Question on: WAEC Mathematics - 2005
From a point R, 300m north of P, man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degreeee
A
026o
B
045o
C
210o
D
240o
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Correct Option: D
(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}<br />
\theta = cos^{-1}(0.5000)=60^{\circ})
The bearing of P from (Q = \theta + 180 = 60 + 180 = 240^{\circ})
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