已知向量 $\overrightarrow{B A}=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \overrightarrow{B C}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, 则 $\angle A B C=$
解析:根据向量的夹角公式求解.

$\because \overrightarrow{B A}=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \overrightarrow{B C}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$,

$\therefore|\overrightarrow{B A}|=1,|\overrightarrow{B C}|=1, \overrightarrow{B A} \cdot \overrightarrow{B C}=\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}}{2} ，$

$\therefore \cos \angle A B C=\cos \langle\overrightarrow{B A}, \overrightarrow{B C}\rangle=\frac{\overrightarrow{B A} \cdot \overrightarrow{B C}}{|\overrightarrow{B A}| \cdot|\overrightarrow{B C}|}=\frac{\sqrt{3}}{2}$.

$\because\langle\overrightarrow{B A}, \overrightarrow{B C}\rangle \in\left[0^{\circ}, 180^{\circ}\right], \therefore \angle A B C=\langle\overrightarrow{B A}, \overrightarrow{B C}\rangle=30^{\circ}$.

答案: A