65. In triangle $ABC$ with angle $\widehat{ABC}=60^{\circ}$, the bisector of angle $A$ intersects $BC$ at point $M$. On side $AC$, a point $K$ is taken such that $\widehat{AM} K=30^{\circ}$. Find $\widehat{OKC}$, where $O$ is the center of the circumcircle of triangle $AMC$.
65. If $\widehat{B A C}=2 \alpha$, it is easy to find that $\widehat{K M C}=\widehat{M K C}=$ $=30^{\circ}+\alpha$, i.e., $|M C|=|K C|$. Extend $M K$ until it intersects the circle at point $N$; $\triangle K M C$ is similar to $\triangle K A N$, so $|A N|=|K N|=R$ - the radius of the circle (since $\widehat{A M N}=30^{\circ}$).

Points $A, K$ and $O$ lie on a circle with center at $N$, $\widehat{A N O}=60^{\circ}$; therefore, $\widehat{A K O}=30^{\circ}$ or $150^{\circ}$, depending on whether angle $A M C$ is obtuse or acute.

Answer: $30^{\circ}$ or $150^{\circ}$.