Answer
$A\approx41^\circ, B\approx71^\circ, C\approx68^\circ$
Work Step by Step
Step 1. Using the distance formula with the given numbers, we have
$a=BC=\sqrt {(4-1)^2+(5-3)^2}\approx3.6$, $b=AC=\sqrt {(1)^2+(-5)^2}\approx5.1$, $c=AB=\sqrt {4)^2+(-3)^2}=5$
Step 2. Using the Law of Cosines, we have
$b^2=a^2+c^2-abc\ cosB$ or $5.1^2=3.6^2+5^2-2(3.6)(5)cosB$
which gives
$cosB\approx0.3319$ and $B=acos(0.3319)\approx71^\circ$
Step 3. Using the Law of Sines, we have
$\frac{sinC}{c}=\frac{sinB}{b}$ and $sinC=\frac{5sin(71^\circ)}{5.1}\approx0.9270$
thus
$C=asin(0.9270)\approx68^\circ$ and $A\approx180^\circ-71^\circ-68^\circ=41^\circ$
Step 4. The solutions are
$A\approx41^\circ, B\approx71^\circ, C\approx68^\circ$
