12.446 Side $BC$ of triangle $ABC$, lying at the base of the inclined prism $ABC A_1 B_1 C_1 \left(A A_1 \left\| B B_1 \right\| C C_1\right)$, is equal to $a$, and the angles adjacent to it are $\beta$ and $\gamma$. Find the angle between the lateral edge and the plane of the base, if the volume of the prism is $V$ and $A A_1 = A_1 B = A_1 C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-316.jpg?height=305&width=368&top_left_y=415&top_left_x=82)

Let $A_{1} K$ be the height of the given prism $A B C A_{1} B_{1} C_{1}, \angle B=\beta, \angle C=\gamma$.

From the equality $A A_{1}=A_{1} B=A_{1} C$ it follows that $K$ is the center of the circumcircle of $\triangle A B C$ (equal inclined lines have equal projections) with radius $A K$. By the Law of Sines for $\triangle A B C$,

$$
A B=\frac{a \cdot \sin \gamma}{\sin \left(180^{\circ}-\beta-\gamma\right)}, \quad A C=\frac{a \cdot \sin \beta}{\sin \left(180^{\circ}-\beta-\gamma\right)}
$$

Moreover, $S_{A B C}=\frac{A B \cdot B C \cdot A C}{4 \cdot A K}$. From this, we have:

$$
\operatorname{tg} \angle A_{1} A K=\frac{A_{1} K}{A K}=\frac{V}{S_{A B C} \cdot A K}=\frac{4 V}{A B \cdot A C \cdot B C}=\frac{4 V \cdot \sin ^{2}(\beta+\gamma)}{a^{3} \cdot \sin \beta \cdot \sin \gamma}
$$

Answer: $\quad \operatorname{arctg} \frac{4 V \cdot \sin ^{2}(\beta+\gamma)}{a^{3} \cdot \sin \beta \cdot \sin \gamma}$.