Page:Spherical Trigonometry (1914).djvu/44

26 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 $$\sin B$$.

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

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

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: