Answer
$\angle A=63.0^\circ$, $\angle B=15.5^\circ$, $\angle C=101.5^\circ$
Work Step by Step
Given $a=40, b=12, c=44$,
use the Law of Cosines for angle C, $c^2=a^2+b^2-2ab\cdot cos(C)$, $44^2=40^2+12^2-2\times40\times12cos(C)$ and $cos(C)=-0.2$ which gives $\angle C=101.5^\circ$
Similarly, for angle $A$, $a^2=c^2+b^2-2bc\cdot cos(A)$, $40^2=44^2+12^2-2\times44\times12cos(A)$ and $cos(A)=0.4545$ which gives $\angle A=63.0^\circ$
Angle B can be obtained as $\angle B=180^\circ-\angle A -\angle C=15.5^\circ$
