12.057. The base of a right prism is a right-angled triangle with an acute angle $\alpha$. The diagonal of the larger lateral face is $d$ and forms an angle $\beta$ with the lateral edge. Find the volume of the prism.
Solution.

Let $\triangle ABC$ be a right triangle, $\angle C=90^{\circ}, \angle BAC=\alpha$ (Fig. 12.59). By the condition, $BA_1=d, \angle BA_1A=\beta$. Then, from $\triangle BA_1A$, the height of the prism $AA_1=d \cos \beta$, and $AB=d \sin \beta$, since $AB$ is the hypotenuse of $\triangle ABC$, then $AC=AB \cos \alpha=d \sin \beta \cos \alpha$. The area of the base

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-757.jpg?height=582&width=580&top_left_y=65&top_left_x=92)

Fig. 12.59

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-757.jpg?height=334&width=518&top_left_y=289&top_left_x=733)

Fig. 12.60

$S_{\text{base}}=\frac{1}{2} AB \cdot AC \sin \alpha=\frac{1}{2} d \sin \beta \cdot d \sin \beta \cos \alpha \sin \alpha=\frac{1}{4} d^2 \sin 2\alpha \sin^2 \beta$.

Volume $V=S_{\text{base}} \cdot AA_1=\frac{1}{8} d^3 \sin^2 \beta \cos \beta \sin 2\alpha=\frac{1}{8} d^3 \sin \beta \sin 2\beta \sin 2\alpha$.

Answer: $\frac{d^3 \sin \beta \sin 2\beta \sin 2\alpha}{8}$.