SSC CGL 9 Aug 2015 Shift 2 Paper

Question : 123 of 200

Marks: +1, -0

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

Solution:

∠BQP = ∠BAP
∴ ∠BQP =

{∠A}/2

∠BQR = ∠BCR
∴ ∠BQR =

1/2

∠C
∴ ∠BQP + ∠BQR
=

1/2

(∠A + ∠C)
⇒ ∠PQR =

1/2

(180° - ∠B)
= 90° -

B/2

Go to Question:

Prev Question

Next Question