Answer
$b\approx2.95$
$A \approx28.7^\circ$
$C \approx106.3^\circ$
Work Step by Step
1. Use the Law of Cosines, we have $b^2=(2)^2+(4)^2-2(2)(4)cos45^\circ\approx8.6863$, thus $b\approx2.95$
2. Use the Law of Sines, we have $\frac{sinA}{2}=\frac{sinC}{4}=\frac{sin45^\circ}{2.95}$
3. We have $A=sin^{-1}(\frac{2sin45^\circ}{2.95})\approx28.7^\circ$
4. We have $C=sin^{-1}(\frac{4sin45^\circ}{2.95})\approx106.3^\circ$
