14. In $\triangle A B C$, $G$ is the centroid, and the sides opposite to the three interior angles $A, B, C$ are $a$, $b, c$, respectively, and $56 a \overrightarrow{G A}+40 b \overrightarrow{G B}+35 c \overrightarrow{G C}=\mathbf{0}$. Then $\angle B=$ $\qquad$.
$1460^{\circ}$ Hint: Since $\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=\mathbf{0}$, we have
$$
40 b \overrightarrow{G A}+40 b \overrightarrow{G B}+40 b \overrightarrow{G C}=\mathbf{0} \text {. }
$$

Therefore,
$$
\text { (56a } \overrightarrow{G A}-40 b) \overrightarrow{G A}+(35 c-40 b) \overrightarrow{G C}=\mathbf{0} \text {. }
$$

Since $\overrightarrow{G A}, \overrightarrow{G C}$ are not collinear, we have
$$
7 a-5 b=0,7 c-8 b=0 .
$$

Let $a=5 k$, then $b=7 k, c=8 k$, by the cosine rule we get
$$
\cos B=\frac{25 k^{2}+64 k^{2}-49 k^{2}}{2 \times 5 k \times 8 k}=\frac{1}{2} .
$$

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