Let the line m cut AB and CD at point P and Q respectively ∠DOQ=x (exterior angle) Hence, Y+2x (corresponding angle) ∴ y=x ...(i) Also ∠DOO=x (vertically opposite angles) In ΔOCD, sum of the angles =180° ∴ y+2y+2x+x=180° 3x+3y=180° x+y=60 ...(ii) From (i) and (ii) x=y=30=2y=60 ∴ ∠ODS=180−60=120° ∴ θ=180−3x=180−3(30)=180−90=90°. ∴ The required ratio =90:120=3:4.