since the angles B and C are both less than 180°, their sines must always be positive hence, cos B must have the same sign as cos b, and the cos C must have the 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 as c; that is, the sides about the right angle are always of the same species as their opposite angles. From Formula (1), we see that when α is less than 90°, or when COS a is positive, the cosines of b and 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 solu tion, or no solution at all. Let BAC be a right-an gled triangle, in which C' B and b are given. Prolong the sides BA and BC till they meet in B'. Take = 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; that is, A'C' angle A = AC, and the angle A' equal to the hence, the two triangles BAC, right-angled; they have also B'A'C', are C' B A' ly be two solutions, the sides including the given angle, in the one case, being supplements of those which include the given angle, in the other case. Ifb B, the triangle will be bi-rectangular, and there = will be but a single solution. Ifb differs less from 90° than 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. This is done by simply multiplying the radius of the Tables, R, into the middle part. log cot a (105° 17′ 29′′) log cos Clog cot a + log tan b-10; 9.436811 Formula (2), Art. 74, gives for c, middle part, log sin clog sin a+ log sin C-10; log sin a (105° 17′ 29′′) log sin C (102° 41′ 33′′) 9.984346 9.989256 9.973602... c = 109° 46' 32". 2. Given b = 51° 30′, and B = 58° 35', to find c, a, and C. Because b < B, there are two solutions. OPERATION. Formula (7), gives for c, middle part, log sin c = log tan blog cot B-10; Formula (1), gives for 90°-a, middle part, Formula (10), gives for 90° - C, middle part, log cos C = log tan blog cot a 10; In a similar manner, all other cases may be solved. 3. Given a = 86° 51′, and B = 18° 03′ 32′′, to find b, c, and C. Ans. b 18° 01' 50", c = 86° 41′ 14′′, C = 88° 58′ 25′′. 4. Given b = 155° 27' 54", and c = 29° 46′ 08′′, to a = 142° 09′ 13′′, B = 137° 24′ 21′′, C = 54° 01′ 16′′. find Ans. 5. Given find α, b, c = 73° 41′ 35′′, and B = 99° 17′ 33′′, to and C. Ans. a = 92° 42′ 17′′, b = 99° 40′ 30′′, C′ = 73° 54′ 47′′. 6. Given b = 115° 20', and B = 91° 01' 47", to find 7. Given B = 47° 13′ 43′′, and C126° 40' 24", find a, b, and C. to 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. 90° Thus, let α and B be given, to find C. Regarding a, as a middle part, we have, log cos a + (a. c.) log cot Blog cot C; from which C may be found. In like manner, other cases may be treated. |