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 t...
Question 1 Report
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.
Answer Details
Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is
\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]
(a) Distance QR (cosine rule).
\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]
\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]
\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]
(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:
\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]
\[\angle PQR = 36.0^\circ.\]
The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is
\[255^\circ + 36^\circ = 291^\circ.\]
Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).
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