12.449 The side of the base of a regular triangular pyramid is equal to $a$. The lateral face forms an angle $\alpha$ with the base plane. Find the distance between a lateral edge and a non-intersecting side of the base.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-318.jpg?height=289&width=341&top_left_y=337&top_left_x=76)
Let $FO$ be the height of the given regular pyramid $FABC$, and $K$ be the midpoint of $AC$. Then $AC \perp FK, AC \perp BK \Rightarrow AC \perp FBK$ (Theorem 11) $\Rightarrow$ $AC \perp KP$, where $KP$ is the height of $\triangle FBK$, i.e., $KP$ is the common perpendicular of $AC$ and $FB$. $\triangle KPB \sim \triangle FBO \Rightarrow \frac{KP}{FO} = \frac{KB}{FB}$. But $FO = KO \cdot \tan \angle FKO = \frac{a}{2\sqrt{3}} \cdot \tan \alpha, KB = \frac{a\sqrt{3}}{2}$,

$$
FB = FC = \sqrt{FK^2 + KC^2} = \sqrt{\frac{KO^2}{\cos^2 \alpha} + \frac{a^2}{4}} = \frac{a}{2\sqrt{3}} \sqrt{\tan^2 \alpha + 4}
$$

Answer: $\quad KP = \frac{a \sqrt{3} \cdot \tan \alpha}{2 \sqrt{4 + \tan^2 \alpha}}$