3.45. The diagonals of the axial section of a truncated cone are divided by the point of intersection in the ratio $2: 1$. The angle between the diagonals, facing the base of the cone, is $\alpha$. The length of the diagonal is $l$. Find the volume of the truncated cone.
3.45. Let $A A_{1} B_{1} B$ be the axial section of a truncated cone, where $A A_{1}$ and $B B_{1}$ are its generators, $A B$ and $A_{1} B_{1}$ are the diameters of its bases, $F$ and $K$ are the centers of these bases, and $A B_{1} \cap A_{1} B=O$ (Fig. 3.47). According to the condition, $A O: O B_{1}=2: 1, \angle A O B=\alpha, A B_{1}=l$. We have $A O=\frac{2}{3} l, O B_{1}=\frac{1}{3} l, \angle O A B=\angle O B A=\frac{\pi}{2}-\frac{\alpha}{2}$. Draw $B_{1} E \perp A B$ and from $\triangle A B_{1} E$ we find $B_{1} E=l \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=l \cos \frac{\alpha}{2}$. Then from $\triangle O F A$ we get that $A F=A O \sin \frac{\alpha}{2}=\frac{2}{3} l \sin \frac{\alpha}{2}$, and finally from $\triangle O K B_{1}$ we get that $K B_{1}=O B_{1} \sin \frac{\alpha}{2}=\frac{1}{3} l \sin \frac{\alpha}{2}$. Therefore,

$$
\begin{gathered}
V_{\text {truncated cone}}=\frac{\pi}{3} l \cos \frac{\alpha}{2}\left(\frac{4}{9} l^{2} \sin ^{2} \frac{\alpha}{2}+\frac{2}{9} l^{2} \sin ^{2} \frac{\alpha}{2}+\frac{1}{9} l^{2} \sin ^{2} \frac{\alpha}{2}\right)= \\
=\frac{7 \pi}{27} l^{3} \cos \frac{\alpha}{2} \sin ^{2} \frac{\alpha}{2}=\frac{7 \pi l^{3}}{54} \sin \alpha \sin \frac{\alpha}{2}
\end{gathered}
$$