Circum-centre of ΔABC is O. If ∠BAC = 75° and ∠BCA = 80°, then what is the value (in degrees) of ∠OAC?

Given : O is the circum-centre of triangle ABC. ∠BAC = $$75^\circ$$, ∠BCA = $$80^\circ$$

To find : ∠OAC = $$\theta$$ = ?

Solution : In $$\triangle$$ ABC

=> $$\angle$$ A + $$\angle$$ B + $$\angle$$ C = $$180^\circ$$

=> $$75^\circ+\angle B+80^\circ=180^\circ$$

=> $$\angle B=180-155=25^\circ$$

In a circle, angle subtended by an arc at the centre is double the angle subtended by the same arc on any other point on the circle

=> $$\angle AOC = 2 \times \angle ABC$$

=> $$\angle$$ AOC = $$2\times25=50^\circ$$

Also, in $$\triangle$$ AOC, OA = OC (radii of circle), => $$\angle$$ OAC = $$\angle$$ OCA = $$\theta$$

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

=> $$50^\circ+\theta+\theta=180^\circ$$

=> $$2\theta=180-50=130^\circ$$

=> $$\theta=\frac{130}{2}=65^\circ$$

=> Ans - (B)