since the angles B and C are both less than 180°, their : hence, cos B must have the cos C must have the sines must always be positive the same sign as cos b, and same sign as COS C. This can only be the case when B is of the same species as b, and C of the same species c; that is, the sides about the right angle are always of the same species as their opposite angles. as From Formula (1), we see that when α is less than 90°, or when cos a is positive, the cosines of band C will have the same sign; that is, b and c will be of the same species. When a is greater than 90°, or when cos a is negative, the cosines of b and c will be contrary; that is, b and c will be of different species: hence, when the hypothenuse is less than 90°, the two sides about the right angle, and consequently the two oblique angles, will be of the same species; when the hypothenuse is greater than 90°, the two sides about the right angle, and consequently the two oblique angles, will be of different species. These two principles enable us to determine the nature of the part sought, in every case, except when an oblique angle and the opposite side are given, to find the remaining parts. In this case, there may be two solutions, one solution, or no solution at all. C' B'A' = BA, B'C' = BC, and join A' and C" by the arc of a great circle: then, because the triangles BAC and B'A'C' have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the remaining parts will be equal, each to each; are that is, A'C' AC, and the angle A' equal to the angle A hence, the two triangles BAC, B'A'C', right-angled; they have also one oblique angle and the opposite side, in each, equal. Now, if b < B, there will evidently be two solutions, the sides including the given B C C' a B' angle, in the one case, being supplements of those which in clude the given angle, in the other case. If b = B, the triangle will be bi-rectangular, and there will be but a single solution. If b> B, the triangle cannot be constructed, that is, there will be no solution. SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. 76: In a right-angled spherical triangle, the right angle is always known. If any two of the other parts are given, the remaining parts may be found by Napier's rules for circular parts. Six cases may arise. There may be given, I. The hypothenuse and one side. II. The hypothenuse and one oblique angle. IV. One side and its adjacent angle. V. One side and its opposite angle. In any one of these cases, we select that part which is either adjacent to, or separated from, each of the other given parts, and calling it the middle part, we employ that one of Napier's rules which is applicable. Having determined a third part, the other two may then be found in a similar manner. It is to be observed, that the formulas employed are to be rendered homogeneous, in terms of R, as explained in Art. 30. The method of proceeding will be readily understood from a few examples. From Formula (10), Art. 74, we have, log cos C = log cot a + log tan 6 — 10; From Formula (2), Art. 74, we have, log sin c = log sin a log sin C 10; log cos B = log sin C + log cos b — 10; Ans. c = 109° 46′ 32′′, B = 40° 29′ 50′′, C = 102° 41′ 33′′. 2. Given b = 51° 30′, and B = 58° 35', to find Because b < B, there are two solutions. OPERATION. From Formula (7), we have, log sin c = log tan b + log cot B - 10; a In a similar manner, all other cases may be solved. 4. Given b = 155° 27′ 54′′, and c— 29° 46′ 08′′, a, B, and C. find Ans. find to a = 142° 09′ 13′′, B = 137° 24′ 21′′, C = 54° 01′ 16′′. 5. Given α, b, c = 73° 41′ 35′′, and B = 99° 17′ 33′′, to and C. and Ans. a = 92° 42′ 17′′, b = ̧99° 40′ 30′′, C′ = 73° 54′ 47′′. 115° 20′, and B = 91° 01′ 47′′, to find 7. Given B = 47° 13′ 48′′, and C 126° 40′ 24′′, to find a, b, and C. Ans. a = 133° 32′ 26′, b = 32° 08′ 56′′, c = 144° 27′ 03′′. In certain cases, it may be necessary to find but a single part. This may be effected, either by one of the formulas given in Art. 74, or by a slight transformation of one of them. C. Regarding Thus, let a and B be given, to find C. 90° a, as a middle part, we have, from which C may be found. In like manner, other cases may be treated. |