# Area of a Spherical Triangle from Side Lengths

I am currently working on a proof involving finding bounds for the f-vector of a simplicial $3$-complex given an $n$-element point set in $\mathbb{E}^3$, and (for a reason I won't explain) am needing to find the answer to the following embarrassingly easy (I think) question.

What is the area of a spherical triangle with all equal side lengths equal to $\pi / 3$?

I have tried using L'Juilier's Theorem, but the first term $$\tan(\frac{1}{2}s)=\tan(\frac{1}{2}(\frac{3\pi}{3}))=\tan(\frac{\pi}{2})$$ is undefined, where $s$ is the semiperimeter (perimeter divided by 2).

Any ideas for how to compute this?

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I'm not big on spherical geometry, but I know every spherical triangle has angles adding up to strictly more than $\pi$; your "side lengths" add up to exactly $\pi$; could it be there's no such triangle? – Gerry Myerson May 14 '12 at 23:34
You're forgetting to divide your semi-perimeter by $2$. – Blue May 15 '12 at 0:34
@DayLateDon: That is exactly it. Thank you. – Samuel Reid May 15 '12 at 1:02

Let a spherical triangle have angles $A$, $B$, and $C$ (measured in radians at the vertices along the surface of the sphere) and let the sphere on which the spherical triangle sits have radius $R$. Then the surface area $\Delta$ of the spherical triangle is $$\Delta=R^2(A+B+C-\pi).$$
Using the spherical Law of Cosines ($\cos c=\cos a\cos b+\sin a\sin b\cos C$, where $a$, $b$, and $c$ are the side lengths of the triangle), for your triangle, $A=B=C=\arccos\frac{1}{3}$.