$4-$  [ Volume of a Tetrahedron and Pyramid $\quad$]

In the triangular pyramid $A B C D$, it is known that $A B=8, C D=12$, the distance between the lines $A B$ and $C D$ is 6, and the volume of the pyramid is 48. Find the angle between the lines $A B$ and $C D$.
Let's complete the pyramid $ABCD$ to a parallelepiped by drawing three pairs of parallel planes through the opposite edges. Then the volume of the pyramid is one-third of the volume of the parallelepiped. As the planes of the bases of the parallelepiped, we take the parallel planes passing through the lines $AB$ and $CD$. Then the height of the parallelepiped is equal to the distance between these lines, the area of the base is equal to the area of the parallelogram, the diagonals of which are equal to the segments $AB$ and $CD$, and the angle $\varphi$ between the diagonals is equal to the angle between these lines. Let $V$ be the volume of the pyramid, $d$ be the distance between the lines $AB$ and $CD$. Then

$$
V=\frac{1}{3} \frac{1}{2} AB \cdot CD \cdot \sin \varphi \cdot d=\frac{1}{3} \frac{1}{2} 8 \cdot 12 \cdot \sin \varphi \cdot 6=96 \cdot \sin \varphi=48
$$

from which $\sin \varphi=1 / 2$. Therefore, $\varphi=30^{\circ}$.

## Answer

$30^{\circ}$.

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