For, by means of (188), tan (a + b) cos ≥ (A + B) : tanc cos (A — B), in which the second member is always positive, since and (AB) are each less than 90°, so that the factors of the first member, tan (a + b) and cos (A + B) must have the same sign. Therefore, (a + b) and 1⁄2 (A + B) are of the same species. 180. In the solution of oblique-angled spherical triangles, there are six cases, the data in them being, respectively, I. Two sides and an angle opposite one of them. II. Two angles and a side opposite one of them. III. Two sides and the included angle. IV. Two angles and the included side. V. The three sides. VI. The three angles. CASE I. 181. Given two sides and an angle opposite one of them. Let there be given, in the obliqueangled spherical triangle ABC, the sides a and b, and the angle A; to solve the triangle. To find B. We have, from (148), A C B a log sin Blog sin blog sin a log sin A. (238) To find C and c. We have, by Napier's analogies, (186) and (188), cos(a+b) cos (a - b) tan (4B), or, by logarithms, log cot Clog cos (a+b) — log cos (a - b) +log'tan -- (4+B), (239) log tan c = log cos (A+B) — log cos (A—B) ≥ ≥ +log tan (a+b), which determine 1⁄2 C and 1 c, and thence C and c. (240) In this case, since B is found from its sine, it will sometimes admit of two values, the one supplementary to the other. When B has two values, C and c must each have two corresponding values. Whether both values of B are admissible must be determined by one of the propositions of Art. 179. Thus (Prop. VI.), if b differs more from 90° than a, B must be of the same species as b, and there can be but one solution; but if b differs less from 90° than a, there may be two solutions. Or (Prop. VIII.), if only one of the supplementary values of B makes (A+B) of the same species as (a+b), there can be but one solution; but if both values of B fulfil that condition, there will be two solutions. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the side a equal to 63° 50', the side b equal to 80° 19', and the angle A equal to 51° 30'; to solve the triangle. Solution. By (238) we have a = 63° 50' b = 80° 19' A = 51° 30' ar. co. log sin 0.046958 B = 59° 15′ 57′′, or 120° 44′ 3′′ log sin 9.993768 log sin 9.893544 log sin 9.934270 As b differs less from 90° than a, both values of B are admissible, and we have 1 (b—a) — 8° 14′ 30′′, ≥ (a+b)=72° 4′ 30′′, (A+B)=55° 22′58′′ or 86° 7′2′′, and (B—A) = 3° 52′ 58′′ or 34° 37' 2". The cosines of (b-a) and (B—A) are the same as those of (a-b) and (A-B), respectively, by Art. 79. Hence, by (239) and (240), (B-A) ar. co. log cos+ 0.000998 or ar.co.log cos+ 0.084618 2. Given, in an oblique-angled spherical triangle, two sides equal to 99° 40′ 48′′ and 64° 23′ 15′′, and an angle opposite to the first of these equal to 95° 38′ 4′′; to find the other side and angles. Ans. Side, 100° 49′ 30′′; angles, 65° 33′ 10′′ and 97° 26′ 30′′. CASE II. 182. Given two angles and a side opposite one of them. Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A and B, and the side a; to solve the triangle. To find b. We have, from (148), To find C and c. We use equations (239) and (240), as in the last article. This case is exactly analogous to Case I., and gives rise to the same ambiguities, as may be shown by passing to the polar triangle. If B differs more from 90° than A, b must be of the same species as B, and there can be but one solution; but if B differs less from 90° than A, there may be two solutions. (Prop. VII. Art. 179.) Or, if only one of the supplementary values of b makes (a+b) of the same species as (A+B), there can be but one solution; but if both values of b fulfil that condition, there will be two solutions. (Prop. VIII. Art. 179.) EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the angle A equal to 135°, the angle B equal to 60°, and the side a equal to 155°; to find the other parts. Ans. C, 98° 3' 4" or 16° 57' 1"; b, 31° 10′ 17′′ or 148° 49' 43"; c, 143° 42′ 57′′ or 10° 2′ 6′′. 2. Given, in an oblique-angled spherical triangle, two angles equal to 97° 26′ 30′′ and 65° 33' 10", and the side opposite to the first equal to 100° 49′ 30′′; to find the other parts. CASE III. 183. Given two sides and the included angle. Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the sides a and b, and the included angle C; to solve the triangle. To find A and B. By means of Napier's analogies (186) and (187), we have b) cot C, tan (AB) = sin(a - b) cotC; or, by logarithms, log tan (A+B)= log cos (a-b) - log cos(a+b) 1⁄2 1 log tan (A-B) = log sin (ab) — log sin (a+b) which determine (A+B) and (A-B). The sum of these values gives A, and the second subtracted from the first gives B. To find c. We use equation (240), as in the first two cases. The value of c might also be obtained by (147) or (149); but as it is thus determined from its sine, it would be necessary to remove the ambiguity by means of the principles contained in Art. 179. As A, B, and c may all be found by means of tangents, there can be but one value for each. It will be observed that ≥ (A+B) must always be of the same species as (a+b). (Prop. VIII. Art. 179.) EXAMPLES. 1. Given, in an oblique-angled spherical triangle ABC, the side a equal to 70°, the side b equal to 38° 30′, and the included angle equal to 31° 34′ 26′′; to solve the triangle. (a + b) Solution. = 54° 15', (a — b) = 15° 45', and C= 15° 47′ 13′′; then, Ans. Angle A, 130° 3' 11"; angle B, 30° 28' 11"; side c, 40°. 2. Given, in an oblique-angled spherical triangle, an angle equal to 48° 36', and the two adjacent sides equal to 112° 22′ 58′′ and 89° 16' 53"; to find the other parts. |