Point O - the center of the circle, A and B - points of the circle, angle AOB = 128 degrees. Calculate the magnitude of the angle OBC when C is a point on the circle and the angle CAO = 20 degrees.

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 \( 

This problem involves calculating the magnitude of angle OBC in a circle given that the central angle AOB is 128 degrees and the angle CAO is 20 degrees. The solution uses the Inscribed Angle Theorem and the Angle Sum Theorem. Suitable for students in Grades 9-10.

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