3.28. The base of the 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.
3.28. Given that $A B C A_{1} B_{1} C_{1}$ is a right prism, $A C=C B, \angle A C B=\alpha, A_{1} B=l$, $\angle A_{1} B A=\beta$ (Fig. 3.30), 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 get $D C=A D \operatorname{ctg} \frac{\alpha}{2}=\frac{1}{2} l \cos \beta \operatorname{ctg} \frac{\alpha}{2}$.

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-516.jpg?height=563&width=541&top_left_y=1051&top_left_x=612)

Fig. 3.30

$$
\begin{gathered}
V_{\text {pr }}=\frac{1}{2} l \cos \beta \cdot \frac{1}{2} l \cos \beta \operatorname{ctg} \frac{\alpha}{2} \cdot l \sin \beta= \\
=\frac{1}{4} l^{3} \cos ^{2} \beta \operatorname{ctg} \frac{\alpha}{2} \sin \beta=\frac{1}{8} l^{3} \sin 2 \beta \cos \beta \operatorname{ctg} \frac{\alpha}{2} .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-517.jpg?height=538&width=542&top_left_y=106&top_left_x=44)