In the following diagram, angle POQ = 55°, OQ = QP and the bearing of Q from O is 060°. Find the bearing of (a) O from Q. (b) P from Q.

The bearing of Q from O is $060^\circ$.

To find the bearing of O from Q, we use the rule for back bearings. If the bearing is less than $180^\circ$, add $180^\circ$. If it's greater than $180^\circ$, subtract $180^\circ$.
Since $060^\circ < 180^\circ$, we add $180^\circ$.

The bearing of O from Q is:

Since OQ = QP, triangle OQP is an isosceles triangle with base OP. This means the angles opposite the equal sides are equal: $\angle QPO = \angle QOP$.
We are given $\angle POQ = 55^\circ$, so $\angle QOP = 55^\circ$.
Therefore, $\angle QPO = 55^\circ$.

The sum of angles in a triangle is $180^\circ$.

We know the bearing of O from Q is $240^\circ$. This is the angle measured clockwise from the North line at Q to the line segment QO.
From the diagram, the line segment QP is clockwise from the line segment QO.
Therefore, to find the bearing of P from Q, we add $\angle OQP$ to the bearing of O from Q.

The bearing of P from Q is: