10,11

In the tetrahedron $ABCD$, the dihedral angles at edges $BC$ and $CD$ are right angles. The length of one of the edges of the tetrahedron is three times the length of a non-intersecting edge. The vertex of the cone coincides with one of the vertices of the tetrahedron, and the circle of the base of the cone is circumscribed around one of the faces. Find the angle in the axial section of the cone.
Planes of faces $ABC$ and $ADC$ are perpendicular to the plane of face $BCD$ and intersect along the line $AC$, which means that edge $AC$ is perpendicular to the plane of face $BCD$. Therefore, $\angle ACB = \angle ACD = 90^\circ$. Note that vertex $A$ cannot be the vertex of the cone mentioned in the problem, because the edges emanating from this vertex are not equal to each other ($AD > AC$, the hypotenuse is greater than the leg). Similarly for vertices $B$ and $D$.

Therefore, the vertex of the cone is point $C$. Let $CA = CB = CD = a$. Then $AB = AD = a\sqrt{2}$, so either $BD = 3a$ or $BD = \frac{a}{3}$. Suppose $BD = 3a$. Then

$$
\begin{gathered}
AD = AB = a\sqrt{2} + a\sqrt{2} = 2a\sqrt{2} < 3a = BD,
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-33.jpg?height=83&width=1780&top_left_y=1559&top_left_x=139)

$$
\cos \alpha = \frac{BD}{2AB} = \frac{\frac{a}{3}}{2a\sqrt{2}} = \frac{1}{6\sqrt{2}}, \sin \alpha = \sqrt{1 - \frac{1}{72}} = \sqrt{71}
$$

Let $R$ be the radius of the circumcircle of triangle $ABD$ (the radius of the base of the cone), and $\phi$ be the angle in the axial section of the cone. By the Law of Sines,

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-33.jpg?height=123&width=392&top_left_y=1842&top_left_x=830)

By the Law of Cosines,

$$
\cos \phi = \frac{AC^2 + AC^2 - (2R)^2}{2AC \cdot AC} = \frac{2AC^2 - 4R^2}{2AC^2} = \frac{2a^2 - \frac{111a^2}{72}}{2a^2} = -\frac{1}{71}
$$

## Answer

$\arccos \left(-\frac{1}{71}\right) = 2 \operatorname{arctg} \frac{6}{\sqrt{35}}$.