1. If the $$\angle ABC$$ and $$\angle ACB$$ of triangle ABC is $$80^\circ$$ and $$60^\circ$$ respectively. If the Incenter of the triangle is at point ‘I’ then calculate angle BIC.
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By: anil on 05 May 2019 03.00 pm
Given : I is the incentre of $$ riangle$$ ABC and $$angle$$ ABC = $$80^circ$$ and $$angle$$ ACB = $$60^circ$$ To find : $$angle$$ BIC = $$ heta$$ = ? Solution : Sum of angles of $$ riangle$$ ABC : => $$angle$$ ABC + $$angle$$ ACB + $$angle$$ A = $$180^circ$$
=> $$80^circ+60^circ+$$ $$angle$$ A = $$180^circ$$ => $$angle$$ A = $$180^circ-140^circ=40^circ$$ Incentre of a triangle = $$90^circ+frac{angle A}{2}$$ => $$ heta=90^circ+frac{40^circ}{2}$$ => $$ heta=90^circ+20^circ$$ => $$ heta=110^circ$$ => Ans - (C)
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=> $$80^circ+60^circ+$$ $$angle$$ A = $$180^circ$$ => $$angle$$ A = $$180^circ-140^circ=40^circ$$ Incentre of a triangle = $$90^circ+frac{angle A}{2}$$ => $$ heta=90^circ+frac{40^circ}{2}$$ => $$ heta=90^circ+20^circ$$ => $$ heta=110^circ$$ => Ans - (C)