We are dealing with a circle where:

• $O$ is the center of the circle,
• $A$ and $B$ are points on the circle,
• $\angle AOB = 128^\circ$,
• $C$ is another point on the circle such that $\angle CAO = 20^\circ$.

We need to calculate $\angle OBC$.

Step-by-step solution:

1. Inscribed Angle Theorem: First, note that $\angle AOB$ is a central angle. In a circle, an inscribed angle subtended by the same arc is half of the central angle. So, the angle subtended by arc $AB$ at any other point on the circle (such as $C$) is half of $\angle AOB$.

   Therefore: $\angle ACB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 128^\circ = 64^\circ$

2. Angle Sum in Triangle $OAC$: Now, consider triangle $OAC$. Since $O$ is the center and $A$ and $C$ are points on the circle, the segments $OA$ and $OC$ are radii, meaning triangle $OAC$ is isosceles.

   The sum of angles in a triangle is $180^\circ$. Let $\angle OCA$ be the angle we want to find. We already know that $\angle CAO = 20^\circ$ and $\angle AOC$ is the remainder of the angle at $O$.

   $\angle AOC = 360^\circ - \angle AOB = 360^\circ - 128^\circ = 232^\circ$

   Using the triangle angle sum rule: $\angle OCA + \angle CAO + \angle AOC = 180^\circ$ Substituting the known values: $\angle OCA + 20^\circ + 232^\circ = 180^\circ$ Solving for $\angle OCA$: $\angle OCA = 180^\circ - 232^\circ - 20^\circ = 128^\circ$

Final Answer:

The magnitude of $\angle OBC$ is (