Example 3 In $\triangle A B C$, $D$ is a point on side $B C$, $B D=$ $\frac{1}{2} D C, \angle A D B=120^{\circ}, A D=2$, if the area of $\triangle A D C$ is $3-\sqrt{3}$, then $\angle B A C=$ $\qquad$
Solve: Since $\angle A D B=120^{\circ}$, and $B D=\frac{1}{2} D C$, it is easy to know that $S_{\triangle A B D}=\frac{1}{2} S_{\triangle A D C}=\frac{1}{2}(3-$

Figure 3 $\sqrt{3})$. Also, $S_{\triangle A B D}=\frac{1}{2} B D \cdot A D \cdot \sin 120^{\circ}$, so we have $B D \cdot \frac{\sqrt{3}}{2}=\frac{1}{2}(3-\sqrt{3})$, solving for $B D=\sqrt{3}-1$, hence $D C=2(\sqrt{3}-1)$.

Also, $\angle A D C=180^{\circ}-120^{\circ}=60^{\circ}$, by the cosine rule we get $A C^{2}=D C^{2}+A D^{2}-2 D C \cdot A D \cdot \cos 60^{\circ}$ $=[2(\sqrt{3}-1)]^{2}+4-2 \times 2(\sqrt{3}-1) \times 2 \times \frac{1}{2}=$ $6(\sqrt{3}-1)^{2}=D C \cdot B C$, therefore $\triangle A D C \backsim \triangle B A C$, so $\angle B A C=\angle A D C=60^{\circ}$.