Question 84, Let $G$ be the centroid of $\triangle ABC$, and $\sqrt{7 GA} \sin A + 3 \overrightarrow{GB} \sin B + 3 \sqrt{7 GC} \sin C = \mathbf{0}$, find $\angle ABC$.


Question 84, Solution: Since $\mathrm{G}$ is the centroid of $\triangle \mathrm{ABC}$, we have $\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=\mathbf{0}$. Therefore, we know:
$$
\begin{array}{l}
\sqrt{7} \sin A: 3 \sin B: 3 \sqrt{7} \sin C=1: 1: 1 \\
\Rightarrow \sin A=\frac{3 \sqrt{7}}{7} \sin B, \quad \sin C=\frac{\sqrt{7}}{7} \sin B
\end{array}
$$

According to the cosine rule, we have:
$$
\begin{array}{l}
\sin ^{2} B=\sin ^{2} A+\sin ^{2} C-2 \sin A \cdot \sin C \cdot \cos B \\
\Rightarrow \sin ^{2} B=\left(\frac{3 \sqrt{7}}{7} \sin B\right)^{2}+\left(\frac{\sqrt{7}}{7} \sin B\right)^{2}-2\left(\frac{3 \sqrt{7}}{7} \sin B\right)\left(\frac{\sqrt{7}}{7} \sin B\right) \cdot \cos B \\
\Rightarrow \cos B=\frac{1}{2}
\end{array}
$$

Therefore, $\angle \mathrm{ABC}=60^{\circ}$.