5. Let $H$ be the orthocenter of $\triangle A B C$ and satisfy: $3 \overrightarrow{H A}+4 \overrightarrow{H B}+5 \overrightarrow{H C}=\overrightarrow{0}$, then the value of $\cos \angle A H B$ is
Answer: $-\frac{\sqrt{6}}{6}$
Using the conclusion $\frac{S_{\triangle H B C}}{\overrightarrow{H A}}+\frac{S_{\triangle H C A}}{\overrightarrow{H B}}+\frac{S_{\triangle H A B}}{\overrightarrow{H C}}=\overrightarrow{0}$
After simple calculation, it is easy to get that the area ratio of the corresponding triangles is the tangent ratio,
Let $\frac{\tan A}{3}+\frac{\tan B}{4}+\frac{\tan C}{5}=\lambda$
Then use $\tan A+\tan B+\tan C=\tan A \tan B \tan C$
To deduce that $\lambda=\sqrt{5}$, thus $\tan C=\sqrt{5}$, so $\cos \angle A H B=-\cos \angle C=-\frac{\sqrt{6}}{6}$