4.35. The generatrix of a truncated cone forms an angle $\alpha$ with the base plane. Inside the cone, there are two spheres touching each other and the lateral surface of the cone, with the first sphere touching the lower base of the cone and the second sphere touching the upper base. The distance between the centers of the spheres is $l$. Find the radii of the bases of the cone.
4.35. Let $O A=R$ and $O_{1} B=r$ be the radii of the bases of a truncated cone, and $M O=x$ and $N O_{1}=y(y<x)$ be the radii of the inscribed spheres (Fig. 4.31). By the condition, $x+y=l, \angle A=\alpha$. In $\triangle A O M$ and $\triangle B O_{1} N$ we have

$$
R=x \operatorname{ctg} \frac{\alpha}{2}, r=y \operatorname{ctg}\left(90^{\circ}-\frac{\alpha}{2}\right)=y \operatorname{tg} \frac{\alpha}{2}
$$

Next, in trapezoid $F H N M$, draw $N K \| H F$; since $N M=\ell$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-589.jpg?height=446&width=452&top_left_y=108&top_left_x=101)

Fig. 4.30

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-589.jpg?height=427&width=402&top_left_y=123&top_left_x=689)

Fig. 4.31

$K M=x-y, \angle N M K=\angle B A O=\alpha$, then from $\triangle N K M$ we get $x-y=$ $=l \cos \alpha$. Thus, we arrive at the system

$$
\left\{\begin{array}{l}
x+y=l \\
x-y=l \cos \alpha
\end{array}\right.
$$

from which

$$
x=\frac{l+l \cos \alpha}{2}, y=\frac{l-l \cos \alpha}{2}
$$

Substituting the found values into the equalities (1), we finally obtain

$$
\begin{gathered}
R=x \operatorname{ctg} \frac{\alpha}{2}=\frac{l(1+\cos \alpha) \operatorname{ctg} \frac{\alpha}{2}}{2}=l \cos ^{2} \frac{\alpha}{2} \operatorname{ctg} \frac{\alpha}{2} \\
r=y \operatorname{tg} \frac{\alpha}{2}=\frac{l(1-\cos \alpha) \operatorname{tg} \frac{\alpha}{2}}{2}=l \sin ^{2} \frac{\alpha}{2} \operatorname{tg} \frac{\alpha}{2}
\end{gathered}
$$

Answer: $l \cos ^{2} \frac{\alpha}{2} \operatorname{ctg} \frac{\alpha}{2}, l \sin ^{2} \frac{\alpha}{2} \operatorname{tg} \frac{\alpha}{2}$.