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 degree
A.
026o
B.
045o
C.
210o
D.
240o
Correct answer is D
\(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}\\
\theta = cos^{-1}(0.5000)=60^{\circ}\)
The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)