Answer
two triangles are possible: $C\approx74.6^\circ$, $A =65.4^\circ$, $a \approx2.83$
or $C\approx105.4^\circ$ $A= 34.6^\circ$, $a \approx1.77$,
Work Step by Step
Step 1. Based on the given conditions, use the Law of Sines, we have $\frac{sinC}{3}=\frac{sin40^\circ}{2}$, thus $C=sin^{-1}(\frac{3sin40^\circ}{2})\approx74.6^\circ$ or $180-74.6=105.4^\circ$
Step 2. $C\approx74.6^\circ$, we have $A=180-40-74.6=65.4^\circ$ and $\frac{a}{sin65.4^\circ}=\frac{2}{sin40^\circ}$, thus $a=\frac{2sin65.4^\circ}{sin40^\circ}\approx2.83$
Step 3. $C\approx105.4^\circ$, we have $A=180-40-105.4=34.6^\circ$ and $\frac{a}{sin34.6^\circ}=\frac{2}{sin40^\circ}$, thus $a=\frac{2sin34.6^\circ}{sin40^\circ}\approx1.77$, two triangles are possible.