3.30. The lateral edge of a regular triangular prism is equal to the side of the base. Find the angle between the side of the base and the non-intersecting diagonal of the lateral face.
3.30. According to the condition, in the prism $A B C A_{1} B_{1} C_{1}$ we have $A B=B C=A C=A A_{1}$, i.e., all lateral faces of the prism are equal squares (Fig. 3.32). Let $a$ be the side of these squares; we need to find the angle between the diagonal $A_{1} C$ and the side $A B$. Since $A_{1} B_{1} \| A B$ and $A_{1} B_{1}$ has a common point with $A_{1} C$, the required angle is $\angle C A_{1} B_{1}$. Denote it by $\alpha$ and consider the isosceles triangle $C A_{1} B_{1}$, in which $C A_{1}=C B_{1}=a \sqrt{2}, A_{1} B_{1}=a, \angle A_{1} C B_{1}=180^{\circ}-2 \alpha$. By the sine theorem, we have

$$
\begin{gathered}
\frac{A_{1} B_{1}}{\sin \left(180^{\circ}-2 \alpha\right)}=\frac{C B_{1}}{\sin \alpha} ; \frac{a}{\sin 2 \alpha}=\frac{a \sqrt{2}}{\sin \alpha} ; \frac{1}{2 \cos \alpha}=\sqrt{2} \\
\cos \alpha=\frac{1}{2 \sqrt{2}}=\frac{\sqrt{2}}{4} ; \alpha=\arccos \frac{\sqrt{2}}{4}
\end{gathered}
$$