12.067. The base of a right prism is an isosceles triangle with an angle $\alpha$ at the vertex. The diagonal of the face opposite this angle is $l$ and forms an angle $\beta$ with the base plane. Find the volume of the prism.
Solution.

Let $A B C A_{1} B_{1} C_{1}$ be a right prism, $A C=C B, \angle A C B=\alpha$, $A_{1} B=l, \angle A_{1} B A=\beta$ (Fig. 12.69), we need to find $V_{\text {pr }}=S_{\triangle A B C} \cdot A_{1} A$. From $\triangle A_{1} A B$ we find $A A_{1}=l \sin \beta, A B=l \cos \beta$, and from $\triangle A D C$ we have

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-765.jpg?height=446&width=428&top_left_y=129&top_left_x=174)

Fig. 12.69

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-765.jpg?height=494&width=388&top_left_y=81&top_left_x=762)

Fig. 12.70

$D C=A D \operatorname{ctg} \frac{\alpha}{2}=\frac{1}{2} l \cos \beta \operatorname{ctg} \frac{\alpha}{2}$. Therefore,

$V_{\mathrm{mp}}=\frac{1}{2} l \cos \beta \cdot \frac{1}{2} l \cos \beta \operatorname{ctg} \frac{\alpha}{2} \cdot l \sin \beta=\frac{1}{4} l^{3} \cos ^{2} \beta \operatorname{ctg} \frac{\alpha}{2} \sin \beta=\frac{1}{8} l^{3} \sin 2 \beta \cos \beta \mathrm{ctg} \frac{\alpha}{2}$.

Answer: $\quad \frac{1}{8} l^{3} \sin 2 \beta \cos \beta \operatorname{ctg} \frac{\alpha}{2}$.