12.217. 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.
Solution.

Let trapezoid $ABCD, AB=CD$ (Fig. 12.84) be the axial section of the given truncated cone, $O$ - the center of the inscribed sphere, $M, N, K$ - the points of tangency of the sphere with the diameters $BC$ and $AD$ of the bases of the cone and its generatrix $CD$.

If $r$ and $R$ are the radii of the upper and lower bases of the truncated cone, $l$ is the generatrix, $x$ is the radius of the inscribed sphere, $\angle CDN=\alpha$, then $CM=CK=r, DN=DK=R, l=CD=CK+KD=R+r, MN=2xl$.

The total surface area of the truncated cone

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0966.jpg?height=422&width=810&top_left_y=76&top_left_x=271)

Fig. 12.84

$S=\pi(R+r)l+\pi R^2+\pi r^2=\pi(l^2+R^2+r^2)=\pi(l^2+(R+r)^2-2Rr)=$ $=\pi(2l^2-2Rr)=2\pi(l^2-Rr)$.

Let $CL$ be the height of trapezoid $ABCD, CL=MN=2x$, then from $\triangle CLD l=CD=\frac{CL}{\sin \angle CDN}=\frac{2x}{\sin \alpha}$.

$\angle COD=90^\circ, OK \perp CD, OK=x, CK \cdot KD=OK^2, Rr=x^2 \Rightarrow S=$

$=2\pi\left(\frac{4x^2}{\sin^2 \alpha}-x^2\right)=2\pi x^2\left(\frac{4}{\sin^2 \alpha}-1\right)$.

The surface area of the sphere $S_1=4\pi x^2$.

Since $S=2S_1$, then $2\pi x^2\left(\frac{4}{\sin^2 \alpha}-1\right)=8\pi x^2 \Rightarrow \sin^2 \alpha=\frac{4}{5}; \sin \alpha=$ $=\frac{2}{\sqrt{5}}; \alpha=\arcsin \frac{2}{\sqrt{5}}$.

Answer: $\arcsin \frac{2}{\sqrt{5}}$.