11.168. The base of the pyramid is a rectangle with an area of $S$. Two lateral faces are perpendicular to the base plane, while the other two are inclined to it at angles of 30 and $60^{\circ}$. Find the volume of the pyramid.
Solution.

According to the problem, the base of the pyramid $EABCD$ is a rectangle $ABCD$, and the lateral faces $ABE$ and $CBE$ are perpendicular to the plane of the base (Fig. 11.62). Therefore, their common edge $EB$ is perpendicular to the plane of the base and serves as the height of the pyramid. $BA$ is the projection of $EA$ on the plane of the base, $BA \perp AD$. Then $EA \perp AD$, and consequently, $\angle EAB$ is the angle of inclination of the lateral face $EAD$ to the plane of the base, $\angle EAB=60^{\circ}$. Similarly, $\angle ECB=30^{\circ}$.

Let $EB=H$. From $\triangle ABE\left(\angle ABE=90^{\circ}\right): AB=BE \operatorname{ctg} \angle EAB=\frac{H \sqrt{3}}{3}$.

From $\triangle CBE\left(\angle CBE=90^{\circ}\right): BC=BE \operatorname{ctg} \angle ECB=H \sqrt{3}$.

Since $AB \cdot BC=S$, then $\frac{H \sqrt{3}}{3} \cdot H \sqrt{3}=S, H=\sqrt{S}$. The volume of the parallelepiped $V=\frac{1}{3} S H=\frac{S \sqrt{S}}{3}$.

Answer: $\frac{S \sqrt{S}}{3}$.