...Next, let A and с be given, then to find the other parts, we have by (2), (3), and (8), of Art. 6, sin a = sin с sin A, tan b — tan с cos A, cot В = соя с tan A. Hence a only is given by its sine, and in this case there is only one solution ;...

...Given the hypotenuse с and the oblique angle A. From Formulas [38], [39], and [42] it follows that sin a = sin с sin A, tan b = tan с cos A, cot B= cose i&nA. Here a is determined by sin a, since a and A must be alike in kind (see Exercise XX.,...

...с and an angle A ; to find a, b, B. By (3), (5), and (8) of Art. 85, or by Napier's Rules, we have sin a = sin с sin A, tan b = tan с cos A, cot В = cos с tan A. Since a is found by its sine, it would be ambiguous, but the ambiguity is removed because a and A are...

...and an angle A ; to find a, b, B. By (;5), (5), an(1 (8) of Art- 185, or ЬУ Napier's Rules, we have sin a = sin с sin A, tan b = tan с cos A, cot B = cos с tan A. Since a is found by its sine, it would be ambiguous, but the ambiguity is removed...

...с and an angle A ; to find a, b, B. By (3), (5), and (8) of Art. 85, or by Napier's Bnles, we have sin a = sin с sin A, tan b = tan с cos A, cot B = cos с tan A. Since a is found by its sine, it would be ambiguous, but the ambiguity is removed...

...CA8E V. Given the hypotenuse с and the angle A. From Formulas [39], [40], and [43] it follows that sin a = sin с sin A, tan b = tan с cos A, cot B = cos с tan A. Here a is determined by sin a, since a and A must be alike in kind (see Exercise...

...CASE V. Given the hypotenuse с and the angle A. From Formulas [39], [40], and [43] it follows that sin a = sin с sin A, tan b = tan с cos A, cot П = cos с tan A. Here a is determined by sin a, since a and Л must be alike in kind (see Exercise...

...sin a , tan b ., , , , т, s1n A = — — , cos A = , and cos с = cot A cot B. s1n с tan с Or, sin a = sin с sin A, tan b = tan с cos A, and cot B = cos с tan A. Here the side a is determined from its sine ; but the ambiguity is removed...

...CASE V. Git-en the hypotenuse с and the angle A. From Formulas [39], [40], and [43] it follows that sin a — sin с sin A, tan b = tan с cos A, cot B = cos с tan A. Here a is determined by sin a, since a and Л must be alike in kind (see Exercise...

...CASE V Given the hypotenuse с and the angle A. From Formulas [39], [40], and [43], p. 144, we obtain sin a = sin с sin A, tan b = tan с cos A, cot В = cos с tan A. CASE VI Given the two angles A and В. From Formulas [43] and [41], p. 144, we obtain cos с = cot A cot...