COS AR b2 + c2 — a2 2 bc --- (Art. 69), from which, taking the value of the square of the side opposite the angle in the formula, we have subtracting the former from the latter of these two equations and observing that in the right angled triangles OAD and OAE DO2- AD2: OA' and E02. AE20A2 Finding the value of cos a from this equation, we have DOXEO cos a- RXO A2 COS A= AD XAE substituting for o A its value R for D o its value sec b = R❜ = cos b (Art. 33), or if R = 1 COS a cos b cos c COS A= sin b sin c But the angle a of the plane triangle D A E is the same as the angle A of the spherical triangle (Spher. Geom., Prop. 4) hence, translating the above formula, The cosine of either angle of a spherical triangle is equal to radius square into the cosine of the side opposite, minus radius into the rectangle of the cosines of the adjacent sides, divided by the rectangle of the sines of the adjacent sides. The above formula will serve to calculate one of the angles of a spherical triangle when the three sides are given, if we employ the table of natural sines and cosines; but is unsuitable for the application of logarithms, in consequence of the sign-in the numerator requiring a subtraction to be performed, which operation is impracticable by means of logarithms. We shall therefore derive from this another formula, involving only multiplications, divisions, &c., of the trigonometrical lines contained in it, to which operations logarithms apply. 83. At Art. 74 were derived formulas (2) and (4) for the sum of the sines and difference of the sines of two arcs. In a similar manner two others may be derived for the sum and difference of the cosines. See (7) and (8) of Art. 12, App. I. These four forms would be as follows: The last is read, the difference of the cosines of any two arcs is equal to twice the sine of half their sum into the sine of half their difference. 84. A formula for the cosine of half an angle of a spherical triangle, in terms of the three sides, may now be derived. Resume from Art. 82, * We say either angle, because in the above demonstration no particular angle was selected. + Deduced similarly to (8) of Art. 72, by adding (5) and (6) of that article. i. e. The cosine of half an angle of a spherical triangle is equal to the square root of the sine of half the sum of the three sides into the sine of half the sum minus the side opposite, divided by the rectangle of the sines of the adjacent sides. In a similar manner may be derived a formula for the sine of half an angle in terms of the three sides. Since sin A2 sina cosa, by multiplying (1) and (2) a formula for sin A may be obtained analogous to (2) of Art. 73. Any of the above forms (1), (2), (3) may be employed for the solu tion of a spherical triangle when the three sides are given. (2) is read thus: the sine of half either angle of a spherical triangle is equal to radius* into the square root of the sine of half the sum of three sides of the triangle minus one of the adjacent sides, into the sine of half the sum minus the other adjacent side, divided by the rectangle of the sines of the adjacent sides. * Radius must be introduced as a factor for homogeneity. The blank form of (2) for examples will be the following. EXAMPLE I.-Given a 120° 17', b = 75° 3', c=48° 56'. Prop. 8, Note.) must be less than 90°, since the whole angle cannot exceed 180°. (Spher. Geom., The result is not ambiguous in the application of this formula, for half the angle EXAMPLE II. Given a 81° 17', b=114° 3', c = 59° 12'. Ans. 157° 39' 20', B = 25° 10′ 4′′-2, c19° 22' 56'8 This formula, for the solution of a spherical triangle when the three sides are given, is very convenient for calculation by logarithms. Applied to each of the angles separately, it will serve to determine them all. An interesting astronomical application of this case of solution is to finding the time of day and error of a watch by an altitude of the sun or other heavenly body. See Nautical Astronomy, p. 287. 85. It is proved (Spher. Geom., Prop. 13), that the three angles of a triangle being given, the triangle is determined. A formula for calculating either of the sides when the three angles are given, may be easily derived from that of the last article, by means of the polar triangles. It is necessary first to premise that the polar triangles of the whole range of triangles will include all possible triangles: for as each side of a triangle passes through all values from 180° to 0°, the opposite angle of the polar triangle will pass through all values from 0° to 180°. Wherefore whatever can be proved of the polar triangles of all possible triangles may be considered as proved for all triangles. Resume the equation before (1) of the preceding article. COSA = sin (a+b+c) sin (b + c-a) sin b sin c For the parts of the triangle in this formula, substitute their equivalents in the polar triangle. It will be remembered (Spher. Geom., Prop. 6), that each angle of a spherical triangle is the supplement of the side opposite in the polar triangle, and vice versâ ; hence if a', b', and c' represent the sides, and A', B', c', the angles of the polar triangle, we have A 180°- a' a = 180° A', b = 180° - B' and c 180° - c' putting these values of the letters a, a, b, and c in their places in the formula above, it becomes cos (1800-a') sin(5400 = (A' + B' + c ́)) sin † (180° — (B' + c' - ·A')) sin (180° — B') sin (180° — c ́) : But (180°-a') = (90° — a') and cos (90°-a') = sin a' (Art. 23); also (540° — (A'+ B' + c ) = 270° (A' + B'+C') and · § sin (2700(A'+ B'+ C')) = —cos (A' + B+ c'); also (180°— (B'+c' — A') = 90° — † (B' + c' — A') and sin (900 — † (B' + C' —A')) = cos(B' + c' — A'); also sin (180° — B) = sin B' (Art. 15) and sin (180° — c') = sin c'. |