om a point R, 300m north of P, a man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degree

Question 1 Report

026°

045°

210°

240°

Answer Details

$\sin y = \frac{300 m}{600 m} = \frac{1}{2}$
$y = \sin^{- 1} 0.5$
$y = 30^{\circ}$
$x = 180^{\circ} + (90^{\circ} - 30^{\circ})$
$x = 180^{\circ} + 60^{\circ}$
$x = 240^{\circ}$

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Bearings

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