3.55. A truncated cone is described around a sphere, with the area of one base being 4 times larger than the area of the other. Find the angle between the slant height of the cone and the plane of its base.

## Group B
3.55. The axial section of the figure is an isosceles trapezoid \( L M M_{1} L_{1} \), in which a circle with center \( O \) is inscribed

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-535.jpg?height=327&width=387&top_left_y=102&top_left_x=45)

Fig. 3.57

(Fig. 3.57). By the problem statement, \( L L_{1} \) and \( M M_{1} \) are the diameters of the lower and upper bases of the cone, \( A \) and \( B \) are the centers of these bases. Let the radii of the bases of the truncated cone be \( R \) and \( r \), then \( \pi R^{2}: \pi r^{2}=4 \), or \( R=2 r \). Draw \( M K \| A B \); we have \( M L=R+r=3 r \) (due to the property of segments of tangents drawn from one point to a circle), \( K L=B L-B K=B L- A M=2 r-r=r \). From \( \triangle M K L \) we find \( \cos \angle M L B=\frac{K L}{M L}=\frac{r}{3 r}=\frac{1}{3} \), i.e., \( \angle M L B=\arccos \frac{1}{3} \).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-535.jpg?height=386&width=527&top_left_y=749&top_left_x=46)

Fig. 3.58