...- A', 180° - B', 186° - c', &c. and for the angles A, B, C, 180° - a', 180° - b', &c. we shall have — cos A' = cos B' cos c' — sin B' sin c' cos a'. Here again we have three symmetrical equations applying to any spherical triangle, viz. COSA =...

...1st. Let all the sides be < 180, but A' >180°, Fig. 14. The formula bemg true for the triangle ABC, we have cos a = cos b cos c + sin b sin c cos (360° — A') or in the triangle A'B'C', by PI. Trig. (76), cos a' = cos b' cos c' -f- sin b' sin...

...1800 - C, A1 = 180° - a, B' = 180° - b, C' = 180° - c. Since equations (1) are equally applicable to the polar triangle, we have, cos a' — cos b• cos c' + sin V sin c' cos A' : substituting for a•, b•, c• and A', their values from the polar triangle, we...

...would evidently be similar, only interchanging b and c, B and C. By the fundamental formulae we should have cos a = cos b cos c + sin b sin c cos A sin b cos c — cos b sin c cos A cot (7= sn and denoting the auxiliary angle in this case by %,...

...180° - <7, A' = 180° -a, B' = 180° - 6, C" = 180° - c. Since equations (1) are equally applicable to the polar triangle, we have, cos a' = cos b' cos c' + sin b' sin c' cos A1 : substituting for a', b', c' and A', their values from the polar triangle, we have, — cos A =...

...angle, plus the cosine of the second side into the cosine of the included angle. By (150) and (152) we have cos a = cos b cos c -\- sin b sin c cos A, cos c = cos a cos b -\- sin a sin b cos C; and by means of (147), sin C sin c = sin a --. — -.....

...therefore, cos a = -^OC* + ~W "* A' ' Substituting the functions derived from the triangles 0 A' O A' O, we have cos a = cos b cos c -)- sin b sin c cos A. In like manner may be deduced cos b = cos c cos a -)- sin c sin a cos B, cos c = cos a cos b -)-...

...angle, plus the cosine of the second side into the cosine of the included angle. By (150) and (152) we have cos a = cos b cos c -\- sin b sin c cos A, cos c = cos a cos b -|- sin a sin b cos (7; and by means of (147), sin c sm a sin C sin A* , . ^...

...angle, plus the jcosine of the second side into the cosine of the included angle. By (150) and (152) we have cos a = cos b cos c -{- sin b sin c cos A, cos c = cos a cos b -)- sin a sin b cos (7; and by means of (147), sin c sin C sin a -. — j. sin...

...substituting this value in ( 2 ), we have, (20 and now substituting these values of OE, OZ/, and QP, in ( 1 ), we have, cos a = cos b cos c + sin b sin c cos A • (3.) In the same way, we may deduce, cos b = cos a cos c + sin a sin c cos .7? • • (4.) cos...