12.104. The ratio of the lateral surface area of a regular triangular pyramid to the area of its base is $k$. Find the angle between a lateral edge and the height of the pyramid.
Solution.

By the condition, $SABC$ is a regular triangular pyramid, $S_{\text {side }}: S_{\triangle ABC}=k, SO \perp(ABC)$; we need to find $\angle ASO$ (Fig. 12.104). Let $SO=h$ and $\angle ASO=\alpha$; then in $\triangle SOA$ we have $AO=h \operatorname{tg} \alpha$, from which $AB=AO \sqrt{3}=h \operatorname{tg} \alpha \cdot \sqrt{3}, OD=\frac{1}{2} OA=\frac{1}{2} h \operatorname{tg} \alpha$. Draw $SD \perp AC$ and from $\triangle SOD$ find $SD=\sqrt{SO^{2}+OD^{2}}=\sqrt{h^{2}+\frac{1}{4} h^{2} \operatorname{tg}^{2} \alpha}=\frac{h}{2} \sqrt{4+\operatorname{tg}^{2} \alpha}$. Therefore, $S_{\text {side }}=\frac{3}{2} AB \cdot SD=\frac{3}{2} h \operatorname{tg} \alpha \cdot \sqrt{3} \cdot \frac{h}{2} \sqrt{4+\operatorname{tg}^{2} \alpha}=$ $\left.=\frac{3}{4} h^{2} \operatorname{tg} \alpha \sqrt{3\left(4+\operatorname{tg}^{2} \alpha\right)}\right), S_{\triangle ABC}=\frac{3 h^{2} \operatorname{tg}^{2} \alpha \cdot \sqrt{3}}{4}$. We have the equation

$$
\frac{3 h^{2} \operatorname{tg} \alpha \sqrt{3\left(4+\operatorname{tg}^{2} \alpha\right)} 4}{4 \cdot 3 h^{2} \operatorname{tg}^{2} \alpha \cdot \sqrt{3}}=k ; \frac{\sqrt{4+\operatorname{tg}^{2} \alpha}}{\operatorname{tg} \alpha}=k ; \sqrt{4 \operatorname{ctg}^{2} \alpha+1}=k
$$

$4 \operatorname{ctg}^{2} \alpha=k^{2}-1 ; \operatorname{ctg} \alpha=\frac{\sqrt{k^{2}-1}}{2}$,

from which $\alpha=\operatorname{arcctg} \frac{\sqrt{k^{2}-1}}{2}$.

Answer: $\quad \operatorname{arcctg} \frac{\sqrt{k^{2}-1}}{2}$.