Answer
$C\approx25.7^\circ$, one triangle.
$B\approx34.3^\circ$
$b\approx1.30$
Work Step by Step
1. Use the Law of Sines, we have $\frac{1}{sinC}=\frac{2}{sin120^\circ}=\frac{b}{sinB}$
2. $C=sin^{-1}(\frac{sin120^\circ}{2})\approx25.7^\circ$ only one triangle.
3. Find the third angle $B\approx180-120-25.7=34.3^\circ$
4. $b=\frac{2sin34.3^\circ}{sin120^\circ}\approx1.30$
