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

(i) ∠QOR

(ii) ∠QPR;

given that ∠A = 60°.

ΔABC with its incircle having center O and touching BC, CA and AB at points P, Q and R, respectively is shown in the below figure:

(i) We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠ORA = ∠OQA = 90°.

In quadrilateral AROQ,

∠ORA + ∠OQA + ∠QOR + ∠A = 360° [∵ Sum of interior angles in a quadrilateral = 360°]

⇒ 90° + 90° + ∠QOR + 60° = 360°

⇒ 240° + ∠QOR = 360°

⇒ ∠QOR = 360° - 240°

⇒ ∠QOR = 120°.

Hence, ∠QOR = 120°.

(ii) From figure,

Arc RQ subtends ∠ROQ at center and ∠QPR at the remaining part of the circle.

∴ ∠QPR = $\dfrac{1}{2}$∠QOR

⇒ ∠QPR = $\dfrac{1}{2} \times 120°$ = 60°.

Hence, ∠QPR = 60°.