Demonstration. The first and third terms of the proportion (750), 1⁄2 1⁄2 sin. (AC): sin. (A — C) : : tang. b (769) are, by (496), both positive, since (770) than 180° and b is less than 90°. (A + C) is less The second and fourth terms must then be both positive or both negative at the same time. But as (A— C) and (771) (a c) are both less than 90°, these terms can be negative only when A is less than C and a less than c which agrees with (768). 64. Theorem. Of two sides of a spherical triangle the one which differs most from 90° must be (772) opposite the angle which differs most from 90°; and this side and angle must be both greater or both less than 90°. Demonstration. Let the side a differ more from (773) 90°, than does the side b; then by (507) sin. a is less than sin. b. But, by (598), (774) sin. a sin. b:: sin. A: sin. B. Hence, sin. A is less than sin. B, and by (507) the (775) angle A differs more from 90° than does the angle B which agrees with the first part of (772). Again, if a is also greater than b it must be greater than 90°; and the opposite angle A must, by (768), (776) be greater than the angle B, and, by (775), differing more from 90° must also be greater than 90°. But if a is less than b, it must, by (773), be acute; and A (777) must, by (768), be less than B, and, by (775), it must also be acute; which is the second part of (772). 65. Problem. To solve a spherical triangle when its three angles are given. Solution. Let ABC (figs. 4. and 5.) be the triangle, the angles A, B, and C being given. From B let fall on AC the perpendicular BP. Then, if, in the right triangle PBC, co. a is made the middle part, co. C and co. PBC are the adjacent parts. Hence, by (474), cos. acotan. C cotan. PBC. If, in the right triangle PBC, co. C is the middle part, co. PBC and PB are the opposite parts; and, if, in the triangle ABP, co. BAP is the middle part, co. PBA and PB are the opposite parts. Hence, by (605), But (778) cos. C: cos. BAP:: sin. PBC: sin. PBA. (779) (fig. 4.) BAP = A, and (fig. 5.) BAP = 180° — A ; (780) (783) sin. B cos. PBC. cos. B sin. PBC, (fig. 5.) sin. PBA= sin. (PBC — B) = sin. B cos. PBC + cos. B sin. PBC; whence (779) becomes (fig. 4.) cós. C: cos. A :: sin. PBC (785) : sin. B cos. PBC-cos. B sin. PBC; (786) and, (fig. 5.) cos. C:- cos. A sin. PBC sin. B cos PBC+ cos. B sin. PBC, which becomes the same as (785) by changing the signs of the second and fourth terms. Divide the two terms of the second ratio of (785) by sin. PBC and reduce, by (11), (787) cos. C: cos. A : 1 sin. B cotan. PBC cos. B. Make the product of the means equal that of the extremes, and we have sin. B cos. C cotan. PBC: (788) sin. B cos. C cotan. PBCby transposition, cos. B cos. C= cos. A ; cos. Acos. B cos. C. (789) (790) cotan. C cotan. PBC sin. B sin. C (791) Divide by sin. B sin. C, and reduce by (11) = cos. Acos. B cos. C whence the value of the side a may be calculated, and in the same way either of the other sides. 66. Corollary. The equation (791) may be brought into a form more easy for calculation by logarithms, as follows. (792) 1 Subtract each member from unity and it becomes sin. B sin. C—cos. A cos. B cos. C COS. a sin. B sin. C But, by (104), changing the signs (793) - cos. (B+C) -cos. B cos. C + sin. B sin. C ; which, substituted in the numerator of (792), gives (794) 1 cos. a cos. (B+C) sin. B sin. c cos. A Now we have by (121), changing the signs, and letting S denote half the sum of the angles or S = { (A + B + C) ; M =¿ (A + B + C) = S, N = ( A + B + C) = S — A ; (795) (796) if we make in (795), SM {N= (797) 67. Corollary. The sides b and c may be found. by the two following equations which are readily deduced from (803), 68. Corollary. The equation (791) can be brought into another form equally simple for calculation. Add each member to unity and it becomes (806) 1 + cos. a= But, by (116), cos. A cos. B cos. C + sin. B sin. C· sin. B sin. C (807) cos. (BC) =cos. B cos. C+ sin. B sin. C; which, substituted in (806), gives (809) cos. (M+N) + cos. (MN) 2 cos. M cos. N; |