12.068. The lateral edge of a regular triangular pyramid forms an angle $\alpha$ with the side of the base. Find the angle between the lateral edge and the height of the pyramid and the permissible values of $\alpha$.
Solution.

Let the side of the base be denoted by \(a\) (Fig. 12.70), and the lateral edge by \(l\). Then, from the isosceles triangle, which is a lateral face of the pyramid, we find \(l = \frac{a}{2 \cos \alpha}\). Since the pyramid is regular, the base of the height is the center of the circle circumscribed around the triangular base of the pyramid. The radius of this circle is \(R = \frac{a \sqrt{3}}{3}\). Therefore, \(\sin \beta = \frac{R}{l} = \frac{a \sqrt{3} \cdot 2 \cos \alpha}{3 a} = \frac{2 \cos \alpha}{\sqrt{3}}\). Since \(0 < \sin \beta \leq 1\), we have the system of inequalities:
\[
\left\{\begin{array}{l}
0 < \frac{2 \cos \alpha}{\sqrt{3}} \leq 1 \\
0 < \alpha < \frac{\pi}{2}
\end{array}\right.
\]

Solving this system and considering that \(0 < \alpha < \frac{\pi}{2}\), we get \(\alpha \in \left(\frac{\pi}{6} ; \frac{\pi}{2}\right)\).

Answer: \(\arcsin \frac{2 \cos \alpha}{\sqrt{3}} ; \frac{\pi}{6} < \alpha < \frac{\pi}{2}\).