12.425 The ratio of the volume of a truncated cone to the volume of a sphere inscribed in it is $\boldsymbol{k}$. Find the angle between the slant height of the cone and the plane of its base and the permissible values of $k$.
## Solution.

Let $A B C D$ be the axial section of the given truncated cone. Then $A B C D$ is an isosceles trapezoid, in which a circle of radius $R$ (where $R$ is the radius of the sphere inscribed in the given cone) can be inscribed.

Then, if $B C=2 r_{1}, A D=2 r_{2}, C K$ is the height of $A B C D$, then $C D=r_{2}+\eta$ (Theorem 2),

$$
K D=r_{2}-r_{1} \Rightarrow 2 R=C K=\sqrt{C D^{2}-K D^{2}}=2 \sqrt{r_{1} r_{2}}
$$

Since the volume of the given truncated cone is $\frac{\pi}{3} \cdot C K\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)$, then $k=\frac{r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}}{2 R^{2}}=\frac{\left(r_{2}-r_{1}\right)^{2}+3 R^{2}}{2 R^{2}}=2\left(\frac{r_{2}-r_{1}}{2 R}\right)^{2}+\frac{3}{2}=2 \operatorname{ctg}^{2}(\angle C D K)+\frac{3}{2} \Rightarrow$ $\Rightarrow \operatorname{tg}(\angle C D K)=\frac{2}{\sqrt{2 k-3}}$ for $k>\frac{3}{2}$.

Answer: $\operatorname{arctg} \frac{2}{\sqrt{2 k-3}}$ for $k>\frac{3}{2}$.