3.89. Two cones have concentric bases and the same angle, equal to $\alpha$, between the height and the slant height. The radius of the base of the outer cone is $R$. The lateral surface area of the inner cone is half the total surface area of the outer cone. Find the volume of the inner cone.
3.89. By the condition, $S C D$ and $S_{1} C_{1} D_{1}$ are cones with a common center $O, \angle C S O=\angle C_{1} S_{1} O=\alpha, O C=R$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-557.jpg?height=498&width=415&top_left_y=109&top_left_x=46)

Fig. 3.89

Solving the equation $\frac{2 \pi R^{2}}{\sin \alpha}=\frac{\pi R^{2}(1+\sin \alpha)}{\sin \alpha}$, we get

$$
r^{2}=\frac{R^{2}(1+\sin \alpha)}{2}=R^{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \text {, i.e., } r=R \cos \left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \text {. }
$$

Thus,

$$
V_{S_{1} C_{1} D_{1}}=\frac{1}{3} \pi r^{3} \operatorname{ctg} \alpha=\frac{1}{3} \pi R^{3} \cos ^{3}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \operatorname{ctg} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-557.jpg?height=319&width=375&top_left_y=1338&top_left_x=59)

Fig. 3.90