Let P and Q be two points on a circle with centre O. If two tan gents of the circle through P and O meet at A with `/_PAQ = 48^@`, then `/_APQ` is

Options

A. `96^@`
B. ` 48^@`
C. `66^@`
D. `60^@`

Answer

To solve the problem, we need to find the angle \( \angle APQ \) given that \( \angle PAQ = 48^\circ \) and the tangents from points P and Q to the circle are equal in length.

### Step-by-Step Solution:

1. **Draw the Circle and Tangents:**
- Draw a circle with center O.
- Mark points P and Q on the circle.
- Draw tangents from points P and Q that meet at point A outside the circle.

2. **Identify the Angles:**
- We know that \( \angle PAQ = 48^\circ \).
- Since PA and QA are tangents from point A to the circle, the lengths of the tangents from a point outside the circle to the circle are equal. Thus, \( PA = QA \).

3. **Use the Isosceles Triangle Property:**
- In triangle \( APQ \), since \( PA = QA \), it is an isosceles triangle.
- Therefore, the angles opposite to the equal sides are equal. This means \( \angle APQ = \angle AQP \).

4. **Sum of Angles in Triangle:**
- The sum of angles in triangle \( APQ \) is \( 180^\circ \):
\[ \angle APQ + \angle AQP + \angle PAQ = 180^\circ \]
- Since \( \angle APQ = \angle AQP \), we can denote \( \angle APQ \) as \( x \):
\[ x + x + 48^\circ = 180^\circ \]
\[ 2x + 48^\circ = 180^\circ \]

5. **Solve for \( x \):**
- Subtract \( 48^\circ \) from both sides:
\[ 2x = 180^\circ - 48^\circ \]
\[ 2x = 132^\circ \]
- Divide by 2:
\[ x = 66^\circ \]

6. **Conclusion:**
- Therefore, \( \angle APQ = 66^\circ \).

### Final Answer:
\[ \angle APQ = 66^\circ \]