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
waec 2005mathematics

The correct answer is D. 240o

\(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}\\

\theta = cos^{-1}(0.5000)=60^{\circ}\)

The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)

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