12.461 A cylinder is inscribed in a cone, the axial section of which is a right triangle; its lower base lies in the plane of the cone's axis. The ratio of the lateral surfaces of the cone and the cylinder is $4 \sqrt{2}$. Find the angle between the plane of the base of the cone and the line passing through the center of the upper base of the cylinder and an arbitrary point on the circumference of the base of the cone.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-326.jpg?height=253&width=384&top_left_y=418&top_left_x=95)
Let $\triangle A O B$ be the axial section of the given cone $\left(\angle A O B=90^{\circ}\right)$, and $M N K L$ be the axial section of the cylinder inscribed in it. Let $O P$ be the height of the cone, and $C$ be the center of the upper base of the cylinder. Let $h=K L$ and $R=P B$. Then $O B=R \sqrt{2}\left(\angle P B O=45^{\circ}\right)$, $\frac{\pi R \cdot O B}{\pi \cdot 2 P L \cdot h}=4 \sqrt{2} \quad$ (by the condition) $\Rightarrow$ $\Rightarrow \frac{R^{2} \sqrt{2}}{2 h(R-h)}=4 \sqrt{2} \Rightarrow\left(\frac{R}{h}\right)^{2}-8\left(\frac{R}{h}\right)+8=0 \Rightarrow \operatorname{ctg} \angle C B P=\frac{R}{h}=4 \pm 2 \sqrt{2}$.

Answer: $\operatorname{arcctg}(4 \pm 2 \sqrt{2})$.