12.239. The ratio of the surface area of a sphere inscribed in a cone to the area of the base of the cone is $k$. Find the cosine of the angle between the slant height of the cone and the plane of its base and the permissible values of $k$.
## Solution.

Let $\triangle A B C$ be the axial section of the cone (Fig. 12.106), $D$ be the center of its base, $B D$ be the height, $O$ be the center of the inscribed sphere, $\angle B A D$ be the angle between the generatrix of the cone and the plane of its base, $\angle B A D=\alpha$, $D A=R, O D=r$.

$\angle O A D=\frac{1}{2} \angle B A D=\frac{\alpha}{2}$ and $r=R \operatorname{tg} \frac{\alpha}{2}$.

The surface area of the sphere $S_{1}=4 \pi r^{2}=4 \pi R^{2} \operatorname{tg}^{2} \frac{\alpha}{2}$.

The area of the base of the cone $S_{2}=\pi R^{2}$.

Then $\frac{S_{1}}{S_{2}}=4 \operatorname{tg}^{2} \frac{\alpha}{2} \Rightarrow 4 \operatorname{tg}^{2} \frac{\alpha}{2}=k \Rightarrow 4 \cdot \frac{1-\cos \alpha}{1+\cos \alpha}=k \Rightarrow \cos \alpha=\frac{4-k}{4+k}$.

Since $\alpha$ is an acute angle, then $0<\cos \alpha<1$.

Then $0<\frac{4-k}{4+k}<1 ; 0<k<4$.

Answer: $\frac{4-k}{4+k} ; 0<k<4$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0988.jpg?height=599&width=567&top_left_y=60&top_left_x=96)

Fig. 12.106

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0988.jpg?height=577&width=559&top_left_y=83&top_left_x=710)

Fig. 12.107