The problem involves a circle with center $O$. The angles and points in the circle are labeled as follows:

• $\angle ROQ = 124^\circ$.
• Points $P, Q, R$ are on the circumference, and $O$ is the center.

We need to find $\angle x$, which is at $\angle QPO$.

Analysis

1. Understanding the Circle Properties: Since $O$ is the center, $\angle ROQ$ is a central angle. The arc $RQ$ subtends both $\angle ROQ$ and $\angle RPQ$.

2. Using the Inscribed Angle Theorem: The inscribed angle $\angle RPQ$ is half the measure of the central angle $\angle ROQ$. Thus, $\angle RPQ = \frac{1}{2} \times \angle ROQ = \frac{1}{2} \times 124^\circ = 62^\circ.$

3. Finding $\angle QPO$: Since $OQ$ and $OP$ are radii of the circle, $\triangle OPQ$ is an isosceles triangle. Let $\angle OQP = \angle OPQ = x$.

   Using the angle sum property of triangles: $\angle OQP + \angle OPQ + \angle POQ = 180^\circ.$ Substituting known values: $x + x + 62^\circ = 180^\circ.$ Simplifying: $2x + 62^\circ = 180^\circ.$ $2x = 180^\circ - 62^\circ = 118^\circ.$ $x = 59^\circ.$

So, the value of $x$ is $59^\circ$, but it does not match any of the given options. Could you confirm if the problem's options are correct or if more details are needed?