10,11 |

A sphere is constructed with the height of a cone as its diameter. The surface area of the part of the sphere lying inside the cone is equal to the surface area of the part of the cone lying inside the sphere. Find the angle in the axial section of the cone.
Let the angle at vertex $A$ of the axial section $A B C$ of a cone be $2 \alpha, O$ - the center of a sphere of radius $R$, constructed on the height $A H$ of the cone as a diameter, $E D$ - a diameter lying in the plane $A B C$ of the circle where the sphere and the lateral surface of the cone intersect, $M$ - the center of this circle. Then $\angle A D H=90^{\circ}, \angle H D M=\angle$ $D A H=\alpha$. From the right triangles $A D H$ and $D M H$ we find that

$$
D H=A H \sin \alpha=2 R \sin \alpha, A D=A H \cos \alpha=2 R \cos \alpha
$$

$$
D M=A D \sin \alpha=2 R \cos \alpha \sin \alpha, M H=D H \sin \alpha=2 R \sin 2 \alpha \text {. }
$$

Let $S 1$ be the area of the surface part of the sphere lying inside the cone, $S 2$ - the area of the part of the cone's surface lying inside the sphere. Then

$$
S 1=2 \pi R \cdot M H=2 \pi R \cdot 2 R \sin 2 \alpha=4 \pi R 2 \sin 2 \alpha
$$

$$
S 2=\pi M D \cdot A D=\pi \cdot 2 R \cos \alpha \sin \alpha \cdot 2 R \cos \alpha=4 \pi R 2 \sin \alpha \cos 2 \alpha .
$$

According to the problem, $S 1=S 2$, i.e., $4 \pi R 2 \sin 2 \alpha=4 \pi R 2 \sin \alpha \cos 2 \alpha$, or

From this equation, we find that $\sin \alpha=\frac{\sqrt{5}-1}{2}$. Then

$\cos 2 \alpha=1-2 \sin 2 \alpha=\sqrt{5}-2$.

## Answer

$\arccos (\sqrt{5}-2)$