In the figure,$AOC$ is a diameter of the circle and $\operatorname{arc} AXB = \frac{1}{2} \operatorname{arc} BYC$. Find $\angle BOC$.

$(120^{\circ})$ We know that the angle subtended by an arc at the center is proportional to the length of the arc.
Given that $\operatorname{arc} AXB = \frac{1}{2} \operatorname{arc} BYC$,it follows that $\angle AOB = \frac{1}{2} \angle BOC$.
Since $AOC$ is a diameter,$\angle AOB + \angle BOC = 180^{\circ}$ (linear pair).
Substituting $\angle AOB = \frac{1}{2} \angle BOC$ into the equation,we get:
$\frac{1}{2} \angle BOC + \angle BOC = 180^{\circ}$
$\frac{3}{2} \angle BOC = 180^{\circ}$
$\angle BOC = 180^{\circ} \times \frac{2}{3} = 120^{\circ}$.