由 $\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB} = -\sqrt{3} \overset{\large\rightharpoonup}{OC}$，兩邊平方得 $$|\overset{\large\rightharpoonup}{OA}|^2 + |\overset{\large\rightharpoonup}{OB}|^2 + 2 \overset{\large\rightharpoonup}{OA} \cdot \overset{\large\rightharpoonup}{OB} = 3 |\overset{\large\rightharpoonup}{OC}|^2$$ 因為 $O$ 為單位圓圓心，$|\overset{\large\rightharpoonup}{OA}| = |\overset{\large\rightharpoonup}{OB}| = |\overset{\large\rightharpoonup}{OC}| = 1$，故 $$1 + 1 + 2 \cos \angle AOB = 3 \implies \cos \angle AOB = \dfrac{1}{2} \implies \angle AOB = 60^\circ$$ 同理，$\overset{\large\rightharpoonup}{OA} + \sqrt{3} \overset{\large\rightharpoonup}{OC} = -\overset{\large\rightharpoonup}{OB} \implies 1 + 3 + 2\sqrt{3} \cos \angle AOC = 1 \implies \cos \angle AOC = -\dfrac{\sqrt{3}}{2} \implies \angle AOC = 150^\circ$。 則 $\angle BOC = 360^\circ - 60^\circ - 150^\circ = 150^\circ$。 $\angle BAC$ 為圓周角，其度數為對應圓心角 $\angle BOC$ 之一半，故 $\angle BAC = \dfrac{150^\circ}{2} = 75^\circ$。故選 $(4)$。