Question 139
In an isosceles triangle PQR, ∠P = 130$$^\circ$$. If I is the in-centre of the triangle, then what is the value (in degrees) of ∠QIR?
Solution
Given : I is the incentre of $$\triangle$$ PQR and $$\angle$$ BAC = 130°
To find : $$\angle$$ QIR = $$\theta$$ = ?
Incentre of a triangle = $$90^\circ+\frac{\angle P}{2}$$
=> $$\theta=90^\circ+\frac{130^\circ}{2}$$
=> $$\theta=90^\circ+65^\circ$$
=> $$\theta=155^\circ$$
=> Ans - (C)
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