This is an answer to a separate question from McCquabena Bannor, sent to me in a private message.
I have had to do it this way because I cannot insert an image in a reply to a private message.
P,Q and R are three ships at sea. The bearing of Q from P is 030° and the bearing of P from R is 300°.If the distance between P and Q is 5km and that between P and R is 6km,calculate, correct to three significant figures, the distance between
(i) Q and R
(ii)The bearing of Q from R.
1) The distance between Q and R
By Pythagoras’, QR^2 = PQ^2 + PE^2
QR^2 = 5^2 + 6^2 = 25 + 36
QR^2 = 61
QR = 7.8102
QR = 7.81 to 3 s.f.
2) The bearing of Q from R
tan(PRQ) = 5/6
PRQ = arctan(5/6)
PRQ = 39.80557°
But PR is at a bearing of 300°.
Therefore, QR is at an additional bearing of 39.80557°.
i.e. Bearing for QR is 300 + 39.80557 = 339.80557
Bearing = 340°, to 3 s.f.