11.3. Answer. $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_d30157328a90ee3507f7g-1.jpg?height=503&width=334&top_left_y=1165&top_left_x=290)

Let $O$ be the common center of the given circles. From the condition, it follows that $BO$ and $CO$ are the bisectors of angles $ABC$ and $BCD$ ($O$ is the incenter) and, moreover, $AO = BO = CO$ ($O$ is the circumcenter). Therefore, if $\angle ABO = \alpha$, then $\angle OBC = \alpha, \angle OCB = \angle OBC = \alpha, \angle BCD = 2\alpha$ (CD is the bisector). Hence, $\angle OAC = \angle OCA = 3\alpha$. Thus, the sum of the angles of triangle ABC is $10\alpha$, from which $\alpha = 18^\circ$.