1. Given that $P$ is a point inside the equilateral $\triangle A B C$, $P A=1$, $P B=2$, $\angle A P B=150^{\circ}$. Then $P C=$ $\qquad$
$1 . \sqrt{5}$.
As shown in Figure 5, $\triangle P A B$ is rotated $60^{\circ}$ clockwise around point $B$ to $\triangle Q C B$, and $P Q$ is connected. Then $\triangle P Q B$ is an equilateral triangle,
$$
\begin{array}{c}
P Q=P B=2, \\
C Q=P A=1, \\
\angle P Q C=\angle B Q C-\angle P Q B \\
=\angle A P B-\angle P Q B=90^{\circ} .
\end{array}
$$

By the Pythagorean theorem, $P C=\sqrt{5}$.