cos.b R2 cos B+R cos A cos C, in the case of A=100°, give sin B sin C cos a=] R cos B cos C, and sin C cos b-R cos B, or cot B cot CR cos a, sin C cos b=R cos B. (5) (6) These are the six equations upon which the solution of spheri cal triangles depends. LXXXIII. We shall terminate these principles by demonstrating Napier's Analogies which serve to simplify several cases in the solution of spherical triangles. By combining the values of cos A and cos C expressed in terms of a, b, c, we have already (Art. 80.) obtained the equa tion R cos A sin c-R cos a sin b-cos C sin a cos b. By a simple permutation, this gives R cos B sin c-R cos b sin a-cos C sin b cos a. Hence by adding these two equations, and reducing, we shall have sin c (cos A+cos B)=(R-cos C) sin (a+b). But since sin C sinA sin B we have Asin sin C (sin A+sin B)=sin C (sin a+sin b), and sin C (sin A-sin B)=sin C (sin a-sin b). Dividing these two equations successively by the preceding one; we shall have Hence, two sides a and b with the included angle C being given, the two other angles A and B may be found by the analogies, cos (a+b): cos(a-b): cot C: tang)(A+B) sin(a+b) : sin(a—b): : cotC: tang)(A—B). If these same analogies are applied to the polar triangle of ABC, we shall have to put 200-A, 200°-B, 200°-a, 200°-b, 200°-c, instead of a, b, A, B, C respectively; and for the result, we shall have these two analogies, cos (A+B): cos(A—B) :: tang1⁄2c : tang\(a+b) sinд(A+B) : sin↓(A—B) : : tang1⁄4c: tang(a—b), by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies. SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. The solution of spherical triangles includes six general cases, which we shall now explain in succession. FIRST CASE. LXXXIV. The three sides a, b, c being given, any angle, for example the angle A opposite the side a, will be found by the formula: LXXXV. Given two sides a and b with the angle A opposite to one of them, to find the third side c, and the other two angles B and C. First. The angle B is found from the equation sin B= sin A sin b sin a Secondly. To find the angle C, we must solve the equation, cot A sin C+ cos C cos b = cot a sin b. For this purpose, take an auxiliary angle ?, such that we may of cot A, being substituted in the equation to be solved, gives cos b sin (cos 4 sin C+sino cos C)=cot a sin b; whence we obtain By this artifice, the two unknown terms of the equation are now reduced to one, from which it is easy to find the angle C. Thirdly. The side c will be found by the equation It might also be determined directly, by solving the equation R cos b cos c+cos A sin b sin c = R2 cos a. This second Case may have two solutions, like the analogous Case in rectilineal triangles. THIRD CASE. LXXXVI. Given two sides a and b, with the included angle C, to find the other two angles A and B and the third side c. First. The angles A and B are found by these two equa in which the second members may be reduced to a single term, by means of an auxiliary quantity. It is simpler, however, in this case, to make use of Napier's analogies, which give Secondly. Knowing the angles A and B, the third side c may be computed by the equation sin c=sin a. termining c directly, we have the equation sin C ; but for de sin A R2 cos c = sin a sin b cos C+R cos a cos b. Assume the auxiliary quantity, such that sin b cos C = cos C tang b; we shall have cos b tang 4, or tang = R LXXXVII. Given two angles A and B with the adjacent side c, to find the other two sides a and b with the third angle C. First. The two sides a and b are given by the formulas They may, however, be computed more easily by Napier's analogies, namely, sin C equation Secondly. Knowing a and b, we shall find C by the equation sin c sin A ; but C may also be found directly, by the R2 cos C=cos c sin A sin B-R cos A cos B. This case and the preceding one offer no ambiguity. LXXXVIII. Given two angles A and B, with the side a opposite one of them, to find the other two sides b and c and the third angle C. First. The side b is found by the equation Secondly. The side c depends on the equation -cos B cos ccot A sin B. cot a sin Thirdly. The angle C is found by solving the equation cos a sin B sin C-R cos B cos C=R2 cos A. This fifth Case, like the second, is susceptible of two solutions, as happens in like manner in the analogous Case of rectilineal triangles. SIXTH CASE. LXXXIX. The three angles A, B, C being given, we can find any side, for example the side opposite the angle A, by the formula +C—A)). -cos (A+B+C) cos (B+C- It may be observed, that of these six general Cases, the last three might have been deduced from the first three, by the property of polar triangles; so that, properly speaking, there are but three different cases in the general solution of spherical tri |