Solution. by (561), By (558), B, cotan. 9.55097 n ; a sin. 9.35871; b tang. 9.80774 n. 2. Given, in the spherical right trangle (fig. 2.), h=32° 34′ and A=44° 44', to solve the triangle. Ans. B=50° 8', = 22° 16′, b 24° 24'. 22. Problem. To solve a spherical right triangle, when its hypothenuse and one of its legs are known. Solution. Let ABC (fig. 2.) be the triangle; h the given hypothenuse, and a the given leg. First. To find the opposite angle A; a is the middle part, and co. A and co. h are the opposite parts. Hence, by (475), sin. a = cos. (co. h) cos. (co. A); or sin. a sin. h sin A; and by (7), (565) sin. A sin. a sin. a cosec. h. * The letter n placed after a logarithm indicates it to be the logarithm of a negative quantity, and it is plain that when the number of such logarithms to be added together is even, the sum is the logarithm of a positive quantity; but if odd, the sum is the logarithm of a negative quantity. (566) Secondly. To find the adjacent angle B; co. Bis the middle part, and co. h and a are the adjacent parts. Hence, by (474), (567) or sin. (co. B.) = tang. a tang. (co. h), cos. B tang. a cotan. h. Thirdly. To find the other leg b; co. h is the middle part, and a and b are the opposite parts. Hence, by (475), (568) cos. h = cos. a cos. b; 23. Scholium. The question is impossible by (505), (570) when the given value of the hypothenuse differs more from 90° than that of the leg. (570 24. Solution. When h and a are both equal to 90°, it may be shown, as in (563), that the values of B and b are indeterminate. EXAMPLE. Given, in the spherical right triangle (fig. 2), a=141° 11', and h = 127° 12'; to solve the triangle. 25. Problem. To solve a spherical right triangle, when one of its legs and the opposite angle are known. Solution. Let ABC (fig. 2.) be the triangle; a the given leg, and A the given angle. First. To find the hypothenuse h; a is the middle part, and co. h and co. A are the opposite parts. Secondly. To find the other angle B; co. A is the middle part, and a and co. B are the opposite parts. Thirdly. To find the other leg b; b is the middle part, and a and co. A are the adjacent parts. Hence, by (474), AB and AC, to the point of meeting A', both of which satisfy the conditions of the problem. For the side BC or a, and the angle A, or by art. 2 its equal A', belong to both the triangles. (576) Now ABA' and ACA' are semicircumferences, since the line AA' joining their points of intersection (577) is the line of intersection of their planes, and therefore passes through the centre of the sphere and is a diameter. Hence h', the hypothenuse of A'BC, is the (578) supplement of h; b' is the supplement of b; and A'BC is the supplement of ABC. One set of values, then, of the unknown quantities, given by the tables, as in (562), correspond to the triangle ABC, and the other set to A'BC. 27. Corollary. When the given values of a and A are equal, (572); (574), and (575) become sin. h=1 (579) (580) (582) 28. Corollary. When a and A are equal to 90°, the values of b and B are indeterminate, as in (563). 29. Scholium. The problem is, by (517), impossible, when the given values of the leg and its opposite angle are such that one surpasses 90 while the other (583) does not, or that one is equal to 90° while the other differs from 90°; and, by (519), it is impossible when the given value of the angle differs more from 90° than that of the leg. EXAMPLE. Given, in the spherical right triangle (fig. 2.), a = 35° 44' and A = 37° 28'; to solve the triangle. 30. Problem. To solve a spherical right triangle, when one of its legs and the adjacent angle are known. Solution. Let ABC (fig. 2.) be the triangle; a the given leg, and B the given angle. First. To find the hypothenuse h; co. B is the middle part, and co. h and a are adjacent parts. Hence, by (474), Secondly. To find the other angle A; co. A is the middle part, and co. B and a are opposite parts. Hence, by (475), cos. A cos. a sin. B. Thirdly. To find the other leg b; a is the middle part, and co. B and bare adjacent parts. Hence, by (474), (586) EXAMPLE. Given, in the spherical right triangle, (fig. 2.), a = 118° 54 and B = 12° 19'; to solve the triangle. Ans. h = 118° 20′, A = 95° 55', b= 10° 49′. |