Answer
two triangles are possible, $C \approx53.5^\circ$, $A\approx 86.5^\circ$, $a\approx6.21$
or $C\approx 126.5^\circ$ $A\approx 13.5^\circ$, $a \approx1.45$
Work Step by Step
Step 1. Based on the given conditions, use the Law of Sines, we have $\frac{sinC}{5}=\frac{sin40^\circ}{4}$, thus $C=sin^{-1}(\frac{5sin40^\circ}{4})\approx53.5^\circ$ or $C\approx180-53.5=126.5^\circ$
Step 2. For $C\approx53.5^\circ$, we have $A\approx180-40-53.5=86.5^\circ$ and $\frac{a}{sin86.5^\circ}=\frac{4}{sin40^\circ}$, thus $a=\frac{4sin86.5^\circ}{sin40^\circ}\approx6.21$
Step 3. For $C\approx126.5^\circ$, we have $A\approx180-40-126.5=13.5^\circ$ and $\frac{a}{sin13.5^\circ}=\frac{4}{sin40^\circ}$, thus $a=\frac{4sin13.5^\circ}{sin40^\circ}\approx1.45$ , two triangles are possible.
