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...

Place \(X\) at the origin. \(Y\) is due east, and \(W\) is \(180\,\text{m}\) from \(X\) on bearing \(135^\circ\) (South-East), giving \(W = (180\sin135^\circ,\ 180\cos135^\circ) = (127.28,\ -127.28)\).

(a)(i) Finding \(|XY|\): \(Y = (y, 0)\). From \(W\) the bearing of \(Y\) is \(050^\circ\), so \[\tan50^\circ = \frac{y - 127.28}{0 - (-127.28)} = \frac{y - 127.28}{127.28}.\] \[y - 127.28 = 127.28\tan50^\circ = 151.68 \Rightarrow y = 278.96.\] Therefore \(|XY| \approx \mathbf{279\,\text{m}}\).

(a)(ii) Finding \(|ZW|\): \(Z\) lies on bearing \(046^\circ\) from \(X\): \(Z = t(\sin46^\circ,\cos46^\circ)\). From \(W\), \(Z\) is on bearing \(337^\circ\): \(Z = W + s(\sin337^\circ,\cos337^\circ)\). Solving \[0.7193t + 0.3907s = 127.28,\qquad 0.6947t - 0.9205s = -127.28,\] gives \(t = |XZ| = 72.2\) and \(s = |ZW| \approx \mathbf{193\,\text{m}}\), with \(Z = (51.95,\ 50.17)\).

(b) Bearing of \(Z\) from \(N\): \(N\) is on \(XY\) (the east line) with \(|XZ| = |ZN|\). Since \(Z = (51.95, 50.17)\) and \(|XZ| = 72.2\), \[(n - 51.95)^2 + 50.17^2 = 72.2^2 \Rightarrow n = 103.9\ (\text{taking } n \neq 0).\] So \(N = (103.9, 0)\) and \(\vec{NZ} = (-51.95,\ 50.17)\), which points North-West. \[\text{angle west of north} = \tan^{-1}\!\frac{51.95}{50.17} = 46^\circ \Rightarrow \text{bearing} = 360^\circ - 46^\circ = \mathbf{314^\circ}.\]