Question 83, In $\triangle \mathrm{ABC}$, it is known that $\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}=3 \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}$,
(1) Prove: $\tan B=3 \tan A$;
(2) If $\cos C=\frac{\sqrt{5}}{5}$, find the value of $A$.
Question 83, Solution: (1) According to the given conditions:
$$
\begin{array}{l}
|\overrightarrow{\mathrm{AB}}| \cdot|\overrightarrow{\mathrm{AC}}| \cos \mathrm{A}=3|\overrightarrow{\mathrm{BA}}| \cdot|\overrightarrow{\mathrm{BC}}| \cos \mathrm{B} \\
\Rightarrow \sin \mathrm{B} \cos \mathrm{A}=3 \sin \mathrm{A} \cos \mathrm{B} \\
\Rightarrow \tan \mathrm{B}=3 \tan \mathrm{A}
\end{array}
$$
(2) Method 1: Since $\tan B=3 \tan A$, both $A$ and $B$ are acute angles. Let $\tan A=x$, then $\tan B=3 x$. At this point, $\sin A=\frac{x}{\sqrt{1+x^{2}}}, \cos A=\frac{1}{\sqrt{1+x^{2}}}, \sin B=\frac{3 x}{\sqrt{1+9 x^{2}}}, \cos B=\frac{1}{\sqrt{1+9 x^{2}}}$ . According to the given conditions:
$$
\begin{array}{l}
\cos (A+B)=-\cos C=-\frac{\sqrt{5}}{5} \\
\Rightarrow \cos A \cos B-\sin A \sin B=-\frac{\sqrt{5}}{5} \\
\Rightarrow \frac{1}{\sqrt{1+x^{2}} \sqrt{1+9 x^{2}}}-\frac{3 x}{\sqrt{1+x^{2}} \sqrt{1+9 x^{2}}}=-\frac{\sqrt{5}}{5}
\end{array}
$$

Thus $1-3 x\frac{1}{3}$. Simplifying the above equation gives:
$$
(x-1)\left(x-\frac{1}{3}\right)(x+1)\left(x+\frac{1}{3}\right)=0 \Rightarrow x=1
$$

Therefore, $\tan A=1$, which means $A=\frac{\pi}{4}$.
Method 2: Since $\tan B=3 \tan A$, both $A$ and $B$ are acute angles. Let $\tan A=x$, then $\tan B=3 x$. According to the given conditions, $\tan C=\frac{\sin C}{\cos C}=2$. Since
$$
\begin{array}{l}
\tan A+\tan B+\tan C=\tan A \cdot \tan B \cdot \tan C \\
\Rightarrow 4 x+2=6 x^{2} \\
\Rightarrow(3 x+1)(x-1)=0 \\
\Rightarrow x=1
\end{array}
$$

Therefore, $\tan A=1$, which means $A=\frac{\pi}{4}$.