Ta có:
$\overrightarrow{AB}=(-5+2;-1-4)=(-3;-5)$
$\overrightarrow{AC}=(8+2;-2-4)=(10;-6)$
$\overrightarrow{BC}=(8+5;-2+1)=(13;-1)$
Suy ra :
$AB=|\overrightarrow{AB}|=\sqrt{(-3)^{2}+(-5)^{2}}=\sqrt{34}$
$AC=|\overrightarrow{AC}|=\sqrt{10^{2}+(-6)^{2}}=2\sqrt{34}$
$BC=|\overrightarrow{BC}|=\sqrt{13^{2}+(-1)^{2}}=\sqrt{170}$
Ta có: $\overrightarrow{AB}$ x $\overrightarrow{AC}$= (-3) x 10 + (-5) x (-6) = 0 suy ra $\overrightarrow{AB}$ vuông góc với $\overrightarrow{AC}$ hay $\widehat{BAC}=90^{\circ}$
Ta có $cos(\overrightarrow{AC},\overrightarrow{BC})=\frac{10\times 13+(-6)\times (-1)}{\sqrt{10^{2}+6^{2}}\times \sqrt{13^{2}+(-1)^{2}}}=\frac{136}{2\sqrt{34}\times \sqrt{170}}=\frac{2}{\sqrt{5}}$
Suy ra $\widehat{ACB}\approx 27^{\circ} \Rightarrow \widehat{ABC}=90^{\circ}-\widehat{ACB}\approx 63^{\circ}$