A 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.
Explanation
XWZ = 90° – (45° + 23°) = 22°
WYX = 180° – (45° + 95°) = 40°
< XZW = 180° – (89° + 22°) = 69°
(a)(i) (frac{XY}{sin 95} = frac{180}{sin 40°})
(XY = frac{180 sin 95}{sin 40} = 278.96m)
(approxeq 279m)
(ii) (frac{ZW}{sin 89°} = frac{180}{sin 69°})
(ZW = frac{180 sin 89°}{sin 69°} = 192.76m)
(approxeq 193m)
(b) < ZNX = <ZXN (base angles of an isosceles triangle)
= 44°^.
(therefore) Bearing of Z from N = 270° + 44° = 314°.