39. EXAMPle. Given in the spherical right triangle (fig. 30), a = 1o, and b = 100°; to solve the triangle. 40. Problem. To solve a spherical right triangle, when the two angles are given. Solution. Let ABC (fig. 30) be the triangle, A and B the given angles. First. To find the hypothenuse h; co. h is the middle part, and co. A and co. B are adjacent parts. Hence cos. h = cotan. A cotan. B. Secondly. To find one of the legs, as a; co. A is the middle part, and co. B and a are the opposite parts. Hence 41. Scholium. The problem is, by $ 20, impossible, when the sum of the given values of A and B is less than 90°, or greater than 270°, or when their difference is greater than 90°. 42. EXAMPLE. Given in the spherical right triangle (fig. 30), A 91° 11', and B=111° 11'; to solve the triangle. CHAPTER III. SPHERICAL OBLIQUE TRIANGLES. 43. Theorem. The sines of the sides in any spherical triangle, are proportional to the sines of the opposite angles. [B. p. 437.] Proof. Let ABC (figs. 32 and 33) be the given triangle. Denote by a, b, c, the sides respectively opposite to the angles A, B, C. From either of the vertices let fall the perpendicular BP upon the opposite side AC. Then, in the right triangle ABP, making BP the middle part, co. c and co. BAP are the opposite parts. Hence, by Napier's Rules, sin. BP sin. c sin. BAP — sin. c sin A. For BAP is either the same as A, or it is its supplement, and in either case has the same sine, by (98). Again, in triangle BPC, making BP the middle part, co. a and co. C are the opposite parts. Hence, by Napier's Rules, sin. BP = sin. a sin. C; and, from the two preceding equations, sin. c sin. A = sin a sin C, which may be written as a proportion, as follows: sin. a sin. A sin. c: sin. C. : In the same way, sin. a sin. A sin. b: sin. B. : 44. Theorem. Bowditch's Rules for Oblique Triangles. If, in a spherical triangle, two right triangles are formed by a perpendicular let fall from one of its verticles upon the oppo site side; and if, in the two right triangles, the middle parts are so taken that the perpendicular is an adjacent part in both of them; then The sines of the middle parts in the two triangles are proportional to the tangents of the adjacent parts. But, if the perpendicular is an opposite part in both the triangles, then The sines of the middle parts are proportional to the cosines of the opposite parts. [B. p. 437.] Proof. Let M denote the middle part in one of the right triangles, A an adjacent part, and O an opposite part. Also let m denote the middle part in the other right triangle, a an adjacent part, and o an opposite part, and let p denote the perpendicular. First. If the perpendicular is an adjacent part in both triangles, we have, by Napier's Rules, Secondly. If the perpendicular is an opposite part in both the triangles, we have, by Napier's Rules, 45. Problem. To solve a spherical triangle, when two of its sides and the included angle are given. [B. p. 438.] Solution. Let ABC (figs. 32 and 33) be the triangle, a and b the given sides, and C the given angle. From B let fall on AC the perpendicular BP. Frst. To find PC, we know, in the right triangle BPC, the hypothenuse a and the angle C. Hence, by means of Napier's Rules, tang. PC = cos. C tang. a. Secondly, AP is the difference between AC and PC, that is, (298) (fig. 32) AP = b— PC, or (fig. 33) AP PC-b. - = (299) Thirdly. To find the side c. If, in the triangle BPC, co. a is the middle part, PC and PB are opposite parts; and if, in the triangle APB, co. c is the middle part, BP and AP are the opposite parts. Hence, by Bowditch's Rules, cos. PC: cos. AP = sin. (co. a) : sin. (co. c), or cos. PC: cos. AP — cos. a : cos. c. (300) Fourthly. To find the angle A. If, in the triangle BPC, PC is the middle part, co. C and BP are adjacent parts; and if, in the triangle APB, AP is the middle part, co. BAP and BP are adjacent parts. Hence, by Bowditch's Rules, sin. PC: sin. PA = cotan. C : cotan. BAP; (301) and BAP is the angle A (fig. 32), when the perpendicular falls within the triangle; or it is the supplement of A (fig. 33), when the perpendicular falls without the triangle. Fifthly. B is found by means of § 43, sin. c sin. C — sin. b : sin. B. (302) 46. Scholium. In determining PC, c, and BAP, by (298), (300), and (301), the signs of the several terms must be carefully attended to; by means of Pl. Trig. § 62. But to determine which value of B, determined by (302), is the true value, regard must be had to the following rules, which are proved in Geometry. I. The greater side of a spherical triangle is always opposite to the greater angle. II. Each side is less than the sum of the other two. III. The sum of the sides is less than 360o. IV. Each angle is greater than the difference between 180°, and the sum of the other two angles. There are, however, cases in which these conditions are all satisfied by each of the values of B. In any such case this angle can be determined in the same way in which the angle A was determined, by letting fall a perpendicular from the vertex A on the side BC. But this difficulty can always be avoided, by letting fall the perpendicular upon that of the two given sides which differs the most from 90°. 47. Corollary. By (299), (111), and (35), we have cos. AP = cos. (b — PC) cos. b cos. PC + sin. b sin. PC, (303) which, substituted in (300), gives cos. PC: cos. b cos. PC + sin. b sin. PC = cos. a : cos. c. Dividing the two terms of the first ratio by cos. PC, we have by (7), 1: cos. b + sin. b tang. PC = cos. a : cos. c. (304) The product of the means being equal to that of the extremes, we have cos. c = cos. a cos. b + sin. b cos. a tang. PC. But, by (298), (305) cos. c = cos. a cos. b + sin. a sin. b cos. C, (307) which is one of the fundamental equations of Spherical Trigonometry. |