3. In tetrahedron $ABCD$, $AB=CD=15$, $BD=AC=20$, $AD=BC=\sqrt{337}$. Then the angle between $AB$ and $CD$ is $\qquad$ .
3. $\arccos \frac{7}{25}$.

As shown in Figure 3, the tetrahedron $A B C D$ is extended to form the rectangular prism $A E B F-H C G D$.
Let $B E=x, B F=y, B G=z$.
$$
\begin{array}{l}
\text { Then } x^{2}+y^{2}=225, y^{2}+z^{2}=400, z^{2}+x^{2}=337 \\
\Rightarrow x=9, y=12, z=16 .
\end{array}
$$

The angle between $A B$ and $C D$ is equal to the angle between $C D$ and $G H$. By the cosine rule, the angle is $\arccos \frac{7}{25}$.