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To solve for \(\angle OBC\), we can use the following observations and steps based on the properties of circles and angles.

### Key Information:
- The diagram is a circle with center \(O\), and points \(A\), \(B\), and \(C\) lie on the circumference.
- Given angles: \(\angle OAB = 24^\circ\) and \(\angle ACB = 31^\circ\).
  
### Step-by-Step Analysis:

1. **Recognizing that \(OA\) and \(OB\) are radii**:
   Since \(OA\) and \(OB\) are radii of the circle, triangle \(OAB\) is isosceles, meaning \(\angle OBA = \angle OAB = 24^\circ\).

2. **Finding \(\angle AOB\)**:
   The sum of the angles in triangle \(OAB\) must equal 180°. Therefore,
   \[
   \angle AOB = 180^\circ - \angle OAB - \angle OBA = 180^\circ - 24^\circ - 24^\circ = 132^\circ.
   \]
   
3. **Using properties of cyclic quadrilaterals**:
   Since points \(A\), \(B\), and \(C\) are on the circumference, quadrilateral \(AOCB\) is cyclic. The angle subtended by the same arc on the opposite side of the circle is supplementary. That is, 
   \[
   \angle AOB + \angle ACB = 180^\circ.
   \]
   Substituting the values we know:
   \[
   132^\circ + \angle ACB = 180^\circ \quad \Rightarrow \quad \angle ACB = 48^\circ.
   \]

4. **Conclusion for \(\angle OBC\)**:
   Since \(OB\) is a radius and \(\angle OBC\) is the angle subtended by \(OB\), and we have already used cyclic properties, \(\angle OBC\) is related to the geometry of the diagram directly. With angle calculations depending

Learn how to solve for angle OBC in a cyclic quadrilateral using angle properties and circle theorems. Given angles include ∠OAB = 24° and ∠ACB = 31°, and the solution uses properties of isosceles triangles and cyclic quadrilaterals. Suitable for students in Grades 9-11.

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