Page:Spherical Trigonometry (1914).djvu/44

thus $PEO$ is a right angle. Therefore $PE=OP\sin P{\widehat {O}}E=OP\sin c$; and $PD=PE\sin P{\widehat {E}}D=PE\sin B=OP\sin c\sin B$.

Similarly, $PD=OP\sin b\sin C$; therefore $$OP\sin c\sin B=OP\sin b\sin C;$$

therefore ${\frac {\sin B}{\sin C}}={\frac {\sin b}{\sin c}}$.

The figure supposes $b$, $c$, $B$, and $C$ each less than a right angle; it will be found on examination that the proof will hold when the figure is modified to meet any case which can occur. If, for instance, $B$ alone is greater than a right angle, the point $D$ will fall beyond $OB$ instead of between $OB$ and $OC$; then $P{\widehat {E}}D$ will be the supplement of $B$, and thus $\sin PED$ is still equal to $\sin B$.

Case III. ― Two sides, the included angle, and another angle.

48. To shew that $\cot a\sin b=\cot A\sin C+\cos b\cos C$.

We have $${\begin{aligned}\cos a&=\cos bcosc+\sin b\sin c\cos A,\\\cos c&=\cos a\cos b+\sin a\sin b\cos C,\\\sin c&=\sin a{\frac {\sin C}{\sin A}}.\end{aligned}}$$

Substitute the values of $\cos c$ and $\sin c$ in the first equation; thus $$\cos a=(\cos a\cos b+\sin a\sin b\cos C)\cos b+{\frac {\sin a\sin b\cos A\sin C}{\sin A}};$$

by transposition $$\cos a\sin ^{2}b=\sin a\sin b\cos b\cos C+\sin a\sin b\cot A\sin C;$$ divide by $\sin a\sin b$; thus $$\cot a\sin b=\cos b\cos C+\cot A\sin C.\dots (7)$$

49. By interchanging the letters five other formulae, like that in the preceding Article, may be obtained; the whole six formulae will be as follows: