 | John Bonnycastle - 1818 - 488 pàgines
...c = sin b sin c r* cos b — r cos a cos c sin a sin c r2 cos c — r cot a cos b sin a sin b „ r cos b cos c + sin b sin c cos A Cos a = ~r~ ~— f-, , r cos a cos c + sin a sin c cos B Cos b = —, H r* f, r cos a cos b + sin a sin b... | |
 | Adam von Burg - 1851 - 696 pàgines
...(tt) lang <$ = tatig A cos b, so wird, wie leicht zu sehen : (|3) sin (C-}- c) = tang b cot a sin y . Ferner ist (zur Berechnung von c) aus der 1. der obigen Gleichungen : cos b cos с -j- sin b sin с cos A = cos a, oder für (a') tang ш = tang b cos A auch : , „ ч / , COS О... | |
 | John Hymers - 1858 - 324 pàgines
...adjacent parts ; sine of the middle part = product of the cosines of the opposite parts. cos A = — cos B cos C + sin B sin C cos. a, cos a = cos b cos с + sin b sin с cos.4, since A = 90° and therefore cos .4 = 0, sin A = 1, become cos... | |
 | Isaac Todhunter - 1859 - 156 pàgines
...A the opposite angle of one triangle, and a' and A' similar quantities for the other. Then cos a = cos b cos c + sin b sin c cos A, cos a'= cos b cos c + sin 6 sin c cos A! ; therefore cos a - cos a = sin 6 sin c (cos A — cos A") ; that... | |
 | 1859 - 698 pàgines
...wenn man nur für eine aufser z auch x und x' gemessen hat. Winkel WMN durch A, so hat man: cos a — cos b cos c + sin b sin c cos A cos a' — cos b cos c — sin b sin c cos A oder Bezeichnet man den Bogen MN durch b und den (3) , l a'-Hfl... | |
 | 1859 - 778 pàgines
...gemessen hat. Bezeichnet man den Bogen M. N durch b und den Winkel WMN durch A, so hat man: cos a = cos b cos c + sin b sin c cos A cos a' — cos 6 cos c — sin & sin c cos A oder (3) . * . • , l a'+a a' — a COS 0 = COS - COS — -=... | |
 | Isaac Todhunter - 1863 - 182 pàgines
...A the opposite angle of one triangle, arid a and A' similar quantities for the other. Then cos a = cos b cos c + sin b sin c cos A, cos a' = cos b cos c + sin b sin c cos A'; therefore cos a - cos a' = sin b sin c (cos A — cos A') ; that... | |
 | Lefébure de Fourcy (M., Louis Etienne) - 1868 - 350 pàgines
...sides b and c, the opposite parts; and formula (11), Art. 157, and formula (1), Art. 149, cos A = — cos B cos C + sin B sin C cos a cos a = cos b cos c + sin b sin c cos A, become, since A = 90°, and therefore cos A = 0 and sin A = 1, oosa... | |
 | Aaron Schuyler - 1864 - 506 pàgines
...180° — C. or — A < 180" — (B + C). . . . A > B + C — 180°. By article 136, we have cos a = cos b cos c -(- sin b sin c cos A. cos a — cos 6 cos c cos .4 = sin b sin r But sin b sin e is positive, since b and c are each less than... | |
 | 1882 - 452 pàgines
...transposing, 1 — cos a cos b cos c = sin* a + sin b sin c cos A cos a; (1) also cosM = — cos A cos B cos C -\- sin B sin C cos a cos A, whence 1 + cos A cos B cos C '= sin1 A + sin B sin Ccos acos A. (2) Dividing (2) by (1), we have the... | |
| |
That is, cos a' = cos b' cos c' + sin b' sin c' cos A'. .: cos (180° -A) = cos (180° - B) cos (180° - C) + sin (180° — B) sin (180° - (7) cos (180° - a). [Art. 16. d.] ... — cos A = (— cos .B) (— cos C) + sin B sin C(— cos a). 
