Given $\triangle{ABC}$ with $\angle{B}=60^{\circ}$ and $\angle{C}=30^{\circ}$, let $P,Q,R$ be points on the sides $BA,AC,CB$ respectively such that $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP=QR$.\\
Find the maximum possible value of $\frac{2[ABC]}{[BPQR]}$ where $[S]$ denotes the area of any polygon $S$.