Problem Analysis:

• We are given an acute angled triangle $\triangle PQR$.
• O is the orthocentre and C is the circumcentre.
• We know $\angle QCR = 138^\circ$ and $\angle PQR = 52^\circ$.
• PM is a perpendicular from P to QR, meaning $\angle PMR = 90^\circ$.
• We need to find the measure of $\angle RPM$.

Finding Angle QPR using Circumcentre Property

The angle subtended by an arc (like QR) at the circumcentre (C) is double the angle subtended by the same arc at any point on the circumference (like P).

Using the property $\angle QCR = 2 \times \angle QPR$:

$ \angle QPR = \frac{\angle QCR}{2} $

Substituting the given value:

$ \angle QPR = \frac{138^\circ}{2} = 69^\circ $

Calculating Angle PRQ

The sum of angles in any triangle is $180^\circ$. For $\triangle PQR$:

$ \angle QPR + \angle PQR + \angle PRQ = 180^\circ $

Substitute the known angles:

$ 69^\circ + 52^\circ + \angle PRQ = 180^\circ $

$ 121^\circ + \angle PRQ = 180^\circ $

Solving for $\angle PRQ$:

$ \angle PRQ = 180^\circ - 121^\circ = 59^\circ $

Determining Angle RPM

PM is perpendicular to QR, forming a right-angled triangle $\triangle PMR$ at M.

In a right-angled triangle, the sum of the two non-right angles is $90^\circ$. Here, $\angle PRM$ is the same as $\angle PRQ$.

$ \angle RPM + \angle PRM = 90^\circ $

$ \angle RPM + \angle PRQ = 90^\circ $

Substitute the value of $\angle PRQ$:

$ \angle RPM + 59^\circ = 90^\circ $

Solving for $\angle RPM$:

$ \angle RPM = 90^\circ - 59^\circ = 31^\circ $

Therefore, the measure of $\angle RPM$ is $31^\circ$.