2. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and $G$ is the centroid of $\triangle A B C$. If
$$
a \overrightarrow{G A}+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarrow{G C}=\mathbf{0} \text {, }
$$

then $\angle A=$ . $\qquad$
2. $30^{\circ}$.

Notice that,
$$
\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=\mathbf{0} \Rightarrow \overrightarrow{G A}=-\overrightarrow{G B}-\overrightarrow{G C} \text {. }
$$

Then $a(-\overrightarrow{G B}-\overrightarrow{G C})+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarrow{G C}=0$
$$
\Rightarrow(b-a) \overrightarrow{G B}=\left(a-\frac{\sqrt{3}}{3} c\right) \overrightarrow{G C} \text {. }
$$

Since $G, B, C$ are not collinear, we have,
$$
\begin{array}{l}
a-b=0, a-\frac{\sqrt{3}}{3} c=0 \\
\Rightarrow a=b, c=\sqrt{3} a \\
\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{\sqrt{3}}{2} \\
\Rightarrow \angle A=30^{\circ} .
\end{array}
$$