Given the bearing of P from Q is 228^ and QPR = 72^, and R is due south of Q, find the bearing of Q from P and the bearing of P from R.

The bearing of P from Q is $228^\circ$.

To find the bearing of Q from P, we use the rule for back bearings: if the bearing of A from B is $\theta$, then the bearing of B from A is $\theta \pm 180^\circ$. Since $228^\circ > 180^\circ$, we subtract $180^\circ$.

The bearing of Q from P is $\boxed{048^\circ}$.

We are given that R is due south of Q. This means the line segment QR lies on the North-South line passing through Q.
The bearing of P from Q is $228^\circ$. This angle is measured clockwise from the North line at Q to the line QP.
The angle from the North line at Q to the South line at Q is $180^\circ$.
Therefore, the angle between the South line (QR) and the line QP, which is $\angle PQR$, is:

We know the sum of angles in a triangle is $180^\circ$.
Given $\angle QPR = 72^\circ$.
From Step 1, $\angle PQR = 48^\circ$.
So, $\angle PRQ = 180^\circ - \angle QPR - \angle PQR$

Since R is due south of Q, the line segment RQ points North from R. This means the North line at R is along RQ.
The bearing of P from R is the angle measured clockwise from the North line at R (which is RQ) to the line segment RP.
This angle is exactly $\angle PRQ$.
Therefore, the bearing of P from R is $60^\circ$. As a three-figure bearing, it is $060^\circ$.
The bearing of P from R is $\boxed{060^\circ}$.