Case III. 34. Of these four things, viz. two sides a and c, and two angles в and c, one opposite, the other adjacent ; three being given, to find the fourth. This case presents four problems. 1. Given a, c, B; to find c. Determine an arc 4' by this condition, cot ccot cos B, or cot c COS B' cot '.... (15.) Substitute this value of cot c for it in the third of equa. (4); it will become sin a cot 'cos B = cos α COS B + sin в cot C; (cot 'sin a whence cot c= cos a) cos B sin B Note. The equa. (15) is akin to equa. (8); showing that the operation here performed is equivalent to letting fall a perpendicular arc from the angle A to the base a; the subsidiary arc being the segment adjacent to the angle B. 2. Given B, C, c; to find a. Here' must be found by equa. (15), and then from equa. (16) we have cot c sin ' .... cot B (17.) Substitute this value of cot c for it in the third of equa. (4), and it will reduce to 4. Given a, c, c; to find в. Determine " from equa. (18); then 35. Of these four things, viz. the three angles and a side, suppose c, any three being given, to find the fourth. This comprises three problems. 1. Given the three angles; to find a side. Suppose the first of equa. (11) to be applied to the solution of the supplemental triangle, by changing a, b, c, and c, into a', b, c, and c'. Then, to bring it back to the triangle proposed, let there be substituted for a', b', c', and c', the corresponding values 180° — A, 180° B, 180° c, 180°. --- C. Those equations will then be transformed into the following, applicable to the present problem. cot a = cotb= cot c -B+ C) cos (A + B c) cos & (A COS (B+C - A) COS (A + B (A + C (21.) ~COS (A + B + C) COS 1⁄2 (A + B − The following are the expressions for the sines of the half angles, viz. Note. In these expressions, although the denominators are negative, the whole fractions under the radical are always positive. The expressions for the tangents and cosines are omitted, to save room, 2. Given A, B, c; to find c. Here, by applying in like manner, the equations (12), (13), to the supplemental triangle, we shall have (23.) from which the subsidiary angle 4 may be determined; and thence On the Analogies of Napier. 36. These are four simple and elegant formulæ discovered by the celebrated inventor of logarithms, of which two serve to determine any two angles of a spherical triangle, by means of the two opposite sides and their included angle; while the other two serve to determine any two sides, by means of the opposite angles and their contained side. Thus, therefore, they together with equa. (1), will serve for the solution of all the cases of oblique spherical triangles. The investigation of these analogies may be given, as below. If from the first of equa. (2), cos c be exterminated, there will result, after a little reduction, cos A sin c = cos a sin b - cos c sin a cos b: cos c sin b cos a: and by a simple permutation of letters, sin a sin b = sin A sin B and reducing, we have cos c) sin (a+b). sin c sin c Freeing these equations from their denominators, and respectively adding and subtracting them, there results, sinc (sin A+ sin B) = sin c (sin a + sin b), and sin c (sin A- sin B) = sin c (sin a sin b). Dividing each of these two equations by the preceding, there will be obtained, The above equations expressed as analogies, are cos(a+b): cos (a b) :: cote: tan§ (A + B) sin (a+b): sin (a - b) :: cot c: tan (A B). These analogies being applied to the supplemental triangle, by putting 180°-A, 180°-B, &c. for a, b, &c. we have b). COS (A + B): cos (A — B) :: tan c: tan (a + b) sin (A + B): sin (A — B) :: tan c: tan (a 37. From a due consideration of the four results, analogies it 1st. That (AB).< 90°, or that the difference of two angles of a spherical triangle is less than 180°. 2dly. That (a + b) and (A + B) are always of the same affection. 3dly. That the difference of two sides is always less than 180°. 4thly. That (ab) and (AB) have always the same sign; whence it follows, that the greatest angle is opposite to the greatest side, and reciprocally. To these it may be added, 5thly. That the least angle is opposite to the least side, and the mean angle, to the mean side. One or other of these observations will serve to remove the ambiguity in the doubtful cases, where either a, b, and в, or A, B, and b, are given. 38. We may now collect the most commodious theorems, and present in one place all that will be usually required in the solution of oblique angled spherical triangles. |