# Deriving the Surface Area of a Spherical Triangle

A triangle on a sphere is composed of points $A$, $B$ and $C$. The $\alpha$, $\beta$ and $\gamma$ denote the angles at the corresponding points of the triangle:

The Girard's theorem states that the surface area of any spherical triangle:

$$A = R^2 \cdot E$$

where $R$ is the radius of the sphere and $E$ is the excess angle of $(\alpha + \beta + \gamma - \pi)$

I'm wondering how to derive this formula.

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–  Inquest Feb 16 '12 at 18:03
TeX tip: Notice the difference between $\alpha$ + $\beta$ + $\gamma$ - $\pi$ and $\alpha + \beta + \gamma - \pi$. In the former, the minus sign looks like a hyphen instead of a minus sign. There's a reason why that's not standard usage. I fixed this in the posting. –  Michael Hardy Feb 16 '12 at 18:06
@Nunoxic, no. That question is about how to determine the angles between ABC, ACB and BAC. –  ezpresso Feb 16 '12 at 18:06
There is a nice proof at math.rice.edu/~pcmi/sphere/gos4.html –  Michael E2 Jan 24 at 22:37

Consider the following three parts of the sphere: let $P_A$ be the lune created from the triangle $ABC$ plus the triangle adjacent across the $BC$ line segment, plus the opposite lune (on the opposite side of the sphere), and similarly for parts $P_B$ and $P_C$.

The area of $P_A$ is $4\alpha R^2$: the total area of the sphere is $4\pi R^2$, and the area of $P_A$ is certainly proportional to $\alpha$.

Notice now that $P_A\cup P_B\cup P_C$ is the entire sphere, and that $P$'s intersect at the triangle + the opposite triangle. We thus have:

area($P_A$) + area($P_B$) + area($P_C$) = area of the sphere + 2 area of the triangle +2 area of the opposite triangle.

As the two triangles have the same area, you get your formula.

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You really should explain what they cut the circle to 4 parts actually means. The first part is the made up of the lunes $ABA'C$ and $A'B'AC'$ put together, for example. Otherwise this is a good argument by symmetry. –  anon Feb 16 '12 at 19:53
@anon: I meant "they cut the sphere"; thanks for spotting it! –  user8268 Feb 16 '12 at 19:57
I've edited it to reflect what I believe you were saying. Additionally, if we regard $A'$, $B\,'$ and $C\,'$ respectively as the points opposite to $A$, $B$ and $C$ on the sphere, we have the decomposition $$\begin{array}{} P_A=ABA'C\;\cup\;A'B\,'AC\,' \\ P_B= BCB\,'A\;\cup\; B\,'C\,'BA \\ P_C= CAC\,'B\;\cup\; C\,'A'CB\,' \end{array}$$ –  anon Feb 16 '12 at 20:18
$P_B = BCB'A \cup B'C'BA'$? –  ezpresso Feb 16 '12 at 21:02
Yes, forgot that last accent mark, thanks. –  anon Feb 17 '12 at 16:22