12.383. The vertex of the cone is located at the center of the sphere, and the base of the cone touches the surface of the sphere. The total surface area of the cone is equal to the surface area of the sphere. Find the angle between the slant height and the height of the cone.
Solution.

Consider the axial section of the given set of bodies (Fig.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1146.jpg?height=361&width=626&top_left_y=1059&top_left_x=638)

Fig. 12.233 12.233). Let the generatrix of the cone $O A=l$, the radius of its base $C A=r$, the radius of the sphere $O C=R, \angle A O C$ between the generatrix and the height of the cone is $\alpha$.

$$
\text { In } \triangle A C O\left(\angle A C O=90^{\circ}\right): r=R \operatorname{tg} \alpha, l=\frac{R}{\cos \alpha}
$$

The total surface area of the cone $S_{1}=\pi r(r+l)=\pi R^{2} \operatorname{tg} \alpha\left(\operatorname{tg} \alpha+\frac{1}{\cos \alpha}\right)$, the surface area of the sphere $S_{2}=4 \pi R^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1147.jpg?height=609&width=553&top_left_y=63&top_left_x=79)

Fig. 12.234

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1147.jpg?height=515&width=616&top_left_y=158&top_left_x=643)

Fig. 12.235

Since $S_{1}=S_{2}$ and $0^{\circ}<\alpha<90^{\circ}$, then

$$
\begin{aligned}
& \pi R^{2} \operatorname{tg} \alpha\left(\operatorname{tg} \alpha+\frac{1}{\cos \alpha}\right)=4 \pi R^{2} \Leftrightarrow \operatorname{tg} \alpha\left(\operatorname{tg} \alpha+\sqrt{1+\operatorname{tg}^{2} \alpha}\right)=4 \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg} \alpha \sqrt{1+\operatorname{tg}^{2} \alpha}=4-\operatorname{tg}^{2} \alpha \Leftrightarrow \operatorname{tg}^{2} \alpha\left(1+\operatorname{tg}^{2} \alpha\right)=\left(4-\operatorname{tg}^{2} \alpha\right)^{2} \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg}^{2} \alpha+\operatorname{tg}^{4} \alpha=16-8 \operatorname{tg}^{2} \alpha+\operatorname{tg}^{4} \alpha \Leftrightarrow 9 \operatorname{tg}^{2} \alpha=16 \Rightarrow \operatorname{tg} \alpha=\frac{4}{3} ; \alpha=\operatorname{arctg} \frac{4}{3} .
\end{aligned}
$$

Answer: $\operatorname{arctg} \frac{4}{3}$.