1. 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?
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By: anil on 05 May 2019 03.03 pm
Given : I is the incentre of $$ riangle$$ PQR and $$angle$$ BAC = 130° To find : $$angle$$ QIR = $$ heta$$ = ? Incentre of a triangle = $$90^circ+frac{angle P}{2}$$ => $$ heta=90^circ+frac{130^circ}{2}$$ => $$ heta=90^circ+65^circ$$ => $$ heta=155^circ$$ => Ans - (C)
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