Answer
$C\approx28.8^\circ$, $A=111.2^\circ$, $c\approx5.80$. Only one triangle is possible.
Work Step by Step
Step 1. Based on the given conditions, use the Law of Sines, we have $\frac{sinC}{3}=\frac{sin40^\circ}{4}$, thus $C=sin^{-1}(\frac{3sin40^\circ}{4})\approx28.8^\circ$ or $180-28.8=151.2^\circ$ (discard this one as B+C>180)
Step 2. We have $A=180-40-28.8=111.2^\circ$
Step 3. We have $\frac{a}{sin111.2^\circ}=\frac{4}{sin40^\circ}$, and $c=\frac{4sin111.2^\circ}{sin40^\circ}\approx5.80$. Only one triangle is possible.
