one letter 
 throughout the formula. See § 12, Plane Trig. 
 
l6 UNIVERSITY EXTENSION. 
 
 §14. From II. derive in a similar way 
 
 x/ A / — cos S cos (S — A) /T7-TT \ 
 
 tan ^ « = \/ ^^ — (VII.) 
 
 r cos (S — 
 
 (S — B) cos (S-C) 
 A + B + C 
 
 where S 
 
 2 
 § 15. Dividing (VI.) 
 
 T/ A i / sin is — F) sin is — c) , 
 
 tsixi }^ A = A/ 1 -1 ^ by 
 
 r sin s sin {s — a) 
 
 , _ ^ / sin (^ ^ — c) sin {s — a) (obtained by advanc- 
 ~y sin s sin {s — ^) ing one letter in VI.) 
 
 tan 3^ A sin (s — ^) 
 
 we have = 
 
 tan }4 B sin {s — a) 
 
 By composition and division 
 
 tan >^ A + tan }^ B sin (^ — i>) + sin (^ — a) 
 
 tan ^ A — tan >^ B sin (s — ^) — sin {s — a) 
 
 which reduces to 
 
 sin >^ (A + B) tan }^ c 
 
 sm 
 
 ^ (A — B) tan ^i {a — b) 
 Work out all the steps. 
 
 (VIII.) 
 
SPHERICAL TRIGONOMETRY. I7 
 
 Multiplying tan >^ A by tan ^ B, values as above given, and 
 reducing, we obtain 
 
 cos 3^ (A + B) tan 3^ c 
 
 cos >^ (A — B) tan Y^ [a ^ b) 
 
 Work out all the steps. 
 
 § 16. By using the principle of Polar triangles 
 Formulas (VIII.) and (IX.) reduce to 
 
 sin yi {a ^ b) cot Y^ C 
 
 (IX.) 
 
 sin Yiij^ — b) tan >^ (A — B) 
 and 
 
 cos Y^ifi ^ b) cot Y^ C 
 
 (X.) 
 
 (XI.) 
 
 COS Y^iP' — b) tan >^ (A + B) 
 
 Formulas (VIII.), (IX.), (X.) and (XL) are called Napier's 
 Analogies. 
 
 Formulas III., VI., VII., with the Napier's Analogies, are suffi- 
 cient to solve all cases of spherical triangles by logarithms. 
 
 These are collected and renumbered in the following summary: 
 
 SUMMARY. 
 
 § 17. [There are two additional formulas for each one of the following, and 
 these may be obtained as indicated, § 12, Plane Trig.] 
 
 sin b sin c 
 
 I. = 
 
 sin B sin C 
 2 
 
1 8 UNIVERSITY EXTENSION. 
 
 5- 
 
 . 1/ A - i / ^^" (-^ ~~ ^^ ^^"^ (^ --f) 
 2. tan >^ A ~ /I/ ; —- 
 
 f sm s sm {s — a) 
 
 cos S cos (S — A) 
 tan J^ « = 
 
 cos (S — B) cos (S — C) 
 sin >^ (A + B) tan >^ c 
 
 sin >^ (A — B) tan y2 (a — b) 
 
 cos >^ (A + B) tan )^ c 
 
 cos >^ (A — B) tan y2 (a + b) 
 
 sin 3^ (^ + /^) cot >^ C 
 
 sin y2 [a — b) tan >^ (A — B) 
 
 cos Yz {a -^ b) cot >^ C 
 
 cos y2{a — b) tan ^ (A + B) 
 
 CASES. 
 § 1 8. Given : 
 
 1. Two sides and angle opposite one of them i, 4, i. 
 
 2. Two angles and side opposite one of thein i, 4, i. 
 
 3. Three sides 2, 2, 2. 
 
 4. Three angles 3» 3' 3; 
 
 5. Two sides and included angle 6, 7, i. 
 
 6. Two angles and included side 4? 5? i- 
 
 The proof in each case may be obtained by substituting the 
 three quantities obtained in the complete solution in some formula. 
 
SPHERICAL TRIGONOMETRY. I9 
 
 EXAMPLES. 
 
 § 19. I. In the spherical triangle ABC let a, b, c be the sides 
 respectively, it is required to find the remaining parts given. 
 
 (a) x\ =: 50°, b = 60°, and a ~ 40°. 
 
 (b) a = 50° 45' 20", b — 69° 12' 40", and A = 44° 22' 10". 
 How many solutions are there ? 
 
 (c) A = 129° 05' 28", B = 142° 12' 42", C = 105° 08' 10". 
 
 (d) a = 124° 53', b = 31° 19', and c = 171° 48' 42". 
 
 2. " Find the shortest distance in miles on the earth's surface from 
 
 Berlin, latitude 52° 31' 13" N., longitude 13° 23' 52" E., to 
 Alexandria, Egypt, latitude 31° 13' N., longitude 29° 55' E., 
 the earth being considered a sphere whose radius is 3,962 
 miles." 
 
 3. Find the length of the shortest day in New Haven, Conn., lati- 
 
 tude 42° 18' N., it being assumed that the centre of the sun 
 at rising or setting is 90° 50' from the zenith, and the 
 declination of the sun 23° 28' N. 
 
 4. Find the time when twilight begins on the above day at New 
 
 Haven, the sun being 18° below the horizon at that time. 
 
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