12.268. One of the faces of a triangular prism inscribed in a cylinder passes through the axis of the cylinder. The diagonal of this face forms angles $\alpha$ and $\beta$ with the adjacent sides of the base of the prism. Find the volume of the prism if the height of the cylinder is $H$.
Solution.

Let the face $A A_{1} B_{1} B$ of the prism $A B C A_{1} B_{1} C_{1}$ pass through the axis $O O_{1}$ of the cylinder in which it is inscribed (Fig. 12.132), $A A_{1}=H, \angle A B A_{1}=\alpha$, $\angle C B A_{1}=\beta$, the center $O$ of the base of the cylinder is the midpoint of the side $A B$ of the triangle $A B C$, inscribed in this base. Hence, $\angle A C B=90^{\circ}$.

$A C$ is the projection of $A_{1} C$ onto the plane

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1022.jpg?height=484&width=504&top_left_y=83&top_left_x=725)

Fig. 12.132 $A B C, A C \perp B C$. Therefore, $A_{1} C \perp B C$.

In $\triangle A_{1} A B\left(\angle A_{1} A B=90^{\circ}\right)$:

$A B=A A_{1} \operatorname{ctg} \angle A B A_{1}=H \operatorname{ctg} \alpha ; A_{1} B=\frac{A A_{1}}{\sin \angle A B A_{1}}=\frac{H}{\sin \alpha}$.

In $\triangle A_{1} C B\left(\angle A_{1} C B=90^{\circ}\right): B C=A_{1} B \cos \angle C B A_{1}=\frac{H \cos \beta}{\sin \alpha}$.

From $\triangle A B C\left(\angle A C B=90^{\circ}\right)$:

$$
\begin{aligned}
& A C=\sqrt{A B^{2}-B C^{2}}=\sqrt{H^{2} \operatorname{ctg}^{2} \alpha-\frac{H^{2} \cos ^{2} \beta}{\sin ^{2} \alpha}}=\frac{H}{\sin \alpha} \sqrt{\cos ^{2} \alpha-\cos ^{2} \beta}= \\
& =\frac{H}{\sin \alpha} \sqrt{\frac{1+\cos 2 \alpha}{2}-\frac{1+\cos 2 \beta}{2}}=\frac{H}{\sin \alpha} \sqrt{\sin (\beta+\alpha) \sin (\beta-\alpha)}
\end{aligned}
$$

Volume of the prism:

$$
\begin{aligned}
& V=S_{\triangle A B C} \cdot A A_{1}=\frac{1}{2} A C \cdot B C \cdot A A_{1}=\frac{1}{2} \cdot \frac{H}{\sin \alpha} \sqrt{\sin (\beta+\alpha) \sin (\beta-\alpha)} \times \\
& \times \frac{H \cos \beta}{\sin \alpha} \cdot H=\frac{H^{3} \cos \beta}{2 \sin ^{2} \alpha} \sqrt{\sin (\beta+\alpha) \sin (\beta-\alpha)} \\
& \text { Answer: } \frac{H^{3} \cos \beta}{2 \sin ^{2} \alpha} \sqrt{\sin (\beta+\alpha) \sin (\beta-\alpha)}
\end{aligned}
$$