I have Sine Law, I have Cosine Law, Ah~~, Problem Solved

From sine rule ,

$\begin{aligned} \frac b{\sin B} & = \frac a{\sin A} \\ b\sin A & = a \sin B \\ 2b \sin A & = 2a \sin B = \sqrt 2 a \\ \implies \sin B & = \frac 1{\sqrt 2} \end{aligned}$

For a triangle, $B = 45^\circ$ and $135^\circ$ . By cosine rule :

$\begin{aligned} b^2 & = a^2 + c^2 - 2ac \cos B \\ & = 18 + 4 - 2(3\sqrt 2)(2) \left(\pm \frac 1{\sqrt 2}\right) \\ & = 22 \mp 12 \\ \implies b & = \boxed{\sqrt{10} \text{ and }\sqrt{34}} \end{aligned}$

The equations $\sqrt{2}a = 2b \sin A$ and $a = 3 \sqrt{2}$ solve to $b = \frac{3}{\sin A}$ .

The area of $\triangle ABC$ is $A_{\triangle ABC} = \frac{1}{2}bc \sin A = \frac{1}{2} \cdot \frac{3}{\sin A} \cdot 2 \sin A = 3$ .

By Heron's formula the area of $\triangle ABC$ is also $A_{\triangle ABC} = \frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)} = \frac{1}{4}\sqrt{(18 + b^2 + 4)^2 - 2(324 + b^4 + 16)} = 3$ , which solves to $\boxed{b = \sqrt{10} \text{ and } b = \sqrt{34}}$ .

Since $\sqrt 2 a = 2 b \sin A$ $\\$ $\\$

By Sine Law that $\frac {a} {\sin A} = \frac {b} {\sin B}= \frac {c} {\sin C} = 2R$ (R = double the circumradius) $\\$ $\sqrt 2 \sin A = 2 \sin B \sin A$ $\\$ $\frac {\sqrt 2}{2} = \sin B$ $\\$ also since $\angle B$ is inside a triangle mieaning $0 < \angle B < 2\pi$ $\\$ $\angle B = \pi /4 \quad or \quad 3 \pi / 4$ $\\$ $\\$

By Cosine Law $b = \sqrt {a^2 + c^2 - 2 a c \cos B}$ $\\$ $b_{1} = \sqrt {18 + 4 - 12 \sqrt2 \cdot \frac {1}{\sqrt2}} = \sqrt {10}$ $\\$ $b_{1} = \sqrt {18 + 4 + 12 \sqrt2 \cdot \frac {1}{\sqrt2}} = \sqrt {34}$ $\\$