The chord AB of a circle of centre O subtends an angle `theta` with the tangent at A to the circle. Then measure of `/_ABO `is :

Options

A. `theta`
B. `90^@ - theta`
C. `90^@ + theta`
D. `2(180^@ - theta)`

Answer

To solve the problem, we need to find the measure of the angle \( \angle ABO \) given that the chord \( AB \) of a circle with center \( O \) subtends an angle \( \theta \) with the tangent at point \( A \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a circle with center \( O \).
- There is a chord \( AB \) in the circle.
- A tangent line at point \( A \) touches the circle.
- The angle between the chord \( AB \) and the tangent at point \( A \) is given as \( \theta \).

2. **Using the Alternate Segment Theorem**:
- According to the alternate segment theorem, the angle \( \angle BAQ \) (where \( Q \) is a point on the tangent line) is equal to the angle \( \angle BCA \) (where \( C \) is a point on the arc \( AB \)).
- Therefore, we can say:
\[ \angle BCA = \theta \]

3. **Finding the Angle at the Center**:
- The angle subtended at the center \( O \) by the chord \( AB \) is twice the angle subtended at any point on the circumference. Thus:
\[ \angle BOA = 2 \times \angle BCA = 2\theta \]

4. **Analyzing Triangle \( OAB \)**:
- In triangle \( OAB \), \( OA \) and \( OB \) are radii of the circle, hence they are equal.
- Let \( \angle OAB = x \) and \( \angle OBA = x \) (since \( OA = OB \)).
- The sum of angles in triangle \( OAB \) is \( 180^\circ \):
\[ x + x + 2\theta = 180^\circ \]
\[ 2x + 2\theta = 180^\circ \]

5. **Solving for \( x \)**:
- Dividing the entire equation by 2:
\[ x + \theta = 90^\circ \]
- Rearranging gives:
\[ x = 90^\circ - \theta \]

6. **Conclusion**:
- Therefore, the measure of angle \( \angle ABO \) is:
\[ \angle ABO = x = 90^\circ - \theta \]

### Final Answer:
The measure of \( \angle ABO \) is \( 90^\circ - \theta \).