Determine the interior angle ABC.

Here's how to solve the problem:

a) Calculate the length of AC: Step 1: Determine the interior angle $\angle ABC$. The bearing of B from A is $053^\circ$. This means the angle from the North line at A to AB is $53^\circ$. The bearing of C from B is $161^\circ$. This means the angle from the North line at B to BC is $161^\circ$. Since the North lines at A and B are parallel, the alternate interior angle to $53^\circ$ (angle NAB) is the angle between the line BA and the North line at B, which is $53^\circ$. Therefore, $\angle ABC = 161^\circ - 53^\circ = 108^\circ$.

Step 2: Apply the Cosine Rule to triangle ABC. Given $AB = 15\, \text{m}$, $BC = 18\, \text{m}$, and $\angle ABC = 108^\circ$. $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$ $AC^2 = 15^2 + 18^2 - 2(15)(18)\cos(108^\circ)$ $AC^2 = 225 + 324 - 540(-0.309017)$ $AC^2 = 549 + 166.869$ $AC^2 = 715.869$ $AC = \sqrt{715.869}$ $AC = 26.7555\, m$ Correct to three significant figures, the length of AC is $\boxed{26.8\, m}$.

b) Calculate the bearing of C from A: Step 1: Use the Sine Rule to find $\angle BAC$. $\frac{BC}{\sin(\angle BAC)} = \frac{AC}{\sin(\angle ABC)}$ $\frac{18}{\sin(\angle BAC)} = \frac{26.7555}{\sin(108^\circ)}$ $\sin(\angle BAC) = \frac{18 \times \sin(108^\circ)}{26.7555}$ $\sin(\angle BAC) = \frac{18 \times 0.951056}{26.7555}$ $\sin(\angle BAC) = 0.63908$ $\angle BAC = \arcsin(0.63908)$ $\angle BAC = 39.739^\circ$

Step 2: Calculate the bearing of C from A. The bearing of B from A is $053^\circ$. The bearing of C from A is the bearing of B from A plus $\angle BAC$. $Bearing of C from A = 53^\circ + 39.739^\circ = 92.739^\circ$ Correct to three significant figures, the bearing of C from A is $\boxed{092.7^\circ}$.

c) Calculate the height of the pole BT: Step 1: Identify the right-angled triangle ABT. The pole BT is vertical, so $\angle ABT = 90^\circ$. Given $AB = 15\, \text{m}$ and $\angle ATB = 58^\circ$.

Step 2: Use the tangent trigonometric ratio. $\tan(\angle ATB) = \frac{Opposite}{Adjacent} = \frac{AB}{BT}$ $\tan(58^\circ) = \frac{15}{BT}$ $BT = \frac{15}{\tan(58^\circ)}$ $BT = \frac{15}{1.60033}$ $BT = 9.3730\, m$ Correct to three significant figures, the height of the pole BT is $\boxed{9.37\, m}$.

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