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...
Question 1 Report
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 vertical pole with top T is erected at B such that < ATB = 58°. Calculate, correct to three significant figures,
(a) the length of AC.
(b) the bearing of C from A ;
(c) the height of the pole BT.
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:
\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]Download The App On Google Playstore
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