Answer
two triangles are possible, $C\approx30.9^\circ$, $A\approx129.1^\circ$, $a \approx9.07$
or $C\approx149.1^\circ$ $A\approx10.9^\circ$, $a \approx2.20$
Work Step by Step
Step 1. Based on the given conditions, use the Law of Sines, we have $\frac{sinC}{6}=\frac{sin20^\circ}{4}$, thus $C=sin^{-1}(\frac{6sin20^\circ}{4})\approx30.9^\circ$ or $180-30.9=149.1^\circ$
Step 2. For $C\approx30.9^\circ$, we have $A=180-20-30.9=129.1^\circ$ and $\frac{a}{sin129.1^\circ}=\frac{4}{sin20^\circ}$, thus $a=\frac{4sin129.1^\circ}{sin20^\circ}\approx9.07$
Step 3. For $C\approx149.1^\circ$, we have $A=180-20-149.1=10.9^\circ$ and $\frac{a}{sin10.9^\circ}=\frac{4}{sin20^\circ}$, thus $a=\frac{4sin10.9^\circ}{sin20^\circ}\approx2.20$, two triangles are possible.
