The diagram below shows a regular pentagon ABCDE, centre O. The line OA lies on the north line, ON. Calculate (a) the size of AOB, (b) the bearing of E from O.

A regular pentagon has 5 equal sides and 5 equal interior angles. When lines are drawn from the center O to each vertex, 5 congruent isosceles triangles are formed.

The sum of the angles around the center O is $360^\circ$.

Since there are 5 equal angles at the center, each angle is $360^\circ$ divided by 5.

The size of $\angle AOB$ is $\boxed{\text{72}^\circ}$.

Bearing is the angle measured clockwise from the North line.

The problem states that the line OA lies on the North line, ON. So, the direction of OA is North ($0^\circ$).

The vertices of the pentagon are labeled A, B, C, D, E in a counter-clockwise direction. Each central angle (like $\angle AOB$) is $72^\circ$.
Starting from the North line (OA), we measure the angle clockwise to OE.

• The bearing of B from O is $72^\circ$.
• The bearing of C from O is $72^\circ + 72^\circ = 144^\circ$.
• The bearing of D from O is $144^\circ + 72^\circ = 216^\circ$.
• The bearing of E from O is $216^\circ + 72^\circ = 288^\circ$. Bearings are typically written with three digits. The bearing of E from O is $\boxed{\text{288}^\circ}$.