In the diagram, three points A, B and C are on the same horizontal ground. B is 15m from A, on a bearing of 053°, C is 18m from B on a bearing of 161°. A ve...

Answer Details

From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).

(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):

\[053^\circ+180^\circ=233^\circ\]

The bearing of C from B is \(161^\circ\), so

\[\angle ABC=233^\circ-161^\circ=72^\circ\]

Apply the cosine rule to triangle ABC:

\[AC^2=AB^2+BC^2-2\,(AB)(BC)\cos(\angle ABC)\]\[AC^2=15^2+18^2-2(15)(18)\cos72^\circ\]\[AC^2=225+324-540(0.30902)=382.13\]\[AC=19.5\text{ m (3 s.f.)}\]

(b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):

\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\]\[\sin\angle BAC=\frac{18\sin72^\circ}{19.548}=\frac{17.119}{19.548}=0.87575\]\[\angle BAC=61.1^\circ\]

C lies clockwise of B as seen from A, so the bearing of C from A is

\[053^\circ+61.1^\circ=114^\circ\ (\text{3 s.f.})\]

(c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:

\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]