Theory question 110
BearingsA surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point, W, a distance of 180m in the south-east direction, he observes the bearing of Z and Y to be 337° and 050° respectively. (a) Calculate, correct to the nearest metre (i) |XY| (ii) |ZW|. (b) If N is on XY such that XZ = ZN, find the bearing of Z from N.
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(a)(i) In triangle XWY: ∠YXW = 45° (east to south-east), ∠XWY = (360 − 315) + 50 = 95° (X is on bearing 315° from W), so ∠XYW = 180 − 45 − 95 = 40°.
Sine rule: XY/sin 95° = 180/sin 40°, XY = 180 × 0.9962/0.6428 ≈ 279 m.
(ii) In triangle XWZ: ∠ZXW = 135 − 46 = 89°, ∠XWZ = 337 − 315 = 22°, so ∠XZW = 69°.
ZW/sin 89° = 180/sin 69°, ZW = 180 × 0.9998/0.9336 ≈ 193 m.
(b) ∠ZXN = 90 − 46 = 44°. XZ = ZN, so ∠ZNX = 44° too. From N, X is due west (270°), so the bearing of Z from N is 270° + 44° = 314°.
Draw a clear sketch with north lines at X and W before finding any angles.