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.
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)