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ABC is a cyclic triangle and the bisectors of ∠BAC, ∠ABC and ∠BCA meet the circle at P, Q, and R respectively. Then the angle ∠RQP is
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90° - B 2 -
90° + B 2 -
90° + C 2 -
90° - A 2
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Correct Option: A
As per the given in question , we draw a figure of a cyclic triangle ABC and the bisectors of ∠BAC, ∠ABC and ∠BCA meet the circle at P, Q, and R 
∠BQP =∠BAP
| ∴ ∠BQP = | |
| 2 |
∠BQR = ∠BCR
| ∴ ∠BQR = | ∠C | |
| 2 |
| ∴ ∠BQP + ∠BQR = | (∠A + ∠C) | |
| 2 |
| ⇒ ∠PQR = | (180° - ∠B) | |
| 2 |
[we know that , ∠A + ∠B + ∠C = 180°]
| ∠PQR = 90° – | |
| 2 |