The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270". Calculate, correct: (a) two significant figures, ...

Answer Details

Take \(X\) as origin. \(Y\) is 15 km due south of \(X\); \(Z\) is 20 km on bearing \(270^{\circ}\), i.e. due west of \(X\). So \(\angle YXZ=90^{\circ}\) (south and west are perpendicular).

(a) \(|YZ|\) by Pythagoras:

\[|YZ|=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25\text{ km (2 s.f.)}.\]

(b) Bearing of \(Y\) from \(Z\)

Place \(Z=(-20,0)\), \(Y=(0,-15)\). The vector \(Z\!\to\!Y=(20,-15)\): from \(Z\), \(Y\) lies east and south. The angle east of south:

\[\tan\theta=\frac{20}{15}=1.333\Rightarrow\theta=53.13^{\circ}\ \text{(east of due south)}.\]

Bearing \(=180^{\circ}-53.13^{\circ}=126.87^{\circ}\approx \mathbf{127^{\circ}}\) (nearest degree).