10,11

All dihedral angles of a trihedral angle are equal to $60^{\circ}$. Find the angles formed by the edges of this trihedral angle with the planes of the opposite faces.
Let $O$ be the vertex of the given trihedral angle. On its edges, we lay off segments $O A, O B$, and $O C$, such that $O A = O B = O C = a$. Then the triangles $O A B, O A C$, and $O B C$ are equilateral. Therefore, $A B = B C = A C = a$. Thus, all faces of the pyramid $O A B C$ are equal equilateral triangles with side $a$. Let $M$ be the orthogonal projection of point $A$ onto the plane $O B C$. Then $A O M$ is the angle between the edge $O A$ and the opposite face $B O C$ of the given trihedral angle. Since $A B = A C = A O$, $M$ is the center of the equilateral triangle $B O C$. From the right triangle $A M O$, we find that

$$
\cos \angle A O M = \frac{\partial M}{\partial A} = \frac{\frac{a \sqrt{3}}{3}}{a} = \frac{\sqrt{3}}{3}
$$

Therefore, $\angle A O M = \arccos \frac{\sqrt{3}}{3}$. It is clear that the other angles are also equal to $\arccos \frac{\sqrt{3}}{3}$.

## Answer

$\arccos \frac{\sqrt{3}}{3}$