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.

A.

217\(^o\)

B.

323\(^o\)

C.

037\(^o\)

D.

053\(^o\)

Correct answer is B

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)