3.104. Find the angle between the generatrix and the base of a truncated cone, the total surface area of which is twice the surface area of a sphere inscribed in it.
3.104. Given, $A B C D-$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-569.jpg?height=450&width=515&top_left_y=577&top_left_x=58)

Fig. 3.104 is a truncated cone, $S$ - the center of the inscribed sphere (Fig. 3.104); $S_{\text {full trunc. cone }}: S_{\text {sphere }}=2 ;$ we need to find $\angle A B O$. Consider the axial section $A B C D$ of the truncated cone, $O O_{1}$ - the axis of the truncated cone. Introduce the notations: $S O=r$, $O B=R_{1}, O_{1} A=R_{2}$. Draw $A M \| O O_{1}$; then $A M=O O_{1}$. According to the property of the sides of a circumscribed quadrilateral, $2 A B=B C+$ $+A D$; we have $A B=R_{1}+R_{2}, B M=$ $=R_{1}-R_{2}$. From $\triangle A M B$ we find $R_{1}+R_{2}=\frac{2 r}{\sin \alpha}, R_{1}-R_{2}=2 r \operatorname{ctg} \alpha$, where $\alpha=\angle A B O$. Solving the system of equations

$$
R_{1}+R_{2}=\frac{2 r}{\sin \alpha}, R_{1}-R_{2}=2 r \operatorname{ctg} \alpha
$$

we get $R_{1}=r \operatorname{ctg} \frac{\alpha}{2}, R_{2}=r \operatorname{tg} \frac{\alpha}{2}$. Then

$$
\begin{gathered}
\frac{S_{\text {full trunc. cone }}}{S_{\text {sphere }}}=\frac{\pi\left(R_{1}^{2}+R_{2}^{2}+\left(R_{1}+R_{2}\right) A B\right)}{4 \pi r^{2}}= \\
=\frac{\pi\left(r^{2} \operatorname{ctg}^{2} \frac{\alpha}{2}+r^{2} \operatorname{tg}^{2} \frac{\alpha}{2}+\frac{4 r^{2}}{\sin ^{2} \alpha}\right)}{4 \pi r^{2}}=\frac{4 \pi r^{2}\left(1+\cos ^{4} \frac{\alpha}{2}+\sin ^{4} \frac{\alpha}{2}\right)}{4 \pi r^{2} \sin ^{2} \alpha}=
\end{gathered}
$$

$$
\begin{aligned}
& =\frac{1+\cos ^{4} \frac{\alpha}{2}+\sin ^{4} \frac{\alpha}{2}}{\sin ^{2} \alpha}=\frac{1+\left(\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}\right)^{2}-\frac{1}{2} \sin ^{2} \alpha}{\sin ^{2} \alpha}= \\
& =\frac{2-\frac{1}{2} \sin ^{2} \alpha}{\sin ^{2} \alpha}=\frac{4-\sin ^{2} \alpha}{2 \sin ^{2} \alpha}=2 ; 5 \sin ^{2} \alpha=4 ; \sin \alpha=\frac{2}{\sqrt{5}}=\frac{2 \sqrt{5}}{5}= \\
& \text { Thus, } \alpha=\arcsin \frac{2 \sqrt{5}}{5} .
\end{aligned}
$$