2. A sphere is inscribed in a cone, such that the area of the sphere and the area of the base of the cone are equal. Calculate the cosine of the angle of the axial section of the cone at the vertex which is also the apex of the cone.
Solution. One cross-section of the cone and the inscribed sphere is an isosceles triangle $ABC$, with the length of the base equal to $2R$, where $R$ is the radius of the base of the cone, in which a circle $k$ with radius $r$ is inscribed. Here, $r$ is the radius of the inscribed sphere. Let the center of $k$ be the point $O$, and the points of tangency with the sides $AB, BC$, and $CA$ be the points $K, L$, and $M$ respectively (see diagram).

From the equation $4 r^{2} \pi = R^{2} \pi$ we get

![](https://cdn.mathpix.com/cropped/2024_06_05_7fab3a3ae5ed2b773d3eg-07.jpg?height=480&width=514&top_left_y=887&top_left_x=952)
$R = 2r$.

The triangles $AKC$ and $OMC$ are similar. If we introduce standard notations $\overline{AK} = R, \overline{OK} = \overline{OM} = r, \overline{CK} = H$, and $\overline{AC} = s$, we get:

$$
\begin{aligned}
& \frac{\overline{OM}}{\overline{OC}} = \frac{\overline{AK}}{\overline{CK}} \\
& \frac{r}{H - r} = \frac{R}{s} = \frac{R}{\sqrt{H^2 + R^2}} = \frac{2r}{\sqrt{H^2 + 4r^2}}
\end{aligned}
$$

so $H = \frac{8}{3} r$, and therefore

$$
\sin \frac{\alpha}{2} = \frac{r}{\frac{8}{3} r - r} = \frac{3}{5}
$$

Now, using the identity $\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}$, it is not difficult to calculate that $\cos \alpha = \frac{7}{25}$.