[ Completing a Tetrahedron to a Parallelepiped ]

The cosine of the angle between the skew lines $A B$ and $C D$ is $\frac{\sqrt{35}}{10}$. Points $E$ and $F$ are the midpoints of segments $A B$ and $C D$ respectively, and the line $E F$ is perpendicular to the lines $A B$ and $C D$. Find the angle $A C B$, given that $A B=2 \sqrt{5}, C D=2 \sqrt{7}, E F=\sqrt{13}$.
Let's complete the tetrahedron $ABCD$ to a parallelepiped $AKBLNDCM$ ($AN\|KD\|BM\|LC$) by drawing pairs of parallel planes through its opposite edges. Then $KL\|DC$ and $MN\|AB$, so the segment $EF$, connecting the centers of the faces $AKBL$ and $NDMC$, is perpendicular to the planes of the bases $AKBL$ and $NDMC$. Since the line $EF$ is parallel to the lateral edges of the parallelepiped, this parallelepiped is a right parallelepiped. Let the angle $\angle BEL$ be acute. Then, using the cosine theorem in triangles $BEL$ and $AEL$, we find that

\[
\begin{aligned}
& BL^2 = BE^2 + LE^2 - 2 \cdot BE \cdot LE \cdot \cos \angle BEL = 5 + 7 - 2 \cdot \sqrt{5} \cdot \sqrt{7} \cdot \frac{\sqrt{35}}{10} = 5 \\
& AL^2 = AE^2 + LE^2 + 2 \cdot AE \cdot LE \cdot \cos \angle BEL = 5 + 7 + 2 \cdot \sqrt{5} \cdot \sqrt{7} \cdot \frac{\sqrt{35}}{10} = 19
\end{aligned}
\]

Using the Pythagorean theorem in the right triangles $BLC$ and $ALC$, we find that

\[
BC^2 = BL^2 + CL^2 = 5 + 13 = 18, \quad AC^2 = AL^2 + CL^2 = 19 + 13 = 32.
\]

Finally, using the cosine theorem,

\[
\cos \angle ACB = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{32 + 18 - 20}{2 \cdot 4 \sqrt{2} \cdot 3 \sqrt{2}} = \frac{5}{8}
\]

## Answer

$\arccos \frac{5}{8}$.