Find an angle not on the circumference of a circle problem. I know this has to be extremely easy but I'm not going to solve this problem.
The task is to find the angle at point $A$.
Thanks!

 A: By extended central angle theorem the angle $A=(175^\circ- 49^\circ)/2=63^\circ $.
A: Notice that $\angle EOF+\angle BOC+49^\circ+175^\circ=360^\circ$. It follows that $\angle EOF+\angle BOC=136^\circ$.
$\angle OCA=\dfrac{180^\circ-\angle BOC}{2}$ and $\angle OEA=\dfrac{180^\circ-\angle EOF}{2}$ because $\triangle OBC$ and $\triangle OEF$ are isosceles respectively.
Now $\angle OAC+\angle OCA+\angle BOC+\angle BOA=180^\circ$ and $\angle OAE+\angle OEA+\angle EOF+\angle FOA=180^\circ$. Summing these two gives $$\begin{align}
(\angle OAC+\angle OAE)+\angle OCA+\angle OEA+(\angle BOC+\angle EOF)+(\angle BOA+\angle FOA)&=360^\circ\\
\angle EAC+\angle OCA+\angle OEA+136^\circ+49^\circ&=\\
\angle EAC+\frac{180^\circ-\angle BOC}{2}+\frac{180^\circ-\angle EOF}{2}+136^\circ+49^\circ&=\\
2\angle EAC+(180^\circ-\angle BOC)+(180^\circ-\angle EOF)+370&=720^\circ\\
2\angle EAC+360^\circ-(\angle BOC+\angle EOF)+370&=\\
2\angle EAC+360^\circ-136^\circ+370&=\\
2\angle EAC&=126^\circ\\
\angle EAC&=63^\circ
\end{align}$$
