In triangle PQR, the sides PQ and PR are produced to A and B respectively. The bisectors of ∠AQR and ∠BRQ intersect at point O. If ∠QOR = 50°, then what is the value (in degrees) of ∠QPR?
Given : O is the excentre of $$\triangle$$ PQR and ∠QOR = 50°
To find : $$\angle$$ QPR = $$\theta$$ = ?
Excentre of a triangle = $$90^\circ-\frac{1}{2} \times $$ (Angle opposite to it)
=> $$50^\circ=90^\circ-\frac{\theta}{2}$$
=> $$\frac{\theta}{2}=90-50=40^\circ$$
=> $$\theta=2\times40=80^\circ$$
=> Ans - (C)
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