If the angles$A=B=10$ and the bow $AM=40$ then find the measure of bow $BN$.($O$ is the center.
My Attempt:The angle $AOM$ equal to $40$ degrees then the angle $AON$ equal to $140$ degrees then angle $ACO$ equal to $30$ degrees. now what I have to do????
Let $x=\angle{BON}$.
As you found, we have $$\angle{ACO}=40^\circ-10^\circ=30^\circ$$ Also, $$\angle{ANO}=\frac 12\angle{AOM}=20^\circ$$ from which $\angle{CAN}=30^\circ-20^\circ=10^\circ$ follows.
By the way, since $\angle{OAC}=\angle{OBC}$, the four points $O,A,B,C$ are concyclic, and so $\angle{CAB}=\angle{COB}=x$ from which $\angle{BAN}=x-10^\circ$ follows.
Thus, since $\angle{BON}=2\angle{BAN}$, we have $$x=2(x-10^\circ)\quad\Rightarrow\quad \angle{BON}=x=\color{red}{20^\circ}$$