12.439 The base of a right prism is a rhombus. One of the diagonals of the prism is equal to a and forms an angle $\alpha$ with the base plane, and an angle $\beta$ with one of the lateral faces. Find the volume of the prism.
## Solution.

A

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-311.jpg?height=296&width=361&top_left_y=1264&top_left_x=235)
Let $\quad T=A B C D A_{1} B_{1} C_{1} D_{1}$

- Given that the prism is a right prism, $B D_{1}=a$, and the base $A B C D$ is a rhombus, then $B B_{1}=a \cdot \sin \alpha, B D=a \cdot \cos \alpha$.
Drop a perpendicular $B K$ to the face $C C_{1} D_{1} D$. Since the prism is a right prism, $B K$ lies in the plane of the base $\Rightarrow$
$\Rightarrow \angle B D_{1} K=\beta, B K=a \cdot \sin \beta \Rightarrow$
$\Rightarrow D K=\sqrt{B D^{2}-B K^{2}}=a \sqrt{\cos ^{2} \alpha-\sin ^{2} \beta}$.
$\Delta B D K \sim \triangle C D O \Rightarrow \frac{D K}{D O}=\frac{B D}{C D} \Rightarrow C D=\frac{B D \cdot \frac{B D}{2}}{D K}=\frac{a \cdot \cos ^{2} \alpha}{2 \sqrt{\cos ^{2} \alpha-\sin ^{2} \beta}} \Rightarrow$
$\Rightarrow S_{A B C D}=C D \cdot B K=\frac{a^{2} \cos ^{2} \alpha \cdot \sin \beta}{2 \sqrt{\cos ^{2} \alpha-\sin ^{2} \beta}}$. From this, we have $V_{T}=S_{A B C D} \cdot B B_{1}$.

Answer: $V_{T}=\frac{a^{3} \cdot \sin 2 \alpha \cdot \cos \alpha \cdot \sin \beta}{4 \sqrt{\cos (\alpha+\beta) \cdot \cos (\alpha-\beta)}}$.