This principle, which, being applied successively to the three angles, affords three equations, is sufficient for solving all the problems of spherical trigonometry: it has the same generality of application in regard to spherical triangles, that Art. 45. has in regard to rectilineal ones. For, since we have always three given elements, by means of which the other three are to be determined, this principle will evidently furnish the equations necessary for solving the problem; equations which it is the province of analysis to develope still farther, in order to deduce from them, according to the different cases, the formulas which are most simple and best adapted for logarithmic calculations. LXXVII. The principle in question being absolutely general, it must include all the other principles relating to spherical triangles, and particularly the principle explained in Art. 75. Of this it will be easy to satisfy ourselves. Accordingly, the equation cos C R2 cos c R cos a cos b sin a sin b -R⭑cos2 c R2 sin2 a sin2 b-R2cos2 a cos2 b+ 2R3 cos a cos b cos c— sin2 a sin2 b Now sin2 a sin2 b=(R2 — cos2 a) (R2 — cos2 b) = R⭑— R2 cos2 a -R2 cos2 b + cos2 a cos2 b. Hence by substituting and extracting the root, we shall have sin C=1 Ꭱ sina sinb (R1-R2 cos2 a-R2 cos2 b-R2 cos2 c+2R cos a cos b cos c). For the sake of brevity, put Z= √( R1 — R2 cos2 a-R2 cos2 b-R2 cos2 c + 2R cos a cos b cos c), we shall then have tity Z does not change when a permutation is made between two of the quantities a, b, c; hence we have LXXVIII. The values we have just found for cos C and sin C, may serve for discovering the angles of a spherical triangle, when its three sides are known; though some other formulas are more convenient for logarithmic calculation. Thus, if in the formula R2- Ř cos C = 2 sin2 1 C, the value of cos C is substituted, we shall have 2 sin2 C R2 cos C cos a cos b+sin a sin b ·R cos c sin a sin b R cos P -R cos c= The numerator of this expression is reducible to R cos (a-b) cos c; now by the formula (Art. 28) R cos q 2sin § (p+q) sin (p q), we find R cos (a—b) · b + a) sin § (c. a+b); hence 2 sin (c We might evidently obtain similar formulas for expressing sin A and sin B, by means of the three sides a, b, c. LXXIX. The general problem of spherical trigonometry consists, as we have already said, in determining three of the six quantities A, B, C, a, b, c, by means of the other three. To effect this, we must have equations among four of those quantities taken in every possible order: now six quantities," when combined four by four, or two by two, give 6×5 1×2 or 15 com binations; hence there will be fifteen equations to form: though considering only such of the combinations as are essentially different, these fifteen equations are reduced to four. Thus, 1. We have the combination a b c A, which, by changing the letters, includes a b c A, abc B, a b c C. 2. The combination a b A B, from which there result a b A B, bc B C, a c A С. 3. The combination a b A C, which includes the six a b A C, a b B C, a c A B, a c BC, b c A B, b c A C. 4. Lastly the combination a A B C, which includes the three a ABC, bA B C, c A B C. In all, therefore, we have fifteen different combinations, but only four of them essentially different. LXXX. The equation cos A = R3 cos a without any change, represents the first combination a b c A, and those which depend upon it. To form the equation corresponding to the combination ab AB, we must eliminate c from the two formulas, which give the values of cos A and cos B. This elimination has already been performed (Art. 77.), and the result was The third combination is formed of the relation which subsists among a, b, A, C. Here having the two equations -R cos b cos c, cos A sin b sin c=R2 cos ɑ— cos C sin b sin a=R2 cos c-R cos b cos a, we shall first eliminate cos c from them; which will give R cos Asinc + cos C sin a cos b = R cos a sin b: then inserting in this cot A sin C+cos C cos b-cot a sin b. Finally, in order to discover the relation between A, B, C, a, we consider that, in the preceding equation, the term cot a sin b -R cosa sin b sin a sin B =R cos a ; hence, multiplying this equation by sin A, we shall have R cos A sin C=R cos A sin B—sin A cos C cos b. If in this equation we mutually change the letters A and B, also a and b, we shall have R cos B sin C=R cos b sin A-sin B cos C cos a. And from these latter two, excluding cos b, we deduce R2 cos A sin C+R cos B sin C cos C=cos a sin B sin2 C. and which is the required relation between A, B, C, a, or the fourth equation requisite for solving spherical triangles. LXXXI. This last equation between A, B, C, a presents a striking analogy with the first between a, b, c, A; the reason of which must be sought for in the properties of polar or supplemental triangles. We have already seen that the triangle having A, B, C for its angles, and a, b, c for its opposite sides, always corresponds to a polar triangle whose sides are 200°-A, 200°-B, 200°-C, its opposite angles being 200°-a, 200°—b, 200°—c. Now the principle of Art. 76, when applied to this latter triangle, gives cos (200°-a)= R2 cos (200°—A)—R cos (200°—B) cos (200°—C). sin (200°-B) sin (200°-C) which may be reduced to cos a R2 cos A+R cos B cos C sin B sin C as we have found by another method. This formula immediately solves the case where it is required to determine a side by means of three angles; but, in order to obtain a formula more suitable for logarithmic calculation, we may substitute the value of cos a in the equation cos a 2 sin2 a 1 which will give us sin2 a = sin B sin C-cos B cos C-R cos A -R cos (B+C)-R cos A 2 sin B sin C 2 sin B sin C And because (Art. 28.) we have generally R cos p+R cos q= 2 cos(p+q) cos (p-q), this equation is reducible to sin2 acos(A+B+C) cos (B+C—A); where it must be observed that the second member, though under a negative form, is nevertheless always positive. For, generally, sin x cos 100°—cos x sin 100° R we have sin (x-100°)— hence —cos (A+B+C)= sin (A+ B+C A+B+C +C_1003), =C08; a quantity which is always positive, because A+B+C being always included between 200° and 600°, the angle (A+B+C) -100°isincluded between zero and 200°: likewise cos (B+C-A) is always positive, because B+C-A cannot exceed 200°; for, in the polar triangle, the side 200°-A is less than the sum of the other two 200°-B, 200°C; hence we have 200°-A 400° -B-C, or B+C-A 200°. Being thus assured that our result will always be positive, for determining a side by means of the angles we shall have the formula sin la R -COS A+B+C COS B+C-A LXXXII. Before proceeding farther, it may be observed, that from these general formulas we might deduce the formulas which relate to right-angled spherical triangles. For this purpose, we shall make A=100°, both in the four principal formulas and in the formulas derived from them by permutation of the letters. And in the first place, by this substitution, the equation cos A sin b sin C-R2 cos a— -R cos b cos c will give The equations derived from this general equation do not contain A, and therefore do not give any new relation in the case of A=100°. The equation cot A sin C+cos C cos b=cot a sin b, in the case of A=100°, gives cos C cos b=cot a sin b, or The derived equation cot C sin A+cos A cos b-cot c sin b, in the same case, gives R cot C=cot c sin b, or Lastly, the fourth principal equation sin B sin C cos a= R2 cos A+R cos B cos C, and its derived equation sin A sin C |