12.308. The base of the prism is an equilateral triangle with side $a$. The lateral edge is equal to $b$ and forms angles, each equal to $\alpha$, with the intersecting sides of the base. Find the volume of the prism and the permissible values of $\alpha$.
## Solution.

Let the equilateral triangle $ABC$ be the base of the prism $ABC A_1 B_1 C_1$ (Fig. 12.170), $AB = a, BB_1 = b, B_1 E$ be the height of the prism.

Since $\angle B_1 B C = \angle B_1 B A = \alpha$, the point $E$ lies on the bisector $BD$ of angle $ABC$.

$\angle B_1 B E$ is the angle of inclination of $B B_1$ to the plane of the base, $\angle B_1 B C$ is the angle between $B B_1$ and the line $BC$ in this plane, $\angle E B C$ is the angle between the line $BC$ and the projection of the line $B B_1$ onto the plane of the base.

Then $\cos \angle B_1 B C = \cos \angle B_1 B E \cdot \cos \angle E B C$;

$$
\cos \angle B_1 B E = \frac{\cos \angle B_1 B C}{\cos \angle E B C} = \frac{\cos \alpha}{\cos 30^\circ} = \frac{2 \cos \alpha}{\sqrt{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1063.jpg?height=623&width=781&top_left_y=76&top_left_x=300)

Fig. 12.171

$$
\begin{aligned}
& \sin \angle B_1 B E = \sqrt{1 - \cos^2 \angle B_1 B E} = \sqrt{1 - \frac{4 \cos^2 \alpha}{3}} = \frac{2}{\sqrt{3}} \sqrt{\frac{3}{4} - \cos^2 \alpha} = \\
& = \frac{2}{\sqrt{3}} \sqrt{\cos^2 30^\circ - \cos^2 \alpha} = \frac{2}{\sqrt{3}} \sqrt{\sin \left(\alpha + 30^\circ\right) \sin \left(\alpha - 30^\circ\right)} \\
& \text{In } \triangle B_1 E B \left(\angle B_1 E B = 90^\circ\right): \\
& B_1 E = B B_1 \sin \angle B_1 B E = \frac{2 b}{\sqrt{3}} \sqrt{\sin \left(\alpha + 30^\circ\right) \sin \left(\alpha - 30^\circ\right)}
\end{aligned}
$$

The volume of the prism $V = S_{\triangle ABC} \cdot B_1 E = \frac{a^2 \sqrt{3}}{4} \cdot \frac{2 b}{\sqrt{3}} \sqrt{\sin \left(\alpha + 30^\circ\right) \sin \left(\alpha - 30^\circ\right)} = \frac{a^2 b}{2} \sqrt{\sin \left(\alpha + 30^\circ\right) \sin \left(\alpha - 30^\circ\right)}$, where $\sin \left(\alpha + 30^\circ\right) \sin \left(\alpha - 30^\circ\right) > 0 \Leftrightarrow$ $\Leftrightarrow \cos 60^\circ - \cos 2 \alpha > 0 \Leftrightarrow \cos 2 \alpha < 0.5 \Leftrightarrow 60^\circ < 2 \alpha < 300^\circ$, i.e., $30^\circ < \alpha < 150^\circ$.

Answer: $\frac{a^2 b}{2} \sqrt{\sin \left(\alpha + 30^\circ\right) \sin \left(\alpha - 30^\circ\right)}, 30^\circ < \alpha < 150^\circ$.