Answer
$B \approx30.7^\circ$, one triangle.
$C\approx99.3^\circ$
$c\approx3.86$
Work Step by Step
1. Use the Law of Sines, we have $\frac{2}{sinB}=\frac{3}{sin50^\circ}=\frac{c}{sinC}$
2. $B=sin^{-1}(\frac{2sin50^\circ}{3})\approx30.7^\circ$ only one triangle.
3. Find the third angle $C=180-50-30.7=99.3^\circ$
4. $c=\frac{3sin99.3^\circ}{sin50^\circ}\approx3.86$
