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

A.

026\(^o\)

B.

045\(^o\)

C.

210\(^o\)

D.

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