ABC triangle have three points are$A ( 3 , - 1 ) , B ( 2,3 )$ and C $= ( 5,1 ) ,$ then $\angle A =$?

$\cos A=\frac{\overrightarrow{AB}.\ \overrightarrow{AC}}{\mid\overrightarrow{AB}\mid.\mid\overrightarrow{AC}\mid}$

$\begin{array}{l}\overrightarrow{A B}=B-A=(2,3)-(3,-1)=(-1,4) \\ |\overrightarrow{A B}|=\sqrt{(-1)^{2}+4^{2}}=\sqrt{17}\end{array}$

$\begin{array}{l}\overrightarrow{A C}=C-A=(5,1)-(3,-1)=(2,2) \\ |\overrightarrow{A C}|=\sqrt{2^{2}+2^{2}}=\sqrt{8}=2 \sqrt{2}\end{array}$

$\begin{array}{l}\cos A=\frac{-2+8}{\sqrt{17} \times 2 \sqrt{2}}=\frac{6}{2 \sqrt{34}} \\ \therefore \angle A=\cos ^{-1}\left(\frac{3}{\sqrt{34}}\right)\end{array}$