Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted

From Mathworld (edited slightly):

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, $A=B=C=\arccos\frac{1}{3}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.