Given : O is the circum-centre of triangle ABC. ∠BAC = $$75^circ$$, ∠BCA = $$80^circ$$ To find : ∠OAC = $$ heta$$ = ? Solution : In $$ riangle$$ 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 => $$angle AOC = 2 imes angle ABC$$ => $$angle$$ AOC = $$2 imes25=50^circ$$ Also, in $$ riangle$$ AOC, OA = OC (radii of circle), => $$angle$$ OAC = $$angle$$ OCA = $$ heta$$ => $$angle$$ AOC + $$angle$$ OAC + $$angle$$ OCA = $$180^circ$$ => $$50^circ+ heta+ heta=180^circ$$ => $$2 heta=180-50=130^circ$$ => $$ heta=frac{130}{2}=65^circ$$ => Ans - (B)