4.42. Find the area of the curvilinear 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.
4.42. Let us first consider a spherical "bigon," a part of the sphere enclosed within a dihedral angle of magnitude $\alpha$, with its edge passing through the center of the sphere. The area of such a figure is proportional to $\alpha$, and when $\alpha=\pi$, it is equal to $2 \pi R^{2}$; therefore, it is equal to $2 \alpha R^{2}$.

Each pair of planes of the given trihedral angle corresponds to two "bigons." These "bigons" cover the given curvilinear triangle and the triangle symmetric to it relative to the center of the sphere in 3 layers, and the rest of the sphere in one layer. Therefore, the sum of their areas is equal to the surface area of the sphere, increased by $4 S$, where $S$ is the area of the desired triangle. Therefore, $S=R^{2}(\alpha+$ $+\beta+\gamma-\pi)$.