Is there a compact formula for the area (excess angle – assuming a unit sphere) of a right spherical triangle given its side lengths $a$ and $b$?
As explained in an answer to an earlier question about the area of a generic spherical triangle, the excess angle $E$ is given by
$$\tan\frac E4=\sqrt{\tan\frac{a+b+c}4\tan\frac{-a+b+c}4\tan\frac{a-b+c}4\tan\frac{a+b-c}4}$$
However, I do not have $c$. Of course, I can use Napier's rules for right spherical triangles to find $c$.
I thought about using integration in spherical coordinates (assuming B is the north pole and equating $a$ and $c$ with polar and azimuthal angles respectively). That approach, however, requires knowing the exact representation of the great circle that connects $A$ to $B$ in the coordinate chart.
Best Answer
In Theorem 9.2 of this document, it is shown that the area of a spherical right triangle, $E$, is given by $$ \tan\left(\frac E2\right)=\tan\left(\frac a2\right)\tan\left(\frac b2\right) $$ where $a$ and $b$ are the two legs of the right spherical triangle.