12.063. Each lateral edge of a quadrilateral pyramid forms an angle $\alpha$ with the height. The base of the pyramid is a rectangle with an angle $\beta$ between the diagonals. Find the volume of the pyramid if its height is $h$.
## Solution.

Since each of the lateral edges of the quadrilateral pyramid forms an angle $\alpha$ with the height, the vertex of the pyramid projects onto the point of intersection of the diagonals of the rectangle lying at the base (Fig. 12.65). Thus, if $d$ is the length of the diagonal, then $\frac{d}{2}=h \operatorname{tg} \alpha$ and the area of the base $S=4 \cdot \frac{1}{2} \cdot \frac{d}{2} \cdot \frac{d}{2} \sin \beta=\frac{1}{2} d^{2} \sin \beta=2 h^{2} \operatorname{tg}^{2} \alpha \sin \beta$.

The volume of the pyramid $V=\frac{1}{3} S \cdot h=\frac{1}{3} 2 h^{3} \operatorname{tg}^{2} \alpha \sin \beta$.

Answer: $\frac{2}{3} h^{3} \mathrm{tg}^{2} \alpha \sin \beta$.