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.
(a) distance between Q and R ;
(b) baering of R from Q.
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Correct Answer: Option n
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°\)
\(\approxeq 291°\)
< 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°\)
\(\approxeq 291°\)