= Solution
For a <spherical triangle> on a sphere of radius $R$, bounded by shorter <great circle> arcs and with interior angles $\alpha,\beta,\gamma$, the <spherical excess formula>, the triangular form of the <Gauss-Bonnet theorem>, is
$$
\boxed{\operatorname{area}=R^2(\alpha+\beta+\gamma-\pi).}
$$
Here is an area proof. On the unit sphere, a <spherical lune> of angle $\alpha$ has area $2\alpha$: rotating around its vertices shows it occupies a fraction $\alpha/(2\pi)$ of the sphere when one uses the full angular width, or directly integrates $\int_0^\alpha\int_0^\pi\sin\theta\,d\theta\,d\varphi$. The three side great circles of the triangle divide the sphere into eight regions, including the triangle $T$ and its antipodal triangle $-T$. Choose, at each vertex, the lune containing $T$, and also its antipodal lune. Each of $T$ and $-T$ is covered three times by these six lunes; each other region is covered once. Thus, with $A$ the area of $T$,
$$
4(\alpha+\beta+\gamma)=4\pi+4A.
$$
This proves $A=\alpha+\beta+\gamma-\pi$; scaling lengths by $R$ scales area by $R^2$.
Radially project each face of the regular <dodecahedron> from its centre onto the unit sphere. Each edge and the origin lie in a plane, so its image is a shorter <great circle> arc. The twelve faces give congruent <convex regular spherical polygons> partitioning the sphere. At each projected vertex three congruent face angles fill the tangent plane. Hence each spherical pentagon has angle $2\pi/3$ at each of its five vertices. Splitting a pentagon into three <spherical triangles> and summing the <spherical excess formula> gives
$$
\operatorname{area}=5\frac{2\pi}{3}-3\pi=\boxed{\frac\pi3},\qquad
\boxed{\text{each angle}=\frac{2\pi}{3}}.
$$
This also agrees with $4\pi/12$, as required by the spherical partition.
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