323. Find the area of the triangle formed by the intersection of a sphere of radius $R$ with a trihedral angle, the dihedral angles of which are equal to $\alpha, \beta$ and $\gamma$, and the vertex coincides with the center of the sphere.
323. First of all, note that the area of the spherical "digon" formed by the intersection of the surface of the sphere with the faces of a dihedral angle of magnitude $\alpha$, whose edge passes through the center of the sphere, is equal to $2 \alpha R^{2}$. This follows from the fact that this area is proportional to the magnitude $\alpha$, and at $\alpha=\pi$ it is equal to $2 \pi R^{2}$.

Each pair of planes forming two faces of the given trihedral angle corresponds to two "digons" on the surface of the sphere. By summing their areas, we obtain the surface of the sphere, increased by $4 S_{\triangle}$, where $S_{\triangle}$ is the area of the desired triangle. Thus,

$$
S_{\Delta}=R^{2}(\alpha+\beta+\gamma-\pi)
$$

The quantity $\alpha+\beta+\gamma-\pi$ is called the spherical excess of the spherical triangle.