11. Let $I$ be the incenter of $\triangle A B C$. Since $90^{\circ}+\alpha / 2=\angle B I C=\angle D I E=$ $138^{\circ}$, we obtain that $\angle A=96^{\circ}$. ! Let $D^{\prime}$ and $E^{\prime}$ be the points symmetric to $D$ and $E$ with respect to $C E$ and $B D$ respectively, and let $S$ be the intersection point of $E D^{\prime}$ and $B D$. Then $\angle B D E^{\prime}=24^{\circ}$ and $\angle D^{\prime} D E^{\prime}=\angle D^{\prime} D E-\angle E^{\prime} D E=24^{\circ}$, which means that $D E^{\prime}$ bisects the angle $S D D^{\prime}$. Moreover, $\angle E^{\prime} S B=\angle E S B=$ $\angle E D S+\angle D E S=60^{\circ}$ and hence $S E^{\prime}$ bisects the angle $D^{\prime} S B$. It follows that $E^{\prime}$ is the excenter of $\triangle D^{\prime} D S$ and consequently $\angle D^{\prime} D C=\angle D D^{\prime} C=$ $\angle S D^{\prime} E^{\prime}=\left(180^{\circ}-72^{\circ}\right) / 2=54^{\circ}$. Finally, $\angle C=180^{\circ}-2 \cdot 54^{\circ}=72^{\circ}$ and $\angle B=12^{\circ}$.