Answer
$b \approx3.32$
$A \approx62.8^\circ$
$C\approx 17.2^\circ$
Work Step by Step
1. Use the Law of Cosines, $b=\sqrt {(3)^2+(1)^2-2(3)(1)cos100^\circ}\approx3.32$
2. Use the Law of Sines $\frac{3}{sinA}=\frac{3.32}{sin100^\circ}=\frac{1}{sinC}$
3. Thus $A=sin^{-1}(\frac{3sin100^\circ}{3.32})\approx62.8^\circ$
4. Find the third angle $C\approx180-100-62.8=17.2^\circ$
