Page:Spherical Trigonometry (1914).djvu/25

§14] $$OB$$. Then $$Ca$$, $$Cb$$, $$OA$$, $$OB$$ are all perpendicular to $$OP$$, because the planes $$aCb$$ and $$AOB$$ are perpendicular to $$OP$$; therefore



$$Ca$$ is parallel to $$OA$$, and $$Cb$$ is parallel to $$OB$$. Therefore the angle $$aCb=$$the angle $$AOB$$ (Euclid, . 10). Hence,

therefore, $$\frac{\mathrm{arc } ab}{\mathrm{arc } AB}=\frac{Ca}{OA} =\frac{Ca}{Oa}=\sin P \widehat Oa$$.