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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