12.238. A sphere is inscribed in a truncated cone. The sum of the lengths of the diameters of the upper and lower bases of the cone is five times the length of the radius of the sphere. Find the angle between the generatrix of the cone and the plane of the base.
## Solution.

Let the isosceles trapezoid $ABCD$ be the axial section of the given truncated cone, $E$ and $F$ be the centers of its lower and upper bases, $O$ be the center of the inscribed sphere, $EF$ be its diameter and the height of the trapezoid $ABCD$ (Fig. 12.105). If $R$ is the radius of the sphere, then $EF = 2R, AD + BC = 5R$.

$BC + AD = AB + CD = 2CD \Rightarrow CD = \frac{5R}{2}$.

$CK$ is the height of the trapezoid $ABCD, CK = EF = 2R$.

$\angle CDK$ is the angle between the generatrix $CD$ of the cone and the plane of the base.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0987.jpg?height=468&width=930&top_left_y=73&top_left_x=189)

Fig. 12.105

From $\triangle CKD \left(\angle CKD = 90^\circ\right): \sin \angle CDK = \frac{CK}{CD} = \frac{4}{5}; \angle CDK = \arcsin \frac{4}{5}$.

Answer: $\arcsin \frac{4}{5}$.