11. Given in acute $\triangle A B C$, $\angle A=$ $50^{\circ}, A B>B C$. Then the range of $\angle B$ is $\qquad$ .
$11.40^{\circ}<\angle B<80^{\circ}$.
As shown in Figure 12, when $BC$ is the shortest, $\angle ABC=40^{\circ}$. Now, with point $B$ as the center and $AB$ as the radius, draw a circle intersecting line $AC$ at point $C_{1}$. When $BC_{1}=AB$, $\angle ABC_{1}=80^{\circ}$. Therefore, $40^{\circ}<\angle B<80^{\circ}$.