Given the following spherical triangle, is it possible to calculate B
given side b, angle A and angle C = 90 degrees?
If so, which formula is it.
I've tried the sine Rule and cosine rule but I need to know more than one side.
I would use formula:
cos A = sin B cos a
(from Smart's book on p. 16 and simplifying since C = 90 deg
However side b is my known and I'm trying to find B. Is it possible to rearrange the formula?
Thank you to @Somos for the link; my answer is in Napier's rules:
The full set of rules for the right spherical triangle is (Todhunter,1 Art.62)
\begin{alignedat}{4}&{\text{(R1)}}&\qquad \cos c&=\cos a\,\cos b,&\qquad \qquad &{\text{(R6)}}&\qquad \tan b&=\cos A\,\tan c,\\&{\text{(R2)}}&\sin a&=\sin A\,\sin c,&&{\text{(R7)}}&\tan a&=\cos B\,\tan c,\\&{\text{(R3)}}&\sin b&=\sin B\,\sin c,&&{\text{(R8)}}&\cos A&=\sin B\,\cos a,\\&{\text{(R4)}}&\tan a&=\tan A\,\sin b,&&{\text{(R9)}}&\cos B&=\sin A\,\cos b,\\&{\text{(R5)}}&\tan b&=\tan B\,\sin a,&&{\text{(R10)}}&\cos c&=\cot A\,\cot B.\end{alignedat}