All I know is QRT~PRQ and PRQ~RST, need help

In the diagram below,
we have \(\angle PQR = \angle PRQ = \angle STR = \angle RST\), \(QR = 8\), and \(QS = 2\).
Find PQ.

\(\text{Let PQ = x }\)

\(\begin{array}{|rcll|} \hline \text{In $\triangle PQR$}: \\ \hline \mathbf{\dfrac{\sin(180^\circ-2A)}{8}} &=& \mathbf{\dfrac{\sin(A)}{x}} \\\\ \dfrac{\sin(2A)}{8} &=& \dfrac{\sin(A)}{x} \\\\ \dfrac{2\sin(A)\cos(A)}{8} &=& \dfrac{\sin(A)}{x} \\\\ \dfrac{2\cos(A)}{8} &=& \dfrac{1}{x} \\\\ \dfrac{\cos(A)}{4} &=& \dfrac{1}{x} \\\\ \dfrac{4}{\cos(A)} &=& x \\\\ \mathbf{x} &=& \mathbf{\dfrac{4}{\cos(A)}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In $\triangle SQR$}: \\ \hline \mathbf{\dfrac{\sin(3A-180^\circ)}{2}} &=& \mathbf{\dfrac{\sin(180^\circ-A)}{8}} \\\\ \dfrac{\sin\Big(-(180^\circ-3A)\Big)}{2} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{-\sin(180^\circ-3A)}{2} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{-\sin(3A)}{2} &=& \dfrac{\sin(A)}{8} \\\\ -\sin(3A) &=& \dfrac{\sin(A)}{4} \small{ \begin{array}{|rcll|} \hline && \mathbf{\sin(3A)} \\ &=&\sin(2A)\cos(A)+\cos(2A)\sin(A)\\ &=&2\sin(A)\cos(A)\cos(A)+\Big(\cos^2(A)-\sin^2(A)\Big)\sin(A) \\ &=&2\sin(A)\cos^2(A)+\Big(\cos^2(A)-\sin^2(A)\Big)\sin(A) \\ &=&2\sin(A)\cos^2(A)+\Big(\cos^2(A)-(1-\cos^2(A))\Big)\sin(A) \\ &=&2\sin(A)\cos^2(A)+\Big(2\cos^2(A)-1\Big)\sin(A) \\ &=&\sin(A)\Bigg( 2\cos^2(A)+\Big(2\cos^2(A)-1\Big) \Bigg) \\ &=&\mathbf{\sin(A)\Big( 4\cos^2(A)-1\Big)} \\ \hline \end{array} } \\\\ -\sin(A)\Big( 4\cos^2(A)-1\Big) &=& \dfrac{\sin(A)}{4} \\ - \Big( 4\cos^2(A)-1\Big) &=& \dfrac{1}{4} \\ -4\cos^2(A)+1 &=& \dfrac{1}{4} \quad | \quad * (-4) \\ 16\cos^2(A)-4 &=& -1 \\ 16\cos^2(A) &=& 3 \quad | \quad \sqrt{()} \\ 4\cos(A) &=& \sqrt{3} \\ \mathbf{\cos(A)} &=& \mathbf{\dfrac{\sqrt{3}}{4}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{4}{\cos(A)}} \\\\ x &=& \dfrac{4}{\dfrac{\sqrt{3}}{4}} \\\\ \mathbf{x} &=& \mathbf{\dfrac{16}{ \sqrt{3} } } \\ \hline \end{array}\)

PQ \(= \mathbf{\dfrac{16}{ \sqrt{3} } = 9.23760430703 }\)