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:

Answer: Option A

From chord BP
∠BAP = ∠BQP = z
(Angle subtended by same chord on circumference)
Similarly from chord RB
∠RCB = ∠RQB = y
In ΔABC
2x + 2y + 2z = 180°
x + y + z = 90°
y + z = 90° - x
∠RQP = 90° - x
∠RQP = $$90^{\circ }-\frac{\angle \text{B}}{2}$$