12.351. A cylinder is inscribed in a cone, the height of which is equal to the diameter of the base of the cone. The total surface area of the cylinder is equal to the area of the base of the cone. Find the angle between the slant height of the cone and the plane of its base.
## Solution.

Let $D O_{2}$ be the height of the given cone, $O_{2}$ be the common center of the base of the cone and the lower base of the cylinder, $O_{1}$ be the center of the upper base of the cylinder, $D A$ be the generatrix of the cone, $C B$ be the generatrix of the cylinder, $C$ belongs to $D A$ (Fig. 12.212), $O_{2} B=r, O_{2} A=R, O_{1} O_{2}=B C=H$. By the condition $H=2 R$.

The total surface area of the cylinder $S=2 \pi r H+2 \pi r^{2}=4 \pi r R+2 \pi r^{2}$.

By the condition

$$
4 \pi r R+2 \pi r^{2}=\pi R^{2} \Rightarrow 2 r^{2}+4 r R-R^{2}=0 ; r=\frac{-2 R \pm R \sqrt{6}}{2}
$$

Since $r>0$, then $r=\frac{R(\sqrt{6}-2)}{2}$.

$\angle D A O_{2}=\alpha$ - the angle between the generatrix of the cone and the plane of its base.

From $\triangle A B C\left(\angle A B C=90^{\circ}\right):$

$$
\begin{aligned}
& \operatorname{ctg} \alpha=\frac{A B}{B C}=\frac{O_{2} A-O_{2} B}{B C}=\frac{R-r}{H}=\frac{R-\frac{R(\sqrt{6}-2)}{2}}{2 R}=\frac{1-\frac{\sqrt{6}-2}{2}}{2}=\frac{4-\sqrt{6}}{4} \\
& \operatorname{tg} \alpha=\frac{4}{4-\sqrt{6}}=\frac{4(4+\sqrt{6})}{(4-\sqrt{6})(4+\sqrt{6})}=\frac{4(4+\sqrt{6})}{10}=\frac{2(4+\sqrt{6})}{5} \\
& \alpha=\operatorname{arctg} \frac{2(4+\sqrt{6})}{5} \\
& \quad \text { Answer: } \operatorname{arctg} \frac{2(4+\sqrt{6})}{5}
\end{aligned}
$$