waecmaths question:
From a point R 300m north of P, a man walks eastwards to a place Q which is 600, from P. Find the bearing of P from Q correct to the nearest degree.
Option A:
026o
Option B:
045o
Option C:
210o
Option D:
240o
waecmaths solution:
$\begin{align} & \sin \theta =\frac{300}{600} \\ & \sin \theta =\frac{1}{2} \\ & \theta ={{\sin }^{-1}}\frac{1}{2}={{30}^{\circ }} \\ & \text{The bearing of }P\text{ from }Q={{180}^{\circ }}+{{60}^{\circ }}={{240}^{\circ }} \\\end{align}$
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