Page:Spherical Trigonometry (1914).djvu/43

Case II.—Two sides and the angles opposite to them.

46. From the value of $\sin A$ in the preceding Article it follows that $${\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}\dots (5)$$ for each of these is equal to the same expression, namely, $${\frac {\surd (1-\cos ^{2}a-\cos ^{2}b-\cos ^{2}c+2\cos a\cos b\cos c)}{\sin a\sin b\sin c}}\dots (6)$$

Thus the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. We shall give an independent proof of this proposition in the following Article.

47. The sines of the angles of a spherical triangle are proportional to the sines of the opposite sides.^[1]

Let $ABC$ be a spherical triangle, $O$ the centre of the sphere. Take any point $P$ in $OA$, draw $PD$ perpendicular to the plane $BOC$, and from $D$ draw $DE$, $DF$ perpendicular to $OB$, $OC$ respectively; join $PE$, $PF$, $OD$.

Since $PD$ is perpendicular to the plane $BOC$, it makes right angles with every straight line meeting it in that plane; hence $$PE^{2}=PD^{2}+DE^{2}=PO^{2}-OD^{2}+DE^{2}=PO^{2}-OE^{2};$$

1. This fundamental theorem of Spherical Trigonometry is found, under a rather different form, in the 3rd book of the Sphaerica of MENELAUS.