Answer
$C \approx28.8^\circ$, one triangle.
$A\approx 111.2^\circ$
$a \approx5.80$
Work Step by Step
1. Use the Law of Sines, we have $\frac{3}{sinC}=\frac{4}{sin40^\circ}=\frac{a}{sinA}$
2. $C=sin^{-1}(\frac{3sin40^\circ}{4})\approx28.8^\circ$ only one triangle.
3. Find the third angle $A\approx180-40-28.8=111.2^\circ$
4. $a=\frac{4sin111.2^\circ}{sin40^\circ}\approx5.80$
