3.24. The plane of the square forms an angle $\alpha$ with the plane passing through one of its sides. What angle does the diagonal of the square form with the same plane?
3.24. Figure 3.26 shows a square $ABCD$ inclined to the plane $\gamma$ passing through its side $CD$. Draw $AL \perp \gamma$; then $DL$ is the projection of $AD$ on the plane $\gamma$. Since $AD \perp DC$, it follows that $LD \perp DC$ and, therefore, $\angle ADL = \alpha$ is the linear angle of the dihedral angle between the plane of the square $ABCD$ and the plane $\gamma$.

Let $a$ be the side of the square. Draw the diagonal $AC$ of the square; then

$AC = a \sqrt{2}$. We need to find $\angle ACL$, where $CL$ is the projection of

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-514.jpg?height=529&width=593&top_left_y=99&top_left_x=563)

Figure 3.26 $AC$ on the plane $\gamma$. From $\triangle ALD$ we have $AL = a \sin \alpha$, and from $\triangle ALC$ we find

$$
\sin \angle ACL = \frac{AL}{AC} = \frac{a \sin \alpha}{a \sqrt{2}} = \frac{\sin \alpha}{\sqrt{2}}
$$

i.e., $\angle ACL = \arcsin \frac{\sin \alpha}{\sqrt{2}}$.