Two tangents are drawn from a point P to a circle at A and B. O is the centre of the circle. If `angleAOP = 60^(@)`, then `angleAPB` is

Options

A. `120^(@)`
B. `90^(@)`
C. `60^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the angle \( \angle APB \) given that two tangents are drawn from point \( P \) to a circle at points \( A \) and \( B \), and that \( \angle AOP = 60^\circ \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have a circle with center \( O \).
- Two tangents \( PA \) and \( PB \) are drawn from point \( P \) to points \( A \) and \( B \) on the circle.
- The angle \( \angle AOP = 60^\circ \).

2. **Understand the Properties of Tangents**:
- The radius of the circle is perpendicular to the tangent at the point of contact. Therefore, \( OA \perp PA \) and \( OB \perp PB \).
- This means \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

3. **Calculate the Angle \( \angle AOB \)**:
- Since \( OA \) and \( OB \) are both radii of the circle, we can find \( \angle AOB \) using the angles around point \( O \).
- The angles around point \( O \) in triangle \( OAPB \) can be expressed as:
\[ \angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ \]
- We know:
- \( \angle OAP = 90^\circ \)
- \( \angle OBP = 90^\circ \)
- \( \angle AOP = 60^\circ \)
- Therefore, \( \angle AOB = 60^\circ \).

4. **Substituting the Known Angles**:
- Substitute the known angles into the equation:
\[ 60^\circ + 90^\circ + 90^\circ + \angle APB = 360^\circ \]
- Simplifying this gives:
\[ 60^\circ + 180^\circ + \angle APB = 360^\circ \]
- This simplifies to:
\[ 240^\circ + \angle APB = 360^\circ \]

5. **Solving for \( \angle APB \)**:
- Rearranging gives:
\[ \angle APB = 360^\circ - 240^\circ = 120^\circ \]

### Final Answer:
Thus, the angle \( \angle APB \) is \( 120^\circ \).