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.

Given : I is the incentre of $$\triangle$$ ABC and $$\angle$$ ABC = $$80^\circ$$ and $$\angle$$ ACB = $$60^\circ$$

To find : $$\angle$$ BIC = $$\theta$$ = ?

Solution : Sum of angles of $$\triangle$$ 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 = $$90^\circ+\frac{\angle A}{2}$$

=> $$\theta=90^\circ+\frac{40^\circ}{2}$$

=> $$\theta=90^\circ+20^\circ$$

=> $$\theta=110^\circ$$

=> Ans - (C)