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