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.

Answer Details

tan θ = oppadj=RQQP=68 $\frac{\mathrm{<br>}}{}$

tan θ = 0.75

θ = tan−1(0.75)=36.87o ${}^{-1}$

${\mathrm{<br>}}^{}$
∴ The bearing of R from P = 360o ${}^{\mathrm{<br>}}$ - 36.87o ${}^{\mathrm{<br>}}$ = 323o ${}^{\mathrm{<br>}}$ (to the nearest degree)