A boat sails 8 km north from P to Q and then sails 6 km west from Q to R. Calculate the bearing of R from P. Give your answer to the nearest degree.
- 217\(^o\)
-
323\(^o\)
- 037\(^o\)
- 053\(^o\)
The correct answer is: B
Explanation
tan θ = \(\frac {opp}{adj} = \frac {RQ}{QP} = \frac {6}{8}\)
tan θ = 0.75
θ = tan\(^{-1} (0.75) = 36.87^o\)
∴ The bearing of R from P = 360\(^o\) - 36.87\(^o\) = 323\(^o\) (to the nearest degree)
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