12.116. The base of a right prism is an isosceles triangle, with the lateral side equal to $a$, and the angle between the lateral sides equal to $\alpha$. Find the volume of the prism if its lateral surface area is $S$.
Solution.

The volume of the prism \( V = S_{\text{base}} \cdot H \) (Fig. 12.116). The area of the base is: \( S_{\text{base}} = \frac{1}{2} a^2 \sin \alpha \). The lateral surface area \( S = P H \), hence \( H = \frac{S}{P} \). Let \( AC = x \). By the cosine theorem:

\[
x^2 = 2a^2 - 2a^2 \cos \alpha = 2a^2 (1 - \cos \alpha). \text{ Then } \quad x = a \sqrt{2(1 - \cos \alpha)} = 2a \sin \frac{\alpha}{2}
\]

Thus, \( P = 2a + 2a \sin \frac{\alpha}{2} = 2a \left(1 + \sin \frac{\alpha}{2}\right) \). Finally, we get:

\[
\begin{aligned}
& V = \frac{1}{2} a^2 \sin \alpha \frac{S}{2a \left(1 + \sin \frac{\alpha}{2}\right)} = \frac{a \sin \alpha}{4 \left(1 + \sin \frac{\alpha}{2}\right)} = \\
& = \frac{a S \cdot 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{4 \left(\cos^2 \frac{\alpha}{4} + \sin^2 \frac{\alpha}{4} + 2 \sin \frac{\alpha}{4} \cos \frac{\alpha}{4}\right)} = \frac{2a S \sin \frac{\alpha}{2} \left(\cos^2 \frac{\alpha}{4} - \sin^2 \frac{\alpha}{4}\right)}{4 \left(\cos \frac{\alpha}{4} + \sin \frac{\alpha}{4}\right)^2} = \\
& = \frac{a S \sin \frac{\alpha}{2} \left(\cos \frac{\alpha}{4} - \sin \frac{\alpha}{4}\right)}{2 \left(\cos \frac{\alpha}{4} + \sin \frac{\alpha}{4}\right)} = \frac{a S \sin \frac{\alpha}{2} \left(1 - \frac{\sin \frac{\alpha}{4}}{\cos \frac{\alpha}{4}}\right)}{2 \left(1 + \frac{\sin \frac{\alpha}{4}}{\cos \frac{\alpha}{4}}\right)} =
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-802.jpg?height=500&width=386&top_left_y=94&top_left_x=207)

Fig. 12.116

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-802.jpg?height=496&width=450&top_left_y=108&top_left_x=735)

Fig. 12.117

\[
= \frac{a S \sin \frac{\alpha}{2} \left(1 - \tan \frac{\alpha}{4}\right)}{2 \left(1 + \tan \frac{\alpha}{4}\right)} = \frac{a S \sin \frac{\alpha}{2} \left(\tan \frac{\pi}{4} - \tan \frac{\alpha}{4}\right)}{2 \left(1 + \tan \frac{\pi}{4} \tan \frac{\alpha}{4}\right)} = \frac{a S}{2} \sin \frac{\alpha}{2} \tan \left(\frac{\pi}{4} - \frac{\alpha}{4}\right)
\]

Answer: \(\frac{a S}{2} \sin \frac{\alpha}{2} \tan \frac{\pi - \alpha}{4}\).