In Figure, if `P R` is tangent to the circle at `P` and `Q` is the centre of the circle, then `/_P O Q=`

Options

A. `110^0`
B. `100^0`
C. `120^0`
D. `90^0`

Answer

To find the angle \( \angle POQ \) in the given scenario, we can follow these steps:

### Step 1: Identify the Components
- We know that \( PR \) is a tangent to the circle at point \( P \).
- \( Q \) is the center of the circle.
- \( OP \) and \( OQ \) are the radii of the circle.

### Step 2: Establish Relationships
- Since \( OP \) and \( OQ \) are both radii of the circle, we have:
\[ OP = OQ \]
- Therefore, triangle \( OPQ \) is isosceles, which means:
\[ \angle OPQ = \angle OQP \]

### Step 3: Use the Tangent Property
- The tangent at any point on a circle is perpendicular to the radius at that point. Therefore:
\[ \angle OPR = 90^\circ \]

### Step 4: Determine Remaining Angles
- We also know that \( \angle RPQ \) is given as \( 60^\circ \).
- Thus, we can find \( \angle OPR \) and \( \angle RPQ \):
\[ \angle OPQ = \angle OQP = \angle OPR - \angle RPQ = 90^\circ - 60^\circ = 30^\circ \]

### Step 5: Apply the Triangle Sum Property
- The sum of angles in triangle \( OPQ \) is \( 180^\circ \):
\[ \angle OPQ + \angle OQP + \angle POQ = 180^\circ \]
- Substituting the known values:
\[ 30^\circ + 30^\circ + \angle POQ = 180^\circ \]

### Step 6: Solve for \( \angle POQ \)
- Simplifying the equation:
\[ 60^\circ + \angle POQ = 180^\circ \]
- Therefore:
\[ \angle POQ = 180^\circ - 60^\circ = 120^\circ \]

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

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