Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075^\circ and the bearing of R from P is 310^\circ, find the :
(a) distance between Q and R ;
(b) baering of R from Q.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
< RPQ = 50^\circ + 75^\circ = 125^\circ
By cosine rule,
\(|QR|^{2} = |RP|^{2} + |PQ|^{2} - 2(|PR|)(|PQ|) \cos 125^\circ\)
= \(90^{2} + 50^{2} + 2(90)(50) \cos 55^\circ\)
= \(8100 + 2500 + 9000 \times 0.5736\)
= \(10600 + 5162.4\)
\(|QR|^{2} = 15762.4\)
\(|QR| = \sqrt{15762.4} = 125.548km\)
(b) By Sine rule,
\(\frac{RQ}{\sin 125^\circ} = \frac{PR}{\sin
Source: Myschool Past Questions Archive (Record #54604)