Given v⃗=⟨−2,23⟩\vec{v} = \langle -2, 2\sqrt{3} \ranglev=⟨−2,23⟩, find the angle θ\thetaθ in [0,2π)[0, 2\pi)[0,2π).
60∘60^\circ60∘
120∘120^\circ120∘
150∘150^\circ150∘
240∘240^\circ240∘