Answer to Question #93300 in Trigonometry for Precious

Question

Two ships P and Q left a port R at the same time on different routes. Q sailed on a bearing of 150° while P sailed on the north side of Q.After a distance of 8km and 10 km by p anq respectively their distance apart was 12km. Find the bearing of P from R.

Answer 1

Find the angle $x$ between RP and RQ using the law of cosines:

$$PQ^2=RP^2+RQ^2-2RP\cdot RQ\cdot cos x$$ $$cos x=\frac{RP^2+RQ^2-PQ^2}{2RP\cdot RQ}$$ $$cos x=\frac{8^2+10^2-12^2}{2\cdot 8\cdot 10}=\frac{64+100-144}{160}=\frac{20}{160}=\frac{1}{8}$$

$$x\approx 83\degree$$

Then the  bearing of P from R is:

$$150\degree-83\degree=67\degree$$