Question 69

In $$\triangle ABC$$ ,$$\angle B=60^\circ$$ $$\angle C=40^\circ$$ I is the incenter then find $$\angle BIC$$

Solution

In $$ \triangle ABC $$

given $$ \angle ABC = 60^\circ $$

$$ \angle ACB = 40^\circ $$

therefore $$ \angle BAC = 80^\circ $$ (180 - (60 + 40))

I is the incenter

incenter is the point where angle bisectors of a triangles meet

$$ \angle IBC = \frac{60}{2} = 30^\circ $$

$$ \angle ICB = \frac{40}{2} = 20^\circ $$

$$ \angle BIC = 180 - (30 + 20) = 180 - 50 = 130^\circ $$

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