jmc

Labeling our vertices will help a great deal, as will drawing a few radii: First of all, we see that $\angle ABC=110^{\circ }$ must be half of the major arc $AEC,$ thus arc $AEC=2\cdot \angle ABC.$ Then, the minor arc $AC$ must be $360^{\circ }-2\cdot \angle ABC=360^{\circ }-2\cdot 110^{\circ }=140^{\circ }.$

Likewise, the minor arc $EA$ must be $360^{\circ }-2\cdot \angle EFA=360^{\circ }-2\cdot 105^{\circ }=150^{\circ },$ and the minor arc $CE$ is $360^{\circ }-2\alpha .$ Now, arc $AC,$ $CE,$ and $EA$ must add up to $360^{\circ },$ which means that $$\begin{array}{rl}360^{\circ }& =(360^{\circ }-2\alpha )+140^{\circ }+150^{\circ }\\ 360^{\circ }& =650^{\circ }-2\alpha \\ 2\alpha & =290^{\circ }\\ \alpha & =\boxed{145^{\circ }}.\end{array}$$