11. In $\triangle A B C$, $A B=2, B C=3$, and there exists a point $D$ inside $\triangle A B C$ such that $C D=2, \angle A D C = 180^{\circ}-\angle B$. For such a triangle, the range of $\cos B$ is
11. Fold $\triangle A C D$ to the exterior of $\triangle A C B$ to get $\triangle A C D^{\prime}$ (as shown in the figure). That is, $C D^{\prime}=2, \angle D^{\prime}+\angle B$ $=\angle A D C+\angle B=180^{\circ}, \therefore A, B, C, D^{\prime}$ are concyclic $\because C D^{\prime}=A B, \quad \therefore A D^{\prime} \parallel B C$, $\triangle B C D^{\prime}$ is an isosceles trapezoid. $\because D$ is inside $\triangle A B C, \therefore \angle D^{\prime} C A=\angle D C A<\angle B C A=\angle C A D^{\prime}$, $\triangle D^{\prime}<C D^{\prime}=2$. So $A D^{\prime}=2-3,0<\delta<2$ . Then $\cos B=\frac{\frac{1}{2}\left(B C-A D^{\prime}\right)}{A B}=\frac{1+B}{4}$.
$\therefore$ The range of $\cos B$ is $\frac{1}{4}<\cos B<\frac{3}{4}$.