12.107. A truncated cone is described around a sphere, with the area of one base being four times the area of the other base. Find the angle between the slant height of the cone and the plane of its base.
Solution.

According to the condition $S_{2}=4 S_{1}$ (Fig. 12.107). From this, $\pi R_{2}^{2}=4 \pi R_{1}^{2} \Rightarrow R_{2}=2 R_{1}$ $\left(A O=2 B O_{1}\right)$. The required angle can be found as follows: $\cos \angle C D O=\frac{F D}{C D}=\frac{B O_{1}}{C D}$. It is known that $C F^{2}=B C \cdot A D=2 B O_{1} \cdot 2 A O=8 B O_{1}^{2}$. From $\triangle C F D\left(\angle C F D=90^{\circ}\right)$ $C F^{2}+F D^{2}=C D^{2}$. Therefore, $C D=\sqrt{8 B O_{1}^{2}+B O_{1}^{2}}=\sqrt{9 B O_{1}^{2}}=3 B O_{1}$.

Finally, we get: $\cos \angle C D O=\frac{B O_{1}}{3 B O_{1}}=\frac{1}{3} \Rightarrow \angle C D O=\arccos \left(\frac{1}{3}\right)$. Answer: $\quad \arccos \left(\frac{1}{3}\right)$.