Answer
$A\approx145^\circ, B\approx13^\circ, C\approx22^\circ$
Work Step by Step
Step 1. Using the distance formula with the given numbers, we have
$a=BC=\sqrt {(3+3)^2+(4+1)^2}\approx7.8$, $b=AC=\sqrt {(3)^2+(-1)^2}\approx3.2$, $c=AB=\sqrt {(-3)^2+(4)^2}=5$
Step 2. Using the Law of Cosines, we have
$a^2=b^2+c^2-2bc\ cosA$ or $7.8^2=3.2^2+5^2-2(3.2)(5)cosA$
which gives
$cosA\approx-0.8222$ and $A=acos(-0.8222)\approx145^\circ$
Step 3. Using the Law of Sines, we have
$\frac{sinC}{c}=\frac{sinA}{a}$ and $sinC=\frac{5sin(145^\circ)}{7.8}\approx0.3677$
thus
$C=asin(0.3677)\approx22^\circ$ and $B\approx180^\circ-145^\circ-22^\circ=13^\circ$
Step 4. The solutions are
$A\approx145^\circ, B\approx13^\circ, C\approx22^\circ$
