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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:
Answer & Solution
Correct 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 = $${90^ \circ } - \frac{{\angle {\text{B}}}}{2}$$
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