## Solution.

Let $S O$ be the height of the regular pyramid $S A B C$ (Fig. 12.151), and $F$ be the midpoint of $B C$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1042.jpg?height=628&width=572&top_left_y=1180&top_left_x=703)

Fig. 12.151

Then $\angle S F O$ is the angle between a lateral face and the base plane, $\angle S A O$ is the angle between a lateral edge and the base plane, and $\angle A S F$ is the angle between the lateral edge $S A$ and the apothem $S F$ of the opposite face.

Let $\angle S F O=\alpha, \angle S A O=\beta, \angle A S F=\gamma$.

By the given condition, $\operatorname{tg} \alpha=k$.

In $\triangle S O F\left(\angle S O F=90^{\circ}\right): S O=O F \operatorname{tg} \alpha$.

In $\triangle S O A\left(\angle S O A=90^{\circ}\right): S O=O A \operatorname{tg} \beta=2 O F \operatorname{tg} \beta \Rightarrow$

$\Rightarrow 2 \operatorname{tg} \beta=\operatorname{tg} \alpha, \operatorname{tg} \beta=\frac{k}{2}$.

In $\triangle A S F: \gamma=180^{\circ}-(\alpha+\beta) \Rightarrow$

$\Rightarrow \operatorname{tg} \gamma=\operatorname{tg}\left(180^{\circ}-(\alpha+\beta)\right)=-\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{\operatorname{tg} \alpha \operatorname{tg} \beta-1}=\frac{k+\frac{1}{2} k}{\frac{1}{2} k^{2}-1}=\frac{3 k}{k^{2}-2}$.

Answer: $\frac{3 k}{k^{2}-2}$.
12.288. The lateral face of a regular triangular pyramid forms an angle with the base plane, the tangent of which is equal to $k$. Find the tangent of the angle between the lateral edge and the apothem of the opposite face.