12.309. The base of the prism is a rectangle. The lateral edge forms equal angles with the sides of the base and is inclined to the plane of the base at an angle $\alpha$. Find the angle between the lateral edge and the side of the base.
Solution.

Let the rectangle $ABCD$ be the base of the prism $ABCD A_1 B_1 C_1 D_1$, and $A_1 E$ be the height of the prism (12.171). Then $\angle A_1 A E$ is the angle of inclination of the edge

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1064.jpg?height=500&width=495&top_left_y=79&top_left_x=120)

Fig. 12.172

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1064.jpg?height=446&width=470&top_left_y=128&top_left_x=758)

Fig. $12.172, \mathrm{a}$

$A A_1$ to the plane of the base, $\angle A_1 A E = \alpha$. Since $\angle A_1 A D = \angle A_1 A B$, point $E$ lies on the bisector $A F$ of angle $BAD$.

Then $\cos \angle A_1 A D = \cos \angle A_1 A E \cdot \cos \angle E A D = \cos \alpha \cos 45^\circ$, $\angle A_1 A D = \arccos \frac{\sqrt{2} \cos \alpha}{2}$

Answer: $\arccos \frac{\sqrt{2} \cos \alpha}{2}$.