From a point R, 300m north of P, a man walks eastwards to a place; Q which is 600m from P. Find the bearing of P from Q correct to the nearest degree
026\(^o\)
045\(^o\)
210\(^o\)
240\(^o\)
Correct answer is D
Cos θ = \(\frac{adj}{hyp}\)
= \(\frac{300}{600}\)
= 0.5
θ = Cos - 10.5
= 60
∠ RPQ = ∠ PQs
So the bearing of P from Q is 180 + 60 = 240\(^o\)
Answer is D
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