79. Let ABC represent any spherical triangle, and the centre of the sphere on Draw the which it is situated. D A B CD and CE, these lines will be respectively perpendicular to OA and OB (B. VI., P. VI.), and the angles CDP and CEP will be equal to the angles A and B respec In the right-angled triangle OLD, OL OD cos DOL = cos b cos c. The right-angled triangle PQD has its sides respectively perpendicular to those of OLD; it is, therefore, similar to it, and the angle QDP is equal to c, and we have, substituting this value in (2), we have, and now substituting these values of OE, OL, and QP, in (1), we have, cos a = cos b cos c + sin b sin c cos A In the same way, we may deduce, cos b = cos a cos c + sin a sin c cos B COS C = cos a cos b + sin a sin b cos C That is, the cosine of either side of a spherical triangle is equal to the rectangle of the cosines of the other two sides plus the rectangle of the sines of these sides into the cosine of their included angle. 80. If we represent the angles of the polar triangle of ABC, by A', B', and C', and the sides by a', b'' and c', we have (B. IX., P. VI.), Substituting these values in Equation (3), of the preceding article, and recollecting that, cos A' = cos B' cos C'sin B' sin C' cos a'; or, changing the signs and omitting the primes (since the preceding result is true for any triangle), cos Asin B sin C cos a cos B cos C (1.) That is, the cosine of either angle of a spherical triangle is equal to the rectangle of the sines of the other two angles into the cosine of their included side, minus the rectangle of the cosines of these angles. 81. From Equation (3), Art. 79, we deduce, If we add this equation, ber 1, and recollect that sin b sin c (1.) member by member, to the num 1+ cos A, in the first member, is equal to 2 cos2 A (Art. 66), and reduce, we have, cos acos (b+c) = 2 sin (a + b + c) sin †(b + c − a), Equation (2) becomes, after dividing both members by 2 and extract the square root of both members, we have, That is, the cosine of one-half of either angle of a spherical triangle, is equal to the square root of the sine of one-half of the sum of the three sides, into the sine of one-half this sum minus the side opposite the angle, divided by the rectangle of the sines of the adjacent sides. If we subtract Equation (1), of the preceding article, member by member, from the number 1, and recollect that, Dividing the preceding value of sin A, by cos A, Is = 270° — †(A'+B'+C'), s — a = 90°—{(B'-+ C'—A'). Substituting these values in (3), Art. 81, and reducing by the aid of the formulas in Table III., Art. 63, we find, sin ta' = Placing cos (A'+B'+C') cos B'+C'-A') sin B' sin C ¿(A'+B′+C') = 8; whence, (B'+C'—A') = †S — A'. Substituting and omitting the primes, we have, In a similar way, we may deduce from (4), Art. 81. 83. From Equation (1), Art. 80, we have, sin 4 cos A+ cos B cos C = sin B sin C cos a = sin C sin b cos a; sin a (1.) since, from Proportion (1), Art. 78, we have, Also, from Equation (2), Art. 80, we have, sin A = cos B+cos 4 cos sin A sin C cos b = sin C sin a cos b. sin a (2.) |