12.441 The angle between the plane of square $A B C D(A B \| C D)$ and a certain plane $P$ is $\alpha$, and the angle between side $A B$ and the same plane is $\beta$. Find the angle between side $A D$ and plane $P$.
Solution. Without loss of generality, we can assume that the given plane $P$ passes through the vertex $A$ of the given square $ABCD$. Drop perpendiculars $BM$ and $DN$ to the plane $P$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-313.jpg?height=314&width=450&top_left_y=147&top_left_x=193)

Let the segment $KL$ in $\triangle ADN$ be such that $KL \| DN$ and $KL = BM$. Then $B K L M$ is a rectangle, and the line $l$ of intersection of $ABCD$ and the plane $P$ is parallel to $BK$ and $ML$ (Theorem 14). Therefore, if $AE$ and $AF$ are the altitudes of $\triangle ABK$ and $\triangle AML$, they are perpendicular to $l$, i.e., $\angle EAF = \alpha$.

Since $ML \perp AE, ML \perp AF$, then $ML \perp EF$ (Theorem 11) $\Rightarrow EF = BM$.

Thus, $AE = \frac{BM}{\sin \alpha}, AB = \frac{BM}{\sin \beta}, AK = \frac{KL}{\sin x}$ (where $x = \angle DAN$),

$BK \cdot AE = \sqrt{AB^2 + AK^2} \cdot AE = AB \cdot AK = 2S_{ABK} \Rightarrow$

$\Rightarrow \sqrt{\frac{1}{\sin^2 \beta} + \frac{1}{\sin^2 x}} = \frac{\sin \alpha}{\sin \beta \cdot \sin x} \Rightarrow \sin^2 x = \sin^2 \alpha - \sin^2 \beta = \sin (\alpha - \beta) \cdot \sin (\alpha + \beta)$.

Answer: $x = \arcsin \sqrt{\sin (\alpha - \beta) \cdot \sin (\alpha + \beta)}$.