In a triangle ABC, the incircle (centre O) touches BC, CA and AB at P, Q and R respectively. Calculate :

(i) ∠QOR

(ii) ∠QPR, given that ∠A = 60°.

(i) From figure,

OQ ⊥ AC and OR ⊥ AB (∵ OQ and OR are the radii and AC and AB are tangents.)

Now in quadrilateral AROQ,

∠A = 60°, ∠ORA = 90° and ∠OQA = 90°.

∠A + ∠ORA + ∠OQA + ∠QOR = 360°
60° + 90° + 90° + ∠QOR = 360°
240° + ∠QOR = 360°
∠QOR = 360° - 240°
∠QOR = 120°.

Hence, the value of ∠QOR = 120°.

(ii) Arc QR subtends ∠QOR at the centre and ∠QPR at the remaining part of the circle.

∴ ∠QOR = 2∠QPR (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)

120° = 2∠QPR
∠QPR = $\dfrac{120°}{2}$ = 60°.

Hence, the value of ∠QPR = 60°.