Deriving the Surface Area of a Spherical Triangle

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.

A bit late to the party, but we can also use Gauss-Bonnet theorem, which can be used for other $2$-dimensional manifolds as well, e.g. the hyperbolic plane. It states that if $M$ is a $2$-dimensional Riemannian manifold with piecewise smooth boundary, then $$\int_M K dA +\int_{\partial M} k_g+\sum_i \alpha_i=2\pi \chi(M) $$ where $K$ is the Gauss curvature, $k_g$ the geodesic curvature (which is zero for geodesics), $\alpha_i$ the angles by which the tangent to the boundary turns at points where $\partial M$ is not smooth, and $\chi(M)$ the Euler characteristic of $M$.

In our case $M$ is the triangle, $K$ is identically $\frac{1}{R^2}$, the angles $\alpha_i$ are $\pi-\alpha,\pi-\beta,\pi-\gamma$ and $\chi(M)=1$. Therefore the formula becomes $$\frac{A}{R^2}+(\pi-\alpha+\pi-\beta+\pi-\gamma) = 2\pi\implies \boxed{A= (\alpha+\beta+\gamma-\pi)R^2}.$$