12.301. The base of a right prism is an isosceles triangle with a lateral side $a$ and an angle $\alpha$ between the lateral sides. The diagonal of the lateral face opposite the given angle forms an angle $\varphi$ with the adjacent lateral face. Find the volume of the prism.
## Solution.

In $\triangle ABC$, which is the base of the right prism $ABC A_1 B_1 C_1$ (Fig. 12.163), $AC = BC = a, \angle ACB = \alpha$. Then $S_{\triangle ABC} = \frac{1}{2} a^2 \sin \alpha$,

$AB = 2a \sin \frac{\alpha}{2}$. In the plane $ABC$, draw the perpendicular $BD$ to $AC$. Since the planes of the base and the lateral face of the right prism are perpendicular, $BD$ is perpendicular to the lateral face $AA_1 C_1 C \Rightarrow A_1 D$ is the projection of $A_1 B$ on the plane of the lateral face $AA_1 C_1 C, \angle BA_1 D$ is the angle between $A_1 B$ and this face, $\angle BA_1 D = \varphi, BD \perp A_1 D$.

In $\triangle BDC \left(\angle BDC = 90^\circ\right): BD = BC \sin \angle BCD = a \sin \alpha$.

In $\triangle A_1DB \left(\angle A_1DB = 90^\circ\right): A_1B = \frac{BD}{\sin \angle BA_1D} = \frac{a \sin \alpha}{\sin \varphi}$.

In $\triangle A_1AB \left(\angle A_1AB = 90^\circ\right)$:

\[
\begin{aligned}
& A A_1 = \sqrt{A_1 B^2 - A B^2} = \sqrt{\frac{a^2 \sin^2 \alpha}{\sin^2 \varphi} - 4a^2 \sin^2 \frac{\alpha}{2}} = \\
& = \sqrt{\frac{4a^2 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} - 4a^2 \sin^2 \frac{\alpha}{2} \sin^2 \varphi}{\sin^2 \varphi}} = \frac{2a \sin \frac{\alpha}{2}}{\sin \varphi} \sqrt{\cos^2 \frac{\alpha}{2} - \sin^2 \varphi} = \\
& = \frac{2a \sin \frac{\alpha}{2}}{\sin \varphi} \sqrt{\frac{1 + \cos \alpha}{2} - \frac{1 - \cos 2\varphi}{2}} = \frac{2a \sin \frac{\alpha}{2}}{\sin \varphi} \sqrt{\cos \left(\frac{\alpha}{2} + \varphi\right) \cos \left(\frac{\alpha}{2} - \varphi\right)}.
\end{aligned}
\]

The volume of the prism

\[
\begin{aligned}
& V = S_{\triangle ABC} \cdot A A_1 = \frac{a^3 \sin \alpha \sin \frac{\alpha}{2}}{\sin \varphi} \sqrt{\cos \left(\varphi + \frac{\alpha}{2}\right) \cos \left(\varphi - \frac{\alpha}{2}\right)} \\
& \text{Answer: } \frac{a^3 \sin \alpha \sin \frac{\alpha}{2}}{\sin \varphi} \sqrt{\cos \left(\varphi + \frac{\alpha}{2}\right) \cos \left(\varphi - \frac{\alpha}{2}\right)}
\end{aligned}
\]