Find the surface area of a spherical triangle of radius R
Find the surface area of a spherical triangle of radius $R$ with interior angles $\alpha,\,\beta,\,\gamma$.
Explanation of Method
A spherical lune of angle $\varphi$ occupies a fraction $\varphi/2\pi$ of the sphere. Pairing the triangle $T$ with its antipodal image $T'$, the three lunes at the vertices of $T$ (together with their antipodal counterparts) cover the whole sphere, but over-count $T$ and $T'$ — a counting argument (Girard's theorem) then isolates the triangle's own area in terms of its angle sum.
- Compute the area of a lune of angle $\varphi$.
- Set up the covering argument with the three vertex lunes and their antipodal images.
- Solve for the triangle's area.
Step 1 · Lune area
A lune of angle $\varphi$ is a fraction $\dfrac\varphi{2\pi}$ of the full sphere area $4\pi R^2$:
$$ A_{\text{lune}}(\varphi) = \frac\varphi{2\pi}\cdot4\pi R^2 = 2R^2\varphi. $$
Step 2 · Covering argument
Let $T$ have angles $\alpha,\beta,\gamma$ and area $S$, and let $T'$ be its antipodal image (also area $S$). At each vertex, the two lunes of angle $\alpha,\beta,\gamma$ together with their antipodal counterparts cover the whole sphere once, except that $T$ and $T'$ each lie in all three lune-pairs and so are counted an extra two times each:
$$ 2(2R^2\alpha)+2(2R^2\beta)+2(2R^2\gamma) = 4\pi R^2 + 4S. $$
Step 3 · Solve for $S$
$$ 4R^2(\alpha+\beta+\gamma) = 4\pi R^2+4S \implies S = R^2\big(\alpha+\beta+\gamma-\pi\big). $$
Additional Notes
- This derivation matches the source exactly. It is the classical Girard's theorem: the area of a spherical triangle is $R^2$ times its spherical excess $E=\alpha+\beta+\gamma-\pi$.
- As $R\to\infty$ (the flat limit), the excess $E\to0$ so the area stays finite, matching the familiar Euclidean fact $\alpha+\beta+\gamma=\pi$.
Result
The surface area of the spherical triangle is: