Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the : (a) distance of Z from Y ; (b) bearing of Z from Y.

Answer Details

\(Y\) is on bearing \(135^\circ\), \(Z\) on bearing \(225^\circ\) from \(X\). The angle between them at \(X\) is \[\angle YXZ = 225^\circ - 135^\circ = 90^\circ,\] so triangle \(YXZ\) is right-angled at \(X\).

(a) By Pythagoras: \[|YZ| = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = \mathbf{100\,\text{km}}.\]

(b) Using coordinates: \(Y = (60\sin135^\circ, 60\cos135^\circ) = (42.43, -42.43)\) and \(Z = (80\sin225^\circ, 80\cos225^\circ) = (-56.57, -56.57).\) \[\vec{YZ} = (-98.99,\ -14.14),\] which points South-West. The bearing satisfies \(\sin B = \dfrac{-98.99}{100},\ \cos B = \dfrac{-14.14}{100}\) (third quadrant), so \[B = 180^\circ + \tan^{-1}\!\frac{98.99}{14.14} = 180^\circ + 81.9^\circ \approx \mathbf{262^\circ}.\] Bearing of \(Z\) from \(Y\) is about \(262^\circ.\)