Find the volume of a right prism whose base is a right triangle with an acute angle $\alpha$, if the lateral edge of the prism is $l$ and forms an angle $\beta$ with the diagonal of the larger lateral face.

#
Let $AB$ be the hypotenuse of the right triangle $ABC$ lying at the base of the right prism $ABC A_1 B_1 C_1$, and $\angle ABC = \alpha$. Then $AA_1 B_1 B$ is the largest lateral face of the prism $ABC A_1 B_1 C_1$. According to the problem, the diagonal $AB_1$ of this face forms an angle $\beta$ with the edge $AA_1$. That is, $\angle A_1 A B_1 = \beta$. Therefore,

$$
AB = A_1 B_1 = AA_1 \operatorname{tg} \angle A_1 A B_1 = l \operatorname{tg} \beta,
$$

$$
\begin{gathered}
BC = AB \cos \angle ABC = l \operatorname{tg} \beta \cos \alpha, AC = AB \sin \angle ABC = l \operatorname{tg} \beta \sin \alpha \\
S_{\triangle ABC} = \frac{1}{2} AC \cdot BC = \frac{1}{2} l \operatorname{tg} \beta \sin \alpha \cdot l \operatorname{tg} \beta \cos \alpha = \frac{1}{1} l^2 \sin 2 \alpha \operatorname{tg}^2 \beta
\end{gathered}
$$

Therefore,

$$
V_{ABC A_1 B_1 C_1} = S_{\triangle ABC} \cdot AA_1 = \frac{1}{1} l^2 \sin 2 \alpha \operatorname{tg}^2 \beta \cdot l = \frac{1}{1} l^3 \sin 2 \alpha \operatorname{tg}^2 \beta
$$

## Answer

$\frac{1}{1} l^3 \sin 2 \alpha \operatorname{tg}^2 \beta$.