Help and much appreciated

By the SAS congruence theorem,  △PQT ≅ △SRT  So...

m∠QPT  =  m∠RST  =  37°

m∠QTP  =  180° - 50°   =   130°

m∠PQT  =  180° - 37° - 130°  =  13°

Since  QT = RT  we can substitute  QT  in for  RT in the next equation.

$RT+TP\ =\ 11\\~\\ QT+TP\ =\ 11\\~\\ QT\ =\ 11 - TP$

Now we can substitute  11 - TP  in for  QT in the next equation.

By the Law of Sines,

$\frac{TP}{\sin13^\circ} \ =\ \frac{QT}{\sin37^\circ}\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ QT\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ 11-TP\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}+TP\ =\ 11\\~\\ TP\cdot(\frac{\sin37^\circ}{\sin13^\circ}+1)\ =\ 11\\~\\ TP\ =\ 11\div(\frac{\sin37^\circ}{\sin13^\circ}+1)\\~\\ TP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}$

Now that we know the length of  TP ,  we can find the length of  QT.

$QT\ =\ 11-TP\\~\\ QT\ =\ 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}$

Finally, we can use the Law of Sines again to find  QP.

$\frac{QP}{\sin130^\circ}\ =\ \frac{TP}{\sin13^\circ}\\~\\ QP\ =\ TP\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}$

So we have found:

$\begin{array}{c} TP&=& \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 2.993\\~\\ QT& =& 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 8.007\\~\\ QP& =&\frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 10.192 \end{array}$

And all lengths are in meters.