[TIKZ]
\coordinate[label=above: P] (P) at (2,3);
\coordinate[label=above:Q] (Q) at (4,6);
\coordinate[label=right:R] (R) at (12, 0);
\coordinate[label=below: S] (S) at (7.333333,0);
\draw (P) -- (S)-- (R)-- (Q)--(P);
\draw (P) -- (R);
\draw (Q) -- (S);
[/TIKZ]
Let $[\text{figure}]$ denote the area of figure. We have
$[PQRS]=[PQR]+[RSP]=[QRS]+[SPQ]$
Let $PQ=p,\,QR=q,\,RS=r,\,SP=s$. The above relations reduce to
$pq\sin\angle PQR+rs\sin \angle RSP=qr\sin\angle QRS+sp\sin \angle SPQ$
Using $p=r$ and $(\sqrt{3}+1)q=s$ and dividing by $pq$, we get
$\sin \angle PQR+(\sqrt{3}+1)\sin \angle RSP=\sin \angle QRS+(\sqrt{3}+1)\sin \angle SPQ$
Therefore, $\sin \angle PQR-\sin \angle QRS=(\sqrt{3}+1)(\sin \angle SPQ-\sin \angle RSP)$
This can be written in the form
$2\sin \dfrac{\angle PQR+\angle QRS}{2}\cos \dfrac{\angle PQR+\angle QRS}{2}=(\sqrt{3}+1)2\sin \dfrac{\angle SPQ-\angle RSP}{2}\cos \dfrac{\angle SPQ+\angle RSP}{2}$
Using the relations
$\cos \dfrac{\angle PQR+\angle QRS}{2}=-\cos \dfrac{\angle SPQ+\angle RSP}{2}$
and
$\cos \dfrac{\angle SPQ-\angle RSP}{2}=-\sin 15^{\circ}=- \dfrac{\sqrt{3}-1}{2\sqrt{2}}$
we obtain
$\sin \dfrac{\angle PQR-\angle QRS}{2}=(\sqrt{3}-1)\left(-\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)=\dfrac{1}{\sqrt{2}}$
This shows that
$\dfrac{\angle PQR-\angle QRS}{2}=\dfrac{\pi}{4}$ or $\dfrac{3\pi}{4}$
Using the convexity of $PQRS$, we can rule out the latter alternative. We obtain
$\angle PQR-\angle QRS=\dfrac{\pi}{2}$