and as the three equations at the end of the last article together contain all the six parts of a triangle, we may obviously deduce from these equations an equation involving any four of these parts, by eliminating the other two. If for instance we wish to obtain an equation involving the three angles A, B, C, and the side b; we may combine the first equation with the second, to obtain an equation free from a, and then combine the first and third equations, to form another equation free from a: having thus found two equations from which a is excluded, we may finally combine these to form an equation from which c also shall be eliminated; and this will be the equation required. Now the only cases that here require to be distinguished, are the following; namely, those in which the four parts concerned are (1) The three sides and one angle (2) Two sides and the opposite angles (3) Two sides, the included angle and one of the opposite angles (4) The three angles and one side. There are six equations belonging to the third case and three to each of the others. The formulæ for the first case have been given in the last article; and from them the formula for the other cases are now to be deduced. 29. To find an equation involving two sides a, b, and their opposite angles A, B, it is sufficient to combine the first two equations at the end of Art. 27, so as to eliminate c, since C is not found in either of them. This might be done by first forming a new equation containing sin c, but free from cos c; and then another equation containing cos c but not sin c. From the former equation the value of sin c, and from the latter that of cos c would be known: and the squares of these values made equal to unity, would give the equation required. But there is another method of procedure, which though less direct and obvious, is more simple, and has the advantage of bringing to view formulæ that are useful for other purposes. It is as follows: From equation (1) Art. 27, cos A: cos a-cos b cos sin b sin c COS a hence 1+cos A = ; ·cos b cos c+sin b sin c cos a-cos.(b+c) (Art. 26, Equat. 11). sin b sin c If for convenience of reduction, we substitute x-y for a, and x+y for b+c, then cos a―cos (b+c)=cos(x − y)—cos(x+y) =2 sin x sin y (Arts. 26, Equats. 11, 12). If we now substitute for x and y their values in terms of a, b, and c, 1+cos A= 2 sin (a+b+c) sin (b+c-a) sin b sin and if we put h for half the sum of the sides a, 2 sin h sin (h-a) sin b sin c cos b cos c+sin b sin c 1-cos A= and by a method similar to the preceding, we find 2 sin (a+b-c) sin (a+c—b) sin b sin c 1-cos A= or (2) 1-cos A 2 sin (h—b) sin (h-c) sin b sin c If we multiply the corresponding members of equations (1) and (2), substitute sin2 A for 1-cos2 A, and extract the square root, we have the equation (3) sin A 2 √sin h sin (h—a) sin (h—b) sin (h—c); sin b sin c and if for the sake of symmetry we divide each side by sin a, we shall obtain (4) sin A__2 √ sin h sin (h—a) sin (h—b) sin (h—c) ̧ sin a sin a sin b sin c As the second member involves all the three sides of the triangle in the same way, it must be equal to We have here equations each involving two sides of a triangle and the opposite angles; as was required. The theorem answering to them is this: The sines of the sides of a spherical triangle are as the sines of the opposite angles. It has been otherwise proved in Art. 15. 30. To find next an equation involving two sides a and b, the included angle C, and one of the opposite angles B: first substitute for cos c in the second equation at the end of Art. 27, its value as given by the third equation; the result will be cos bcos a cos b + cos a sin a sin b cos C+ sin a sin c cos B; and if we transpose cos2 a cos b, and substitute sin2 a for 1-cos2 a, the first member of the equation will be cos b sin2 a; and if each member be divided by sin a, then cos b sin a=cos a sin b cos C+sin c cos B. But by equation (5) of the last article, sin c=sin b sin C sin B substituting this value, dividing by sin b, and putting cotangents for cosines divided by sines, we obtain (1) cot b sin a=cos a cos C+sin C cot B. In like manner may be obtained the following equations: (2) cot a sin b=cos b cos C+sin C cot A, (3) cot sin b=cos b cos A+ sin A cot C, (4) cot b sin c=cos c cos A+sin A cot B, (5) cot a sin c=cos e cos B+sin B cot A, (6) cot c sin a=cos a cos B+sin B cot C; which are all the equations that taken singly involve two sides, the included angle and one of the opposite angles in a triangle. 31. It remains to find an equation involving the three angles and one of the sides of a triangle. If with the equation cos b sin a=cos a sin b cos C+sin c cos B found in the preceding article, we combine the analogous one, cos a sin b=cos b sin a cos C+sin c cos A, (obtained from the former by interchanging the sides a and b, and the angles A and B,) and eliminate sin b, the result is cos b sin a=cos b sin a cos C+ sin cos A cos C + sin c cos B. And if we transpose cos b sin a cos C, substitute sin' C sin a for 1-cos2 C, divide by sin c, and substitute for -sin O sin c its value sin A obtained from equation (4) of the last article, we shall find that cos b sin A sin C=cos A cos C+cos B, and by transposition, cos B=—cos A cos C+sin A sin C cos b: which shows that The cosine of one of the angles of a spherical triangle is equal to the product (taken negatively) of the cosines of the other two angles added to the product of their sines multiplied into the cosine of the intermediate side. There are three equations comprised in this theorem ; namely, (1) cos A=-cos B cos C+sin B sin C cos a (3) cos C=cos A cos B+sin A sin B cos c. 32. The equations which have now been obtained will give the value of either of the six parts of a triangle in terms of the other three; for which purpose fifteen equations are evidently requisite: but all of them except the equations numbered (5) in Art. 29 are ill suited for logarithmic calculations. We must therefore seek for other formulæ more convenient. By Art. 210, Day's Trigon. 1+cos A=2cos2 A, 1-cos A=2 sin2 A. Then by substitution in equations (1) and (2) of Art. 29, and by reduction, we obtain the following results. |