12.429 A sphere is inscribed in a cone. The circle of contact between the spherical and conical surfaces divides the volume of the sphere in the ratio $5: 27$. Find the angle between the generator and the base plane.
## Solution.

Let $A B C$ be the axial section of the given cone,

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-305.jpg?height=270&width=303&top_left_y=546&top_left_x=161)

$O$ - the center of the sphere inscribed in it with radius $R, M$ and $N$ - the points of tangency of the sphere with the generators of the cone $B A$ and $B C, K$ - the point of intersection of $M N$ and $O B$. Clearly, the circle of tangency of the spherical and conical surfaces (with diameter $M N$) divides the sphere into two segments, and the volume $V_{c}$ of the upper segment is $\pi h^{2}\left(R-\frac{1}{3} h\right)$, where $h=R-O K$ is the height of the segment.

Let $\angle B C A=x$. Then $\angle B O N=x, \quad h=R-R \cos x \Rightarrow$

$$
\Rightarrow V_{c}=\frac{\pi R^{3}}{3}(1-\cos x)^{2}(2+\cos x)
$$

By the problem's condition, $\left(\frac{4}{3} \pi R^{3}-V_{c}\right): V_{c}=27: 5 \Rightarrow(1-\cos x)^{2}(2+\cos x)=\frac{5}{8} \Rightarrow$

$$
\Rightarrow(2 \cos x-1)\left(4 \cos ^{2} x+2 \cos x-11\right)=0 \Rightarrow
$$

Answer: $x=\frac{\pi}{3}$.