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

As shown in Figure 3, the tetrahedron $ABCD$ is extended to form the rectangular parallelepiped $AE BF-HCGD$.
Let $BE=x, BF=y, BG=z$.
Then $x^{2}+y^{2}=225, y^{2}+z^{2}=400, z^{2}+x^{2}=337$
$\Rightarrow x=9, y=12, z=16$.
Also, the angle between $AB$ and $CD$ is equal to the angle between $CD$ and $GH$. By the cosine rule, the angle is $\arccos \frac{7}{25}$.