12.460 The radius of the sphere circumscribed around a regular quadrilateral pyramid is to the side of the base as 3:4. Find the angle between the lateral face and the base plane.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-325.jpg?height=348&width=364&top_left_y=1118&top_left_x=207)

Let $O$ be the center of the sphere circumscribed around the given regular pyramid $F A B C D$, $F K$ be the height of the pyramid, $a = A B$, and $P$ be the midpoint of $C D$. According to the problem, $A O = \frac{3}{4} a$.

Consider two cases.

1) $O$ lies between $F$ and $K$. Then

$$
O K = \sqrt{A O^{2} - A K^{2}} = \frac{a}{4}, F K = F O + O K = a \Rightarrow
$$

$$
\Rightarrow \operatorname{tg} \angle F P K = \frac{F K}{K P} = 2
$$

2) $O$ lies on the extension of $F K$. Then $K O = \frac{a}{4}, F K = F O - O K = \frac{a}{2} \Rightarrow$ $\Rightarrow \operatorname{tg} \angle F P K = 1$.

Answer: $\operatorname{arctg} 2$ or $\frac{\pi}{4}$.