3.47. The bases of two cones with a common vertex lie in the same plane. The difference in their volumes is $V$. Find the volume of the smaller cone, if the tangents drawn from an arbitrary point on the circumference of the base of the larger cone to the circumference of the base of the smaller cone form an angle $\alpha$.

保留源文本的换行和格式，直接输出翻译结果。
### 3.47. Let $V_{1}$ be the volume of the larger cone, and $V_{2}$ be the volume of the smaller cone (i.e., the one we are looking for). According to the problem, $V_{1}-V_{2}=V, A B=A C$, where $B$ and $C$ are the points of tangency, $\angle B A C=\alpha$ (Fig. 3.49). Since $A B=A C$, then $\angle O A B = \angle O A C = \frac{\alpha}{2}$, and from $\triangle O A B$ we find $O B = O A \sin \frac{\alpha}{2}$. Further, we have:

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-528.jpg?height=592&width=495&top_left_y=998&top_left_x=660)

Fig. 3.49 $\frac{1}{3} \pi O A^{2} \cdot S O - \frac{1}{3} \pi O A^{2} \sin ^{2} \frac{\alpha}{2} \cdot S O = V ; \frac{1}{3} \pi O A^{2} \cdot S O \left(1 - \sin ^{2} \frac{\alpha}{2}\right) = V ;$

$$
V_{1} \left(1 - \sin ^{2} \frac{\alpha}{2}\right) = V ; V_{1} = \frac{V}{\cos ^{2} \frac{\alpha}{2}}
$$

## Finally, we get

$$
V_{2} = V_{1} - V = \frac{V}{\cos ^{2} \frac{\alpha}{2}} - V = V \operatorname{tg}^{2} \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-529.jpg?height=480&width=417&top_left_y=385&top_left_x=45)

Fig. 3.50