In $\Delta PQR$,
$\begin{array}{rcl}
\tan \angle PRQ & = & \dfrac{PQ}{RQ} \\
RQ & = & \dfrac{PQ}{\tan 47^\circ}
\end{array}$
and
$\begin{array}{rcl}
\sin \angle PRQ & = & \dfrac{PQ}{PR} \\
PR & = & \dfrac{PQ}{\sin 47^\circ}
\end{array}$
In $\Delta PQS$,
$\begin{array}{rcl}
\tan \angle PSQ & = & \dfrac{PQ}{QS} \\
QS & = & \dfrac{PQ}{\tan 53^\circ}
\end{array}$
and
$\begin{array}{rcl}
\sin \angle PSQ & = & \dfrac{PQ}{PS} \\
PS & = & \dfrac{PQ}{\sin 53^\circ}
\end{array}$
By applying the cosine law to $\Delta RQS$, we have
$\begin{array}{rcl}
RS^2 & = & RQ^2 + QS^2 – 2 \times RQ \times QS \times \cos \angle RQS \\
RS^2 & = & \left(\dfrac{PQ}{\tan 47^\circ} \right)^2 + \left( \dfrac{PQ}{\tan 53^\circ} \right)^2 – 2 \times \dfrac{PQ}{\tan 47^\circ} \times \dfrac{PQ}{\tan 53^\circ} \cos 120^\circ \\
RS^2 & = & PQ^2 (2.140\ 128\ 612)
\end{array}$
By applying the cosine law to $\Delta PRS$, we have
$\begin{array}{rcl}
\cos \angle RPS & = & \dfrac{PR^2 + PS^2 – RS^2}{2 \times PR \times PS} \\
\cos \angle RPS & = & \dfrac{\left( \frac{PQ}{\sin 47^\circ}\right)^2 + \left( \frac{PQ}{\sin 53^\circ}\right)^2 – PQ^2 (2.140\ 128\ 612)}{2 \times \frac{PQ}{\sin 47^\circ} \times \frac{PQ}{\sin 53^\circ}} \\
\angle RPS & = & 67.736\ 504\ 64^\circ \\
\angle RPS & \approx & 68^\circ
\end{array}$
