12.443 The distances from the center of the base of a regular quadrilateral pyramid to a lateral face and to a lateral edge are equal to $a$ and $\boldsymbol{b}$, respectively. Find the dihedral angle at the base of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-314.jpg?height=324&width=363&top_left_y=642&top_left_x=76)

Let $O$ be the center of the given regular pyramid $F A B C D$. If $K$ is the midpoint of $C D$, then $C D \perp F K O$ (Theorem 11) $\Rightarrow F K O \perp F C D$ (Theorem 15) $\Rightarrow$ the perpendicular $O N$ to the face $F C D$ lies in the plane $F K O$. Let also $O M \perp A F, x=O K, h=F O$.

Then

$$
\left\{\begin{array}{l}
O M \cdot A F = A O \cdot F O, \\
O N \cdot K F = O K \cdot F O
\end{array} \Rightarrow \left\{\begin{array}{l}
b \cdot \sqrt{h^{2}+2 x^{2}}=\sqrt{2} \cdot x \cdot h, \\
a \cdot \sqrt{h^{2}+x^{2}}=x \cdot h
\end{array} \Rightarrow\right.\right.
$$

$\Rightarrow \frac{h^{2}+2 x^{2}}{h^{2}+x^{2}}=\frac{2 a^{2}}{b^{2}} \Rightarrow 1+\left(\frac{h}{x}\right)^{2}=\frac{b^{2}}{2 a^{2}-b^{2}}$.

But $1+\left(\frac{h}{x}\right)^{2}=1+\operatorname{tg}^{2} \angle F K O \Rightarrow \cos \angle F K O=\frac{\sqrt{2 a^{2}-b^{2}}}{b}$.

Answer: $\quad \arccos \frac{\sqrt{2 a^{2}-b^{2}}}{b}$