The base of a right prism is a rhombus with an acute angle $\alpha$. Find the volume of the prism if its larger diagonal is equal to $l$ and forms an angle $\beta$ with the base plane.

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Let $A B C D A 1 B 1 C 1 D 1$ be the given rectangular parallelepiped, the base of which is a rhombus $A B C D$ with an acute angle $\alpha$ at vertex $A$. Then $A C$ is the longer diagonal of the rhombus. Since $A C$ is the orthogonal projection of the diagonal $A 1 C$ of the parallelepiped onto the base plane $A B C D$, $A 1 C$ is the longer diagonal of the parallelepiped. By the problem's condition, $A 1 C = l$, $\angle A C A 1 = \beta$. From the right triangle $A C A 1$, we find that

$$
A A 1 = A 1 C \sin \angle A C A 1 = l \sin \beta
$$

$$
A C = A 1 C \cos \angle A C A 1 = l \cos \beta
$$

Let $O$ be the center of the rhombus. Then

$$
O B = O A \operatorname{tg} \angle O A B = \frac{1}{2} A C \operatorname{tg} \frac{1}{2} \angle B A D = \frac{1}{2} l \cos \beta \operatorname{tg} \frac{\alpha}{2}
$$

$$
\begin{gathered}
B D = 2 O B = l \cos \beta \operatorname{tg} \frac{\alpha}{2} \\
S_{A B C D} = \frac{1}{2} A C \cdot B D = \frac{1}{2} l \cos \beta \cdot l \cos \beta \operatorname{tg} \frac{\alpha}{2}
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
V_{A B C D A 1 B 1 C 1 D 1} = S_{A B C D} \cdot A A 1 = \\
= \frac{1}{2} l^2 \cos^2 \beta \cdot \operatorname{tg} \frac{\alpha}{2} \cdot l \sin \beta = \frac{1}{2} l^3 \sin \beta \cos^2 \beta \operatorname{tg} \frac{\alpha}{2}
\end{gathered}
$$

## Answer

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