Answer : (b) 110º
PO is joined.
Since the circle isthe incircle for ∆ ABC, PO, QO, RO are the angle bisector of ∠P, ∠Q and ∠R respectively.
∠CPO = ∠APO
= \(\frac{1}{2}\)× 40° = 20°
Also, OA ⊥ PR, OC ⊥ PQ, OB ⊥ QR (Radii ⊥ tangent at point of contact)
⇒ ∠OAP = 90°
⇒ ∠AOP = y = 180° – (90° + 20°) = 70°
Now, y + y + x + x + z + z = 360°
⇒ 2y + 2 (x + z) = 360°
⇒ 2(x + z) = 360° – 2y = 360° – 140° = 220°
⇒ x + z = 110°
⇒ ∠QOR = 110°