Without loss of generality we can let the length of $\mathrm{QR}$ be 1 unit, and take a coordinate system with the origin at $O$ and axes along $\mathrm{OR}$ and $\mathrm{OQ}$. If $\angle \mathrm{PQR}=\alpha$, where $0<\alpha \hspace{1em}<{90}^{\circ }$, then $\mathrm{PQ}=\cos \hspace{1em}\alpha$ and $\mathrm{PR}=\sin \hspace{1em}\alpha .$ Let $\angle \mathrm{QRO}=\theta \hspace{1em}$ where $0\leqq \hspace{1em}\theta \leqq \hspace{1em}{90}^{\circ }$. Then, from the right angled triangles $\mathrm{PSQ}$ and $\mathrm{PTR}$, we have $\angle \mathrm{PRT}=\angle \mathrm{QPS}=\alpha -\theta ,$ and hence we can write down the coordinates of the point $P$.

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{x}& =& \cos \hspace{1em}\alpha \hspace{1em}\cos (\alpha \hspace{1em}-\theta )\\ \multicolumn{1}{c}{y}& =& \sin \hspace{1em}\alpha \hspace{1em}\cos (\alpha \hspace{1em}-\theta ).\end{array}}$

We see that

${\displaystyle \frac{y}{x}=\frac{\sin \hspace{1em}\alpha \hspace{1em}\cos (\alpha \hspace{1em}-\theta )}{\cos \hspace{1em}\alpha \hspace{1em}\cos (\alpha \hspace{1em}-\theta )}=\tan \hspace{1em}\alpha .}$

and so $P$ lies on the straight line $y=x\tan \hspace{1em}\alpha$.

The position $(x,y)$ depends only on $\cos (\alpha \hspace{1em}-\theta )$, $\alpha$ being a constant, and $\theta$ a variable. The distance of the point $P$ from $O$ is given by

${\displaystyle {\mathrm{OP}}^2=x^2+y^2=\cos {}^2(\alpha \hspace{1em}-\theta )(\cos {}^2\alpha \hspace{1em}+\sin {}^2\alpha )=\cos {}^2(\alpha \hspace{1em}-\theta ).}$

Hence $\mathrm{OP}=\cos (\alpha \hspace{1em}-\theta )$ which is a maximum when $\cos (\alpha \hspace{1em}-\theta )=1$, that is when $\alpha \hspace{1em}=\theta$. This occurs when $\mathrm{OQPR}$ is a rectangle as shown in the diagram.

We get an even simpler method of solution by using the fact that the angles $\mathrm{QOR}$ and $\mathrm{QPR}$ are both 90 degrees so that $\mathrm{OQPR}$ is a cyclic quadrilateral with $\mathrm{PR}$ as a chord. We have $\angle \mathrm{POR}=\angle \mathrm{PQR}=\alpha$ because these two angles are subtended by the same chord of the circle. This shows that

$\angle \mathrm{POR}$ is constant and hence that the locus of $P$ is the straight line $y=x\tan \hspace{1em}\alpha .$

What can you say about the locus of $P$ if the triangle $\mathrm{PQR}$ is not a right angled triangle?