Question

In the given figure, ‘O’ is the centre of the semi-circle and ∠MON =48°, Find ∠MRP.

• A. 60°
• B. 66°
• C. 54°
• D. 72°

Answer 1

Answer: (B)

66°

Answer 2

The correct option is (B): 66°

Clearly, ∠PMQ is a semi-circular angle.
So, ∠PMQ = 90°
Again, ∠MPN = ($\frac{1}{2}$) × ∠MON = ($\frac{1}{2}$) × 48 = 24° {The angle subtended
by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle}
In ΔMPR,
∠MRP = 180° – 90° – 24° = 66°