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.
(a) the length of AC.
(b) the bearing of C from A ;
(c) the height of the pole BT.
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Correct Answer: Option n
Explanation:
< PAB = < ABE = 53° (alternate angles)
< CBE = 180° - < DBC = 180° - 161° = 19°
< ABC = < ABE + < CBE
= 53° + 19° = 72°
(a) In \(\Delta ABC\),
\(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \cos < ABC\)
\(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \cos 72\)
= \(225 + 324 - 540 \cos 72\)
= \(549 - 166.869\)
\(AC^{2} = 382.131\)
\(AC = \sqrt{382.131} = 19.548m\)
\(\approxeq 19.5m\)
(b) \(\frac{\sin A}{18} = \frac{\sin 72}{19.548}\)
\(\sin A = \frac{18 \times \sin 72}{19.548}\)
\(\sin A = 0.8757\)
\(A = \sin^{-1} (0.8757) = 61.13°\)
The bearing of C from A = 61.13° + 53° = 114.13° \(\approxeq\) 114°.
(c) In \(\Delta ATB\),
\(\frac{15}{BT} = \tan 58\)
\(BT = \frac{15}{\tan 58}\)
\(9.373 m \approxeq 9.37m\)
< PAB = < ABE = 53° (alternate angles)
< CBE = 180° - < DBC = 180° - 161° = 19°
< ABC = < ABE + < CBE
= 53° + 19° = 72°
(a) In \(\Delta ABC\),
\(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \cos < ABC\)
\(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \cos 72\)
= \(225 + 324 - 540 \cos 72\)
= \(549 - 166.869\)
\(AC^{2} = 382.131\)
\(AC = \sqrt{382.131} = 19.548m\)
\(\approxeq 19.5m\)
(b) \(\frac{\sin A}{18} = \frac{\sin 72}{19.548}\)
\(\sin A = \frac{18 \times \sin 72}{19.548}\)
\(\sin A = 0.8757\)
\(A = \sin^{-1} (0.8757) = 61.13°\)
The bearing of C from A = 61.13° + 53° = 114.13° \(\approxeq\) 114°.
(c) In \(\Delta ATB\),
\(\frac{15}{BT} = \tan 58\)
\(BT = \frac{15}{\tan 58}\)
\(9.373 m \approxeq 9.37m\)