PLEASE HELP ASAP

In the diagram below,
$\angle PQR = \angle PRQ = \angle STR = \angle TSR =\angle A,\\ RQ = 8,~ \text{ and } SQ = 2.$
Find PQ.

$\text{Let $\angle QPR = 180^\circ - 2A$} \\ \text{Let $\angle RQT = 180^\circ - 2A$} \\ \text{Let $\angle TRS = 180^\circ - 2A$} \\ \text{Let $\angle PTQ = 180^\circ - A$}\\ \text{Let $PQ=x$ } \\ \text{Let $QR = QT = 8$ } \\ \text{Let $TR = SR = y$ }$

$\begin{array}{|rcll|} \hline ST&=&QT-SQ \\ ST&=&8-2\\ ST& =& 6 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{In $\triangle [QPT]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{8} &=& \dfrac{\sin(180^\circ-A)}{x} \\\\ \dfrac{\sin(2A)}{8} &=& \dfrac{\sin(A)}{x} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{8}{x} \qquad (1)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In $\triangle [QTR]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{y}{8} \qquad (2)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In $\triangle [QSR]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{6}{y} \qquad (3)\\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (2)=(3):& \dfrac{y}{8} &=& \dfrac{6}{y} \\\\ & y^2 &=& 48 \\ & y^2 &=& 16*3 \\ & \mathbf{y} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (2)=(1):& \dfrac{y}{8} &=& \dfrac{8}{x} \\\\ & xy &=& 64 \\\\ & x &=& \dfrac{64}{y} \\\\ & x &=& \dfrac{64}{4\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{16}{3}\sqrt{3}} \\ \hline \end{array}$

$PQ = \mathbf{\dfrac{16}{3}\sqrt{3}}$