I In the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°. (a) Calculate, correct to 3 significant figure...

Answer Details

From the diagram: \(PQ=8\text{ m}\), \(QR=13\text{ m}\), the bearing of Q from P is \(050^\circ\), and the bearing of R from Q is \(130^\circ\).

(a)(i) Length PR. Find \(\angle PQR\). The bearing of P from Q is the back-bearing of \(050^\circ\):

\[050^\circ+180^\circ=230^\circ\]

The bearing of R from Q is \(130^\circ\), so

\[\angle PQR=230^\circ-130^\circ=100^\circ\]

Apply the cosine rule to triangle PQR:

\[PR^2=PQ^2+QR^2-2\,(PQ)(QR)\cos(\angle PQR)\]\[PR^2=8^2+13^2-2(8)(13)\cos100^\circ\]\[PR^2=64+169-208(-0.17365)=269.12\]\[PR=16.4\text{ m (3 s.f.)}\]

(a)(ii) Bearing of R from P. Use the sine rule to find \(\angle QPR\):

\[\frac{\sin\angle QPR}{QR}=\frac{\sin\angle PQR}{PR}\]\[\sin\angle QPR=\frac{13\sin100^\circ}{16.405}=\frac{12.803}{16.405}=0.78040\]\[\angle QPR=51.3^\circ\]

R lies clockwise of Q as seen from P, so the bearing of R from P is

\[050^\circ+51.3^\circ=101^\circ\ (\text{3 s.f.})\]

(b) Shortest distance from Q to PR, and the area. The area of triangle PQR is

\[\text{Area}=\tfrac12(PQ)(QR)\sin(\angle PQR)=\tfrac12(8)(13)\sin100^\circ\]\[\text{Area}=52(0.98481)=51.2\text{ m}^2\ (\text{3 s.f.})\]

The shortest distance from Q to PR is the perpendicular height h onto base PR. Using \(\text{Area}=\tfrac12(PR)(h)\):

\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]