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Question
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
- ∠QOR
- ∠QPR;
given that ∠A = 60°.
Solution
The incircle touches the sides of the triangle ABC and OP ⊥ BC, OQ ⊥ AC, OR ⊥ AB
i. In quadrilateral AROQ,
∠ORA = 90°, ∠OQA = 90°, ∠A = 60°
∠QOR = 360° – (90° + 90° + 60°)
∠QOR = 360° – 240°
∠QOR = 120°
ii. Now arc RQ subtends ∠QOR at the centre and ∠QPR at the remaining part of the circle.
∴ `∠QPR = 1/2 ∠QOR`
`=> ∠QPR = 1/2 xx 120^circ`
`=>` ∠QPR = 60°
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