10,11 In a triangular pyramid, each lateral edge of which is equal to $a$, one dihedral angle at the vertex is right, and each of the others is $60^{\circ}$. Find the volume of the pyramid.
Let $A B C D$ be a given triangular pyramid, in which

$$
A D=B D=C D=a, \angle B D C=90^{\circ}, \angle A D B=\angle A D C=60^{\circ}
$$

Since triangles $A D B$ and $A D C$ are equilateral, $A B=A C=a$. The lateral edges $D A, D B$, and $D C$ of the triangular pyramid $D A B C$ are equal, so the height drawn from vertex $D$ passes through the center $O$ of the circumcircle of triangle $A B C$. Since triangle $A B C$ is equal to triangle $B D C$, it is a right triangle. Point $O$ is the midpoint of the hypotenuse $B C$. We then find:

$$
\begin{gathered}
D O=\sqrt{B D^{2}-B O^{2}}=\sqrt{\mathbf{a}^{2}-\frac{\alpha^{2}}{2}}=\frac{a \sqrt{2}}{2} \\
S_{\triangle A B C}=\frac{\alpha^{2}}{2}, V_{A B C D}=\frac{1}{3} S_{\triangle A B C} \cdot D O=\frac{1}{3} \cdot \frac{\alpha^{2}}{2} \cdot \frac{a \sqrt{2}}{2}=\frac{\alpha^{3} \sqrt{2}}{12}
\end{gathered}
$$

## Answer

$\frac{a^{3} \sqrt{2}}{12}$