1. Given vectors $\overrightarrow{A P}=(1, \sqrt{3}), \overrightarrow{P B}=(-\sqrt{3}, 1)$, then the angle between vector $\overrightarrow{A P}$ and $\overrightarrow{A B}$ is $\qquad$
$\overrightarrow{A B}=\overrightarrow{A P}+\overrightarrow{P B}=(1-\sqrt{3}, 1+\sqrt{3})$, then $\overrightarrow{A P} \cdot \overrightarrow{A B}=4,|A P|=2$, $|A B|=\sqrt{(1-\sqrt{3})^{2}+(1+\sqrt{3})^{2}}=2 \sqrt{2}$, thus $\cos \langle\overrightarrow{A P}, \overrightarrow{A B}\rangle=\frac{\sqrt{2}}{2} \Rightarrow\langle\overrightarrow{A P}, \overrightarrow{A B}\rangle=\frac{\pi}{4}$. Therefore, the angle between vector $\overrightarrow{A P}$ and $\overrightarrow{A B}$ is $\frac{\pi}{4}$.