12.053. The lateral surface area of a regular triangular pyramid is 5 times the area of its base. Find the plane angle at the vertex of the pyramid.
## Solution.

Let $SABC$ be a regular triangular pyramid, $S_{\text {side}}=5 S_{\triangle ABC}$ (Fig. 12.55); we need to find $\angle ASB$. Let $SA=l$ and $\angle ASB=\alpha$; then $S_{\text {side}}=\frac{3}{2} l^{2} \sin \alpha$. From $\triangle ASB$ using the cosine theorem, we find $AB^{2}=2 l^{2}-2 l^{2} \cos \alpha=4 l^{2} \sin ^{2} \frac{\alpha}{2};$ then $S_{\triangle ABC}=\frac{4 l^{2} \sin ^{2} \frac{\alpha}{2} \sqrt{3}}{4}=$ $=l^{2} \sqrt{3} \sin ^{2} \frac{\alpha}{2}$. Using the condition, we have: $\frac{3}{2} l^{2} \sin \alpha=5 l^{2} \sqrt{3} \sin ^{2} \frac{\alpha}{2}$; $3 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}=5 \sqrt{3} \sin ^{2} \frac{\alpha}{2} ; \operatorname{tg} \frac{\alpha}{2}=\frac{\sqrt{3}}{5} ; \frac{\alpha}{2}=\operatorname{arctg} \frac{\sqrt{3}}{5}$, i.e., $\alpha=2 \operatorname{arctg} \frac{\sqrt{3}}{5}$.

Answer: $2 \operatorname{arctg} \frac{\sqrt{3}}{5}$.