12.364. The base of the pyramid $SABC$ is an equilateral triangle $ABC$. The edge $SA$ is perpendicular to the base plane. Find the angle between the lateral face $SBC$ and the base plane, if the lateral surface area of the pyramid is to the base area as $11: 4$.
## Solution.

Let $E$ be the midpoint of $BC$ (Fig. 12.224). Then $AE \perp BC, SE \perp BC$, and $\angle SEA$ is the angle between the lateral face $SBC$ and the base plane of the pyramid.

Let $\angle SEA = \alpha, AE = 1$.

Then from $\triangle SAE \left(\angle SAE = 90^\circ\right)$:

$AS = AE \operatorname{tg} \angle SEA = \operatorname{tg} \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1124.jpg?height=638&width=604&top_left_y=632&top_left_x=656)

Fig. 12.224

$$
SE = \frac{AE}{\cos \angle SEA} = \frac{1}{\cos \alpha}
$$

The lateral surface area of the pyramid $S_6 = 2 S_{\triangle ASC} + S_{\triangle BSC} = SA \cdot AC +$

$$
\begin{aligned}
& + \frac{1}{2} BC \cdot SE = BC \left(\operatorname{tg} \alpha + \frac{1}{2 \cos \alpha}\right) \text{ the base area } S_{\text{base}} = \frac{1}{2} BC \cdot AE = \\
& = \frac{1}{2} BC \cdot \operatorname{tg} \alpha \text{ since } S_6 = \frac{11}{4} S_{\text{base}}, \text{ then } BC \left(\operatorname{tg} \alpha + \frac{1}{2 \cos \alpha}\right) = \frac{11}{8} BC \Rightarrow \\
& \quad \Rightarrow \operatorname{tg} \alpha + \frac{1}{2 \cos \alpha} = \frac{11}{8} \Leftrightarrow 8 \sin \alpha + 4 = 11 \cos \alpha \Rightarrow
\end{aligned}
$$

$\frac{16 \operatorname{tg} \frac{\alpha}{2}}{1 + \operatorname{tg}^2 \frac{\alpha}{2}} + 4 = 11 \frac{1 - \operatorname{tg}^2 \frac{\alpha}{2}}{1 + \operatorname{tg}^2 \frac{\alpha}{2}} \Rightarrow 16 \operatorname{tg} \frac{\alpha}{2} + 4 + 4 \operatorname{tg}^2 \frac{\alpha}{2} = 11 - 11 \operatorname{tg}^2 \frac{\alpha}{2} \Leftrightarrow$

$\Leftrightarrow 15 \operatorname{tg}^2 \frac{\alpha}{2} + 16 \operatorname{tg} \frac{\alpha}{2} - 7 = 0 \Rightarrow$

1) $\operatorname{tg} \frac{\alpha}{2} = -\frac{7}{5}$;
2) $\operatorname{tg} \frac{\alpha}{2} = \frac{1}{3}$.

Since $0^\circ < \alpha < 90^\circ \Rightarrow \operatorname{tg} \frac{\alpha}{2} > 0 \Rightarrow$

$$
\operatorname{tg} \frac{\alpha}{2} = \frac{1}{3}, \operatorname{tg} \alpha = \frac{2 \operatorname{tg} \frac{\alpha}{2}}{1 - \operatorname{tg}^2 \frac{\alpha}{2}} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{3}{4}, \alpha = \operatorname{arctg} \frac{3}{4}
$$

Answer: $\operatorname{arctg} \frac{3}{4}$.