Answer
$c \approx2.32$
$A \approx16.1^\circ$
$B\approx 123.9^\circ$
Work Step by Step
1. Use the Law of Cosines, $c=\sqrt {(3)^2+(1)^2-2(3)(1)cos40^\circ}\approx2.32$
2. Use the Law of Sines $\frac{1}{sinA}=\frac{3}{sinB}=\frac{2.32}{sin40^\circ}$
3. Thus $A=sin^{-1}(\frac{sin40^\circ}{2.32})\approx16.1^\circ$
4. Find the third angle $B\approx180-40-16.1=123.9^\circ$
