Chords AC and BD of a circle with . centre O intersect at right angles at E. If `/_OAB = 25^@`, then the value of `/_EBC `is

Options

A. `30^@`
B. `25^@`
C. `20^@`
D. `15^@`

Answer

To solve the problem, we need to find the value of angle \( \angle EBC \) given that chords \( AC \) and \( BD \) intersect at right angles at point \( E \) and that \( \angle OAB = 25^\circ \).

### Step-by-Step Solution:

1. **Understanding the Circle and Chords**:
- We have a circle with center \( O \).
- Chords \( AC \) and \( BD \) intersect at point \( E \) at right angles (i.e., \( \angle AEB = 90^\circ \)).
- The angle \( \angle OAB \) is given as \( 25^\circ \).

2. **Finding \( \angle OBA \)**:
- Since \( OA \) and \( OB \) are radii of the circle, triangle \( OAB \) is isosceles.
- Therefore, \( \angle OBA = \angle OAB = 25^\circ \).

3. **Calculating \( \angle AOB \)**:
- The sum of angles in triangle \( OAB \) is \( 180^\circ \).
- Thus, \( \angle AOB = 180^\circ - \angle OAB - \angle OBA = 180^\circ - 25^\circ - 25^\circ = 130^\circ \).

4. **Finding \( \angle EAB \)**:
- Since \( E \) is the intersection of the chords, we can use the property that the angle at the center is double the angle at the circumference.
- Therefore, \( \angle AEB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 130^\circ = 65^\circ \).

5. **Finding \( \angle EBC \)**:
- In triangle \( AEB \), we know \( \angle AEB = 90^\circ \) and \( \angle EAB = 25^\circ \).
- We can find \( \angle EBC \) using the fact that the angles in triangle \( AEB \) sum up to \( 180^\circ \).
- Thus, \( \angle EBC = 90^\circ - \angle EAB = 90^\circ - 25^\circ = 65^\circ \).

### Final Answer:
The value of \( \angle EBC \) is \( 65^\circ \).