In ΔABC, ∠B = 70°and ∠C = 60°. The internal bisectors of the two smallest angles of ΔABC meet at O. The angle so formed at O is

Given : In ΔABC, ∠B = 70°and ∠C = 60°

=> $$\angle A = 180^\circ-70^\circ-60^\circ=50^\circ$$

Thus, the two smallest angles are ∠A and ∠C

To find : $$\angle AOC = \theta$$ = ?

Solution : AO and OC are angle bisectors.

=> $$\angle OAC = \frac{\angle A}{2}=\frac{50}{2}=25^\circ$$

Similarly, $$\angle OCA=30^\circ$$

In $$\triangle$$ AOC

=> $$\angle$$ OAC + $$\angle$$ OCA + $$\angle$$ AOC = $$180^\circ$$

=> $$25^\circ+30^\circ+\theta=180^\circ$$

=> $$\theta=180-55=125^\circ$$

=> Ans - (A)