5. As shown in Figure 2, in $\triangle A B C$, $A C=B C, \angle C=20^{\circ}$, and points $M, N$ are on sides $A C, B C$ respectively, such that $\angle B A N=50^{\circ}, \angle A B M=$ $60^{\circ}$. Then $\angle N M B=$ $\qquad$
5. $30^{\circ}$.

It is easy to know that $A B=B N$.
As shown in Figure 5, take a point $D$ on side $A M$ such that $B D=B A$.
Then $B D=B N$, and
$\angle A B D=20^{\circ}$.
Thus, $\angle B D N=60^{\circ}$,
$\angle M D N=40^{\circ}$.
Therefore, $\angle D M N=70^{\circ}$,
$\angle N M B=30^{\circ}$.