Answer
$\angle A=63.435^{\circ}$
$\angle B=53.130^{\circ}$
$\angle C=63.435^{\circ}$
Work Step by Step
Angle at $A:$
Let
${\bf u}=\vec{AB}=\langle 2+1,1-0\rangle=\langle 3,1\rangle$,
${\bf v}=\vec{AC}=\langle 1+1,-2-0\rangle=\langle 2,-2\rangle$,
$\displaystyle \angle A=\cos^{-1}(\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|})=\cos^{-1}\frac{(3)(2)+(1)(-2)}{\sqrt{3^{2}+1^{2}}\cdot\sqrt{2^{2}+(-2)^{2}}}$
$=\displaystyle \cos^{-1}\frac{4}{\sqrt{10}\cdot\sqrt{8}}=\cos^{-1}\frac{1}{\sqrt{5}}$
$=63.435^{\circ}$
Angle at $B:$
Let
${\bf u}=\vec{BC}=\langle 1-2,-2-1\rangle=\langle-1,-3\rangle$,
${\bf v}=\vec{BA}=-\vec{AB}=\langle-3,-1\rangle$
$\displaystyle \angle B=\cos^{-1}(\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|})=\cos^{-1}(\frac{(-1)(-3)+(-3)(-1)}{\sqrt{10}\cdot\sqrt{10}})$
$=\displaystyle \cos^{-1}\frac{6}{10}=\cos^{-1}(0.6)$
$=53.130^{\circ}$
The sum of interior angles is $180^{\circ}$
$\angle C=180^{\circ}-(\angle A+\angle B) $
$=180^{\circ}-116.565^{\circ}$= $63.435^{\circ}$
