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° and the bearing of R from P is 310°, find the :
(a) distance between Q and R ;
(b) baering of R from Q.
Explanation
< RPQ = 50° + 75° = 125°
By cosine rule,
(|QR|^{2} = |RP|^{2} + |PQ|^{2} – 2(|PR|)(|PQ|) cos 125°)
= (90^{2} + 50^{2} + 2(90)(50) cos 55°)
= (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°} = frac{PR}{sin <PQR})
(frac{125.548}{sin 125°} = frac{90}{sin < PQR})
(sin < PQR = frac{90 times 0.8192}{125.548})
(sin < PQR = 0.5782 implies < PQR = sin^{-1} (0.5872))
= (35.96°)
Bearing of R from Q = 180° + 75° + 35.96° = 290.96°)
(approxeq 291°)