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
$$
a \overrightarrow{G A}+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarrow{G C}=0 \text {, }
$$

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

Notice that,
$$
\begin{array}{l}
\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=\mathbf{0} \\
\overrightarrow{G A}=-\overrightarrow{G B}-\overrightarrow{G C} .
\end{array}
$$

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,
$$
\left\{\begin{array} { l } 
{ a - b = 0 , } \\
{ a - \frac { \sqrt { 3 } } { 3 } c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=b, \\
c=\sqrt{3} a .
\end{array}\right.\right.
$$

By $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{\sqrt{3}}{2} \Rightarrow \angle A=30^{\circ}$.