6. Given that the circumcenter of $\triangle A B C$ is $O$, and $3 \overrightarrow{O A}+4 \overrightarrow{O B}+5 \overrightarrow{O C}=\mathbf{0}$, then the value of $\cos \angle A B C$ is
Assume $|\overrightarrow{O A}|=|\overrightarrow{O B}|=|\overrightarrow{O C}|=1$, then $3 \overrightarrow{O A}+5 \overrightarrow{O C}=-4 \overrightarrow{O B}$ $\Rightarrow 9+25+30 \cos \angle A O C=16 \Rightarrow \cos \angle A O C=-\frac{3}{5}$,
so $2 \cos ^{2} \angle A B C-1=-\frac{3}{5} \Rightarrow \cos \angle A B C=\frac{\sqrt{5}}{5}$.