11.193. The base of the pyramid is a right-angled triangle. The lateral edges of the pyramid are equal, and the lateral faces passing through the legs form angles of 30 and $60^{\circ}$ with the base plane. Find the volume of the cone circumscribed around the pyramid if the height of the pyramid is $h$.
## Solution.

Let $D O$ be the height of the pyramid $D A B C$ (Fig. 11.80), $\angle A C B=90^{\circ}$, $D O=h$. Since the lateral edges of the given pyramid are equal, $O$ is the center of the circle circumscribed around $\triangle A B C$, and since $\triangle A B C$ is a right triangle, $O$ is the midpoint of the hypotenuse $A B$. Drop perpendiculars $O F$ and $O E$ to the legs $A C$ and $B C$ of the base, respectively. Since $O F$ is the projection of $D F$ on the base plane, and $O E$ is the projection of $D E$, then $D F \perp A C, D E \perp B C, \angle D F O$ is the angle of inclination of the face $A D C$ to the base plane, $\angle D F O=60^{\circ}, \angle D E O$ is the angle of inclination of the face $B D C$ to the base plane, $\angle D E O=30^{\circ}$.

From $\triangle D O F\left(\angle D O F=90^{\circ}\right): O F=D O \operatorname{ctg} \angle D F O=h \sqrt{3}$. From $\triangle D O E$ $\left(\angle D O E=90^{\circ}\right): O E=D O \operatorname{ctg} \angle D E O=\frac{h}{\sqrt{3}}$. The radius of the circle circumscribed around $\triangle A B C, R=O C=\sqrt{O F^{2}+O E^{2}}=\sqrt{3 h^{2}+\frac{h^{2}}{3}}=h \sqrt{\frac{10}{3}}$.

The volume of the cone circumscribed around the given pyramid,

$$
\begin{aligned}
& V=\frac{1}{3} \pi R^{2} h=\frac{1}{3} \pi \cdot \frac{10 h^{2}}{3} \cdot h=\frac{10 \pi h^{3}}{9} \\
& \text { Answer: } \quad \frac{10 \pi h^{3}}{9}
\end{aligned}
$$