(662) From the two values of B (656) the true value must be selected by means of (623–627). 45. Scholium. The problem is impossible, by (772), when A differs more from 90° than does C, and when at the same time, one of the two quantities a and A is (663) less than 90°, while the other is greater than 90°. But this case is precisely the same as the impossible case of (659). 46. Problem. To solve a spherical triangle when its three sides are given. Solution. Let ABC (figs. 4. and 5.) be the triangle; a, b, and c being the given sides. From B let fall on AC the perpendicular BP. Then, in the right triangle PBC, if co. C is the middle part, co. a and PC are the adjacent parts. Hence, by (474), cos. C = cotan. a tang. PC. (664) (665) If, in the triangle BPC, co. a is the middle part, BP and PC are the opposite parts; and, if, in the triangle ABP, co. c is the middle part, BP and AP are the opposite parts. Hence, by (605), But cos. a cos. c:: cos. PC; cos. AP. (666) (fig. 4.) AP=b— PC, and (fig. 5.) AP = PC — b. Hence, by (116) and (202), (667) (668) cos. AP cos. (b PC) = cos. (PC—b), cos. b cos. PC + sin. b sin. PC; which, substituted in (665), gives (669) Cos. a: cos. c:: cos. PC: cos. b cos. PC+sin. b sin. PC. Dividing the two terms of the last ratio of this proportion by cos. PC, and reducing by (10), we have (670) cos, a cos. c :: 1: cos. b + sin. b tang. PC. Make the product of the means equal that of the extremes, and we have cos. a cos. b + cos. a sin. b tang. PC = cos. c; by transposition cos. a sin. b tang. PC = cos. c - cos. a cos. b. Divide by sin. a sin. b and reduce the first member by (11), whence the value of the angle C may be calculated, and in the same way either of the other angles. 47. Corollary. The equation (674) may be brought into a form more easy for calculation by logarithms, as follows. Add unity to both its members and it 1+ cos. C cos. (MN) - cos. (M+N) which, substituted in the numerator of (675), gives Now we have (122). · and letting s denote half the sum of the sides or = 2 sin. M sin. N; (678) (679) { N = (a + b —c) = s — c; (680) (687) 48. Corollary. The angles A and B may be found by the two following equations which are easily deduced from (686), 49. Corollary. The equation (674) can be brought into another form equally simple in calculation. Subtract each member from unity (690) cos. (a - b) cos. a cos. b + sin. a sin. b, which substituted in the numerator of (689) gives Now, if in (678), (692) COS. (MN) SM = 1⁄2 (a - b + c) cos. (M + N) (693) (695) and (692) becomes cos. (a - b) — cos. c = 2 sin. (s—a) sin. (s — b) ; which, substituted in (691), gives |