12.452 At the base of a right prism lies a parallelogram with an acute angle $\varphi$ between the diagonals. The diagonals of each of the adjacent lateral faces intersect at angles $\alpha$ and $\beta(\alpha>\beta)$, facing the corresponding sides of the base. Find the volume of the prism, if its height is $\boldsymbol{h}$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-320.jpg?height=203&width=334&top_left_y=220&top_left_x=86)

The lateral face $A A_{1} B_{1} B$ of the given right prism $T=A B C D A_{1} B_{1} C_{1} D_{1} (A B C D$ - its base) is a rectangle, in which $\angle B_{1} O B=\pi-\alpha$ and $B_{1} B=h$ (by condition) $\Rightarrow \angle B_{1} A B=\frac{\pi}{2}-\frac{\alpha}{2} \Rightarrow A B=h \cdot \operatorname{tg} \frac{\alpha}{2}$.

Similarly, $B C=h \cdot \operatorname{tg} \frac{\beta}{2}$.

Thus, the base of the prism is a parallelogram with sides $h \cdot \operatorname{tg} \frac{\alpha}{2}$ and $h \cdot \operatorname{tg} \frac{\beta}{2}$ and an acute angle $\varphi$ between the diagonals. In problem 12.396, the area of such parallelograms was determined:

$$
S_{A B C D}=\frac{A B^{2}-B C^{2}}{2} \cdot \operatorname{tg} \varphi=\frac{h^{2}}{2} \cdot \operatorname{tg} \varphi \cdot\left(\operatorname{tg}^{2} \frac{\alpha}{2}-\operatorname{tg}^{2} \frac{\beta}{2}\right) \text { Hence } V_{T}=S_{A B C D} \cdot h
$$

Answer: $._{T}=\frac{h^{3}}{2} \cdot \operatorname{tg} \varphi \cdot \frac{\sin \frac{\alpha-\beta}{2} \cdot \sin \frac{\alpha+\beta}{2}}{\cos ^{2} \frac{\alpha}{2} \cdot \cos ^{2} \frac{\beta}{2}}$