Question 1 Report
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, |YZ|
(b) The nearest degree, the bearing of Y from Z
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).
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