Spherical Law of Sines

Theorem
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.

Then:
 * $\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$

Proof
Let $X \in \R_{>0}$ such that:
 * $X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

Then from $(1)$:

In a spherical triangle, all of the sides are less than $\pi$ radians.

The same applies to the angles.

From Shape of Sine Function:
 * $\sin \theta > 0$ for all $0 < \theta < \pi$

Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:
 * $X = \dfrac {\sin A} {\sin a}$

Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$:

we arrive at the same point:

where:
 * $X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

as before.

Hence we have:
 * $\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$

Also see

 * Spherical Law of Cosines