12.327. Two cones have a common height; their vertices lie at opposite ends of this height. The generatrix of one cone is equal to $l$ and forms an angle $\alpha$ with the height. The generatrix of the other cone forms an angle $\beta$ with the height. Find the volume of the common part of both cones.
## Solution.

Consider the axial section of the given set of cones (Fig. 12.188): $\triangle A B C$ - the axial section of the cone with slant height $l, \triangle M L K-$ the axial section of the second cone, $B L-$ their common height, $\angle A B L=\alpha$, $\angle M L B=\beta$.

The common part of these cones is two cones with a common base, $\triangle E B F$ and $\triangle E L F$ - their axial sections, $O$ - the center of their common base, $r$ the radius of this base.

Let $B O=h_{1}, L O=h_{2}, B L=H$.

Then the desired volume $V=\frac{1}{3} \pi r^{2} h_{1}+\frac{1}{3} \pi r^{2} h_{2}=\frac{1}{3} \pi r^{2} H$.

In $\triangle E O B\left(\angle E O B=90^{\circ}\right): h_{1}=r \operatorname{ctg} \alpha$.

In $\triangle E O L\left(\angle E O L=90^{\circ}\right): h_{2}=r \operatorname{ctg} \beta$.

Then $H=h_{1}+h_{2}=r(\operatorname{ctg} \alpha+\operatorname{ctg} \beta) \Rightarrow r=\frac{H \sin \alpha \sin \beta}{\sin (\alpha+\beta)}$.

In $\triangle A L B\left(\angle A L B=90^{\circ}\right): H=l \cos \alpha$. Thus,

\[
\begin{aligned}
& V=\frac{1}{3} \pi r^{2} H=\frac{\pi H^{3} \sin ^{2} \alpha \sin ^{2} \beta}{3 \sin ^{2}(\alpha+\beta)}=\frac{\pi l^{3} \cos ^{3} \alpha \sin ^{2} \alpha \sin ^{2} \beta}{3 \sin ^{2}(\alpha+\beta)}= \\
& =\frac{\pi l^{3} \sin ^{2} 2 \alpha \cos \alpha \sin ^{2} \beta}{12 \sin ^{2}(\alpha+\beta)}
\end{aligned}
\]

Answer: $\frac{\pi l^{3} \sin ^{2} 2 \alpha \cos \alpha \sin ^{2} \beta}{12 \sin ^{2}(\alpha+\beta)}$.