Almost A Central Angle

$\color{#20A900}{\boxed{\color{#3D99F6}{\boxed{\angle AOC=2\theta \\ \angle ABC=\theta}}}}$

$\angle BAO+\angle ABC+\angle BCO=\angle AOC~~(\color{#D61F06}{Theorem~~of~~the~~Boomerang})$ $30^\circ+\theta+40^\circ=2\theta$ $\theta=70^\circ$ $\large\color{#20A900}{\boxed{\color{#3D99F6}{\boxed{\angle AOC=2\theta=140^\circ}}}}$

Quad ABCO has to have 360 degrees total. Angle opposite angle O = 360-O. Angle B = O/2 because inscribed angle sub tends the same arc as central angle O. So, we have 40+30+ (360-O) + (O/2) = 360.

Solve for O and you get O/2 = 70, O=140. QED.

By the Thales' Theorem , we have

$\angle AOC = 2(30+40)=$ $\color{#D61F06}\boxed{140}$