4. The side of the base of a regular triangular pyramid is $a$, and the lateral edge is inclined to the base at an angle of $60^{\circ}$. Determine the distance from the centroid of the base to the lateral edge of the pyramid.
Solution. The side face $S A C$ is an equilateral triangle whose height $S M$ with the height $M N$ of the base forms an angle of $60^{\circ}, \measuredangle S B M=60^{\circ}$. The required distance is the segment $T N$, the leg of the right triangle MTN. According to this,

$$
\begin{aligned}
& \overline{M T}=\frac{1}{3} \cdot \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{6}, \quad \sin 60^{\circ}=\frac{\overline{T N}}{\overline{M T}} \\
& \overline{T N}=\frac{a \sqrt{3}}{6} \cdot \frac{\sqrt{3}}{2}=\frac{a}{4}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e3f82bd39eecc817d651g-3.jpg?height=490&width=428&top_left_y=1440&top_left_x=1055)

## Third Year