If PQ and PR be the two tangents to a circle with centre O such that `/_QPR=120^(@)` then `/_POQ` is

Options

A. `90^(@)`
B. `45^(@)`
C. `30^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to find the angle \( \angle POQ \) given that \( \angle QPR = 120^\circ \) and \( PQ \) and \( PR \) are tangents to a circle with center \( O \).

### Step-by-step Solution:

1. **Draw the Diagram**:
- Start by drawing a circle with center \( O \).
- Mark a point \( P \) outside the circle from which two tangents \( PQ \) and \( PR \) are drawn to the circle, touching it at points \( Q \) and \( R \) respectively.

2. **Identify the Angles**:
- Since \( PQ \) and \( PR \) are tangents to the circle, the angles \( \angle OPQ \) and \( \angle OPR \) are both \( 90^\circ \) (tangent to radius).
- We know \( \angle QPR = 120^\circ \).

3. **Use Triangle Properties**:
- In triangle \( PQR \), we can find the angles \( \angle PQR \) and \( \angle PRQ \) using the fact that the sum of angles in a triangle is \( 180^\circ \):
\[ \angle PQR + \angle PRQ + \angle QPR = 180^\circ \]
- Let \( \angle PQR = \angle PRQ = x \). Thus, we have:
\[ x + x + 120^\circ = 180^\circ \]
\[ 2x = 60^\circ \implies x = 30^\circ \]
- Therefore, \( \angle PQR = 30^\circ \) and \( \angle PRQ = 30^\circ \).

4. **Find \( \angle POQ \)**:
- Now, consider triangle \( POQ \):
- We know \( \angle OPQ = 90^\circ \) and \( \angle PQR = 30^\circ \).
- The angle \( \angle POQ \) can be found using the triangle angle sum property:
\[ \angle POQ + \angle OPQ + \angle PQR = 180^\circ \]
\[ \angle POQ + 90^\circ + 30^\circ = 180^\circ \]
\[ \angle POQ + 120^\circ = 180^\circ \]
\[ \angle POQ = 180^\circ - 120^\circ = 60^\circ \]

### Final Answer:
Thus, \( \angle POQ = 60^\circ \).