1. In an isosceles triangle ΔABC, AB = AC and ∠A = 80°. The bisector of ∠B and ∠C meet at D. The ∠BDC is equal to
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By: anil on 05 May 2019 01.48 am
Given : D is the incentre of $$ riangle$$ ABC and $$angle$$ BAC = 80° To find : $$angle$$ BDC = $$ heta$$ = ? Incentre of a triangle = $$90^circ+frac{angle A}{2}$$ => $$ heta=90^circ+frac{80^circ}{2}$$ => $$ heta=90^circ+40^circ$$ => $$ heta=130^circ$$ => Ans - (C)
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