Answer
two triangles.
$C_1\approx30.9^\circ$
$A_1\approx129.1^\circ$
$a_1\approx9.07$
$C_2\approx149.1^\circ$
$A_2\approx10.9^\circ$
$a_2\approx2.20$
Work Step by Step
1. Use the Law of Sines, we have $\frac{6}{sinC}=\frac{4}{sin20^\circ}=\frac{a}{sinA}$
2. $C_1=sin^{-1}(\frac{6sin20^\circ}{4})\approx30.9^\circ$ or $C_2\approx(180-30.9)^\circ=149.1^\circ$ two triangles.
3. Find the third angle $A_1\approx180-20-30.9=129.1^\circ$ or $A_2\approx180-20-149.1=10.9^\circ$
4. $a_1=\frac{4sin129.1^\circ}{sin20^\circ}\approx9.07$ or $a_2=\frac{4sin10.9^\circ}{sin20^\circ}\approx2.20$
