331. The center of a sphere with radius $R$ is located outside a dihedral angle of magnitude $\alpha$ at a distance $a(a<R)$ from its edge and is situated in the plane of one of its faces. Find the area of the part of the sphere's surface that is inside the angle.
331. In Fig. 66, O is the center of the sphere, A and B are the points of intersection of the edge of the dihedral angle with the surface of the sphere, D and C are the midpoints of the arcs $\overline{A D B}$ and $\overline{A C B}$, the plane $A D B$ passes through $O, E$ is the vertex of the segment cut off by the plane $A C B$. The area of the curvilinear triangle $A D C$ is half of the desired area. On the other hand (assuming $\left.\alpha \leqslant \frac{\pi}{2}\right)$,

$$
S_{A D C}=S_{A E C}-S_{A E D^{*}}
$$

Let's find $S_{A E C}$. If $\varphi$ is the angle between the planes $A E O$ and $OEC$, $|E K|=h$, then, obviously, $S_{A E C}=\frac{\varphi}{2 \pi} 2 \pi R h=\varphi R h$; $h$ and $\varphi$ are easily found:

$$
\begin{gathered}
h=|E K|=R-|O K|=R-a \sin \alpha \\
\sin \varphi=\sin \widehat{A K L}=\frac{|A L|}{|A K|}=\frac{\sqrt{R^{2}-a^{2}}}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}} \\
\varphi=\arcsin \frac{\sqrt{R^{2}-a^{2}}}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}}
\end{gathered}
$$

Thus,

$$
S_{A E C}=R(R-a \sin \alpha) \arcsin \frac{\sqrt{R^{2}-a^{2}}}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}}
$$

Now let's find $S_{A E D}$. As is known (see problem 323),

$$
S_{A E D}=R^{2}(\varphi+\psi+\gamma-\pi)
$$

where $\varphi, \psi$ and $\gamma$ are the dihedral angles of the trihedral angle with vertex at $O$ and edges $O E, O A$ and $O D$. The angle $\varphi$ has already been found above.

To determine the angle $\psi$ (the angle at the edge $O A$), we will use the first cosine theorem (problem 166) applied to the trihedral angle with vertex $A$, for which

$$
\widehat{K A L}=\frac{\pi}{2}-\varphi, \sin \widehat{K A O}=\frac{a \sin \alpha}{R}, \sin \widehat{L A O}=\frac{a}{R}
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-157.jpg?height=157&width=911&top_left_y=176&top_left_x=132)

$$
\begin{aligned}
& =\frac{\sqrt{R^{2}-a^{2}}}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}} \sin \alpha
\end{aligned}
$$

It is obvious that $\gamma=\pi / 2$. Therefore,

$$
\begin{aligned}
& S_{A E D}=R^{2}\left[\arcsin \frac{\sqrt{R^{2}-a^{2}}}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}}+\right. \\
& \left.\quad+\arccos \frac{\sqrt{R^{2}-a^{2}} \sin \alpha}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}}-\frac{\pi}{2}\right]
\end{aligned}
$$

Substituting (2) and (3) into (1), we get the answer after simplification. Answer:

$2 R^{2} \arccos \frac{R \cos \alpha}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}}-$

$$
-2 R a \sin \alpha \arccos \frac{a \cos \alpha}{\sqrt{R^{2}-a^{2} \sin ^{2} \alpha}}
$$