3. In tetrahedron $ABCD$,
\[
\begin{array}{l}
\angle ABC=\angle BAD=60^{\circ}, \\
BC=AD=3, AB=5, AD \perp BC,
\end{array}
\]
$M, N$ are the midpoints of $BD, AC$ respectively. Then the size of the acute angle formed by lines $AM$ and $BN$ is \qquad (express in radians or inverse trigonometric functions).
3. $\arccos \frac{40}{49}$.

From $A M=\frac{1}{2}(A B+A D)$, we get
$|A M|^{2}=\frac{1}{4}\left(|A B|^{2}+|A D|^{2}+2 A B \cdot A D\right)=\frac{49}{4}$.
Thus, $|A M|=\frac{7}{2}$.
Similarly, $|B N|=\frac{7}{2}$.
Also, $A M \cdot B N=\frac{1}{4}(A B+A D) \cdot(B A+B C)$
$=\frac{1}{4}\left(-|A B|^{2}-A B \cdot A D-B A \cdot B C+A D \cdot B C\right)$ $=-10$,
which means $|A M| \cdot|B N| \cos \langle A M, B N\rangle=-10$.
Therefore, the size of the acute angle formed by $A M$ and $B N$ is $\arccos \frac{40}{49}$.