12.320. The lateral faces of a regular triangular prism are squares. Find the angle between the diagonal of a lateral face and a non-intersecting side of the base of the prism.
## Solution.

The angle between the lateral diagonal $B_{1} C$ of the lateral face of a regular prism $A B C A_{1} B_{1} C_{1}$ (Fig. 12.181) and the side $A B$ of the base is the angle between $B_{1} C$ and $A_{1} B_{1}$.

If the length of the edge of the prism is 1, then the diagonal of the square $B_{1} C = \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1076.jpg?height=894&width=818&top_left_y=72&top_left_x=237)

Fig. 12.182

Let $D$ be the midpoint of $A_{1} B_{1}$. Then $C D$ is the height and median of the isosceles $\triangle A_{1} C B_{1}$.

In $\triangle C D B_{1} \left(\angle C D B_{1} = 90^{\circ}\right)$:

$\cos \angle A_{1} B_{1} C = \frac{B_{1} D}{B_{1} C} = \frac{1}{2} : \sqrt{2} = \frac{\sqrt{2}}{4} \Rightarrow \angle A_{1} B_{1} C = \arccos \frac{\sqrt{2}}{4}$.

Answer: $\arccos \frac{\sqrt{2}}{4}$.