Please Help With Geo Problem!!!

We are given a side-side-angle triangle, so we can use the Law of Sines to solve this.

$\frac{\sin B}{b}\ =\ \frac{\sin A}{a}\\~\\ \frac{\sin B}{4\sqrt{2}}\ =\ \frac{\sin 60°}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sin 60°}{4\sqrt{3}}\cdot4\sqrt2\\~\\ \sin B\ =\ \sin 60°\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt3}{2}\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt2}{2}\\~\\ \begin{array}{lcl} B\ =\ \arcsin(\frac{\sqrt2}{2})&\quad\text{or}\quad& B\ =\ 180°-\arcsin(\frac{\sqrt2}{2})\\~\\ B\ =\ 45°&\text{or}&B\ =\ 135°\\~\\ &&\text{But}\qquad135°+60°=195°>180°\\~\\ &&\text{So }\qquad B\neq135° \end{array}$

B  =  45°     is the only solution.