 | Charles Hutton - 1812 - 624 pagina’s
...— cos b . cos c -f- sin b . sin c . cos A 1 cos 6 = cos a . cos c -j- sin a . sin c . cos B v (I.) cos c -- cos a . cos b + sin a . sin b . cos c J THEOREM VII. In Every Spherical Triangle, the Sines of the Angles are Proportional to the Sines of... | |
 | Olinthus Gregory - 1816 - 276 pagina’s
...FD = AO cos a cos b + FB = AO cos a cos b + AO sin a sin b cos c. Therefore, dividing by AO, we have cos c = cos a cos b + sin a sin b cos c. Similar relations are deducible for the other sides a and b : hence, generally cas a = cos b cos c... | |
 | Louis Puissant - 1819 - 464 pagina’s
...valeurs de da, l'on trouvera dB- de sîn B tans a ''un autre côté , si l'on différencie l'équation cos c = cos a cos b + sin a sin b cos C, viendra , . » /cos b — cos a cos c\ • , . _ 1n de sm c = da ( -;— j -f- sm « sin b sin Cc?C,... | |
 | Robert Woodhouse - 1819 - 300 pagina’s
...c- n cos. c — cos. a . cos. b , ,, bmce cos. L = - r- - : — - - , page 139, sin. a . sin. b .-. cos. c = cos. a . cos. b + sin. a . sin. b . cos. C; but cos. C=l — ver. sin. C, (ver. sin. stands for versed sine); * For instance, in finding the Moon's... | |
 | Henry Parr Hamilton - 1826 - 354 pagina’s
...Zx'z be an arc of 90° passing through x', and meeting xy in z. Then, since in any spherical triangle cos c = cos a cos b + sin a sin b cos C, we have (1) In the triangle x'xt, cos x', x = cos t, x cos t, x' + sin t, x sin t, x' cos x'y , xy... | |
 | Louis-Benjamin Francœur - 1828 - 608 pagina’s
..., B^ et le côté compris Cj on trouve les deux autres côtes a, b , par les analogies (A - B) — cos C = cos A cos B — sin A sin B cos c (a3) , cos C = cos A cos B (tang A tang B cos c — i). Comme dans les deux derniers cas, on emploie... | |
 | John Radford Young - 1833 - 308 pagina’s
...cos. b cos. c -f- sin. b sin. c cos. A , cos. 4 = cos. a cos. c + sin. a sin. c cos. B > . . . . (A); cos. c = cos. a cos. b + sin. a sin. b cos. C ^ and these equations embody the whole theory of spherical trigonometry and are sufficient to supply... | |
 | Charles William Hackley - 1838 - 338 pagina’s
...seems somewhat lengthy. But a shorter method of solution is deducible from the fundamental formula, cos c = cos a cos b + sin a sin b cos c (1). For substituting cos a tan a for its equal sin a it becomes cos c = cos a (cos b + tan a sin b... | |
 | John Hymers - 1840 - 386 pagina’s
...SPK = cos /, and applying these values and the fundamental formulas of spherical Trigonometry, viz. cos c = cos a cos b + sin a sin b cos C, cot .4 sin C = cot a sin 6 — cos C cos 6, sin a sin B = sin .4 sin b, (where «, 6, c, A, S, C, represent... | |
 | John Hymers - 1841 - 244 pagina’s
...6 sin с cos A (1), cos b -- cos a cos с + sin a sin с cos B, cos с с cos a cos b + sin a sin 6 cos C, three equations which comprehend implicitly...is determined, we proceed to investigate the other three distinct formulae which, in conjunction with (l), will always effect that object. 20. To find... | |
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B . sin c = sin b . sin C cos a = cos b . cos c + sin b . sin c cos b = cos a . cos c + sin a . sin c cos A cos B cos c = cos a . cos b + sin a . sin b . cos C ..2), cotg b . sin c = cos G. 
