Page:Spherical Trigonometry (1914).djvu/25

$OB$. 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, XI. 10). Hence,

${\frac {\mathrm {arc} ab}{\mathrm {radius} Ca}}={\frac {\mathrm {arc} AB}{\mathrm {radius} OA}}$, (Plane Trigonometry, Art. 18);

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