3. In $\triangle A B C$, $A B=B C, \angle A B C=$ $20^{\circ}$. Take a point $M$ on side $A B$ such that $B M=A C$. Then the degree measure of $\angle A M C$ is $\qquad$
3. $\left(30^{\circ}\right)$.

Construct an equilateral triangle $K B C$ outside $\triangle A B C$ with $B C$ as a side, and connect $M K$. It is easy to prove that $\triangle A B C \cong \triangle K B M$. Therefore, $K B=K M=K C . \angle B K M=20^{\circ}$.
With $K$ as the center and $K B$ as the radius, the circle passes through $B, M, C$. $\angle M C B=\frac{1}{2} \angle B K M=10^{\circ}$. Hence, $\angle A M C=30^{\circ}$.