A community library's outreach van sets out from depot \(D\). Branch \(P\) is \(14\) km from \(D\) on a bearing of \(062^{\circ}\), and branch \(Q\) is \(9\...
Question 1 Report
A community library's outreach van sets out from depot \(D\). Branch \(P\) is \(14\) km from \(D\) on a bearing of \(062^{\circ}\), and branch \(Q\) is \(9\) km from \(D\) on a bearing of \(148^{\circ}\), shown below.
- Find the size of angle \(PDQ\). (1)
- Use the cosine rule to find the distance \(PQ\), correct to 3 significant figures. (3)
- Use the sine rule to find angle \(DPQ\), correct to 1 decimal place. (3)
- Hence find the bearing of \(Q\) from \(P\), correct to 1 decimal place. (1)
Answer Details
The angle at \(D\) between the two bearings gives the included angle for the cosine rule, which finds \(PQ\); the sine rule then finds a second angle from the completed triangle, and comparing bearings finally gives the bearing of \(Q\) from \(P\).
- Angle \(PDQ=148^\circ-62^\circ=86^\circ\), the difference between the two bearings measured from \(D\). [1 mark]
- \(PQ^2=14^2+9^2-2(14)(9)\cos86^\circ=196+81-17.58=259.4\), so \(PQ=16.1\) km (3 s.f.). [3 marks]
- \(\dfrac{\sin(\angle DPQ)}{9}=\dfrac{\sin86^\circ}{16.1}\), so \(\sin(\angle DPQ)=\dfrac{9\sin86^\circ}{16.1}=0.558\), giving angle \(DPQ=33.9^\circ\) (1 d.p.). [3 marks]
- The bearing of \(D\) from \(P\) is the back-bearing of \(062^\circ\): \(062^\circ+180^\circ=242^\circ\). Since \(Q\) lies on the side of line \(PD\) nearer south, the bearing of \(Q\) from \(P\) is \(242^\circ-33.9^\circ=208.1^\circ\) (1 d.p.). [1 mark]
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