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^\circ. 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^\circ and 050^\circ 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.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
XWZ = 90^\circ - (45^\circ + 23^\circ) = 22^\circ
WYX = 180^\circ - (45^\circ + 95^\circ) = 40^\circ
< XZW = 180^\circ - (89^\circ + 22^\circ) = 69^\circ
(a)(i) \(\frac{XY}{\sin 95} = \frac{180}{\sin 40^\circ}\)
\(XY = \frac{180 \sin 95}{\sin 40} = 278.96m\)
\(\approxeq 279m\)
(ii) \(\frac{ZW}{\sin 89^\circ} = \frac{180}{\sin 69^\circ}\)
\(ZW = \frac{180 \sin 89^\circ}{\sin 69^\circ} = 192.76m\)
\(\approxeq 193m\)
(b) < ZNX =
Source: Myschool Past Questions Archive (Record #54452)