A taxi depot \(D\) dispatches a driver who covers \(3.8\) km on a bearing of \(065^\circ\), reaching a fuel stop \(P\). The driver then covers a further \(5...
Question 1 Report
A taxi depot \(D\) dispatches a driver who covers \(3.8\) km on a bearing of \(065^\circ\), reaching a fuel stop \(P\). The driver then covers a further \(5.2\) km on a bearing of \(155^\circ\), reaching a customer at \(E\), as shown.
- Work out angle \(DPE\), showing your reasoning. (2)
- Work out the direct distance \(DE\), correct to \(3\) significant figures. (3)
- The only road connecting \(D\) and \(E\) directly is \(7.1\) km long. State, with a reason, whether the driver's two-leg route is closer to the direct distance than the road. (2)
Answer Details
This question finds an angle at a shared point from two bearings, recognises the resulting right angle, and then uses that to justify choosing Pythagoras' theorem rather than the cosine rule.
The bearing of \(D\) from \(P\) is the reverse of the bearing of \(P\) from \(D\):
\[065^\circ + 180^\circ = 245^\circ\]Angle \(DPE\) is the difference between this and the bearing of \(E\) from \(P\):
\[245^\circ - 155^\circ = 90^\circ\][2 marks]
Since angle \(DPE = 90^\circ\), triangle \(DPE\) is right-angled at \(P\), so Pythagoras' theorem applies directly:
\[DE = \sqrt{3.8^2 + 5.2^2} = \sqrt{14.44+27.04} = \sqrt{41.48}\] \[= 6.44 \text{ km (3 s.f.)}\][3 marks]
The direct distance \(DE = 6.44\) km is shorter than the \(7.1\) km road, so the straight-line distance between \(D\) and \(E\) is less than the distance travelled by road; the two-leg route (\(3.8+5.2=9\) km) is actually further than both [2 marks].
Spotting that the angle at \(P\) is exactly \(90^\circ\) allows the simpler Pythagoras' theorem to be used instead of the more general cosine rule, which would give the same answer but requires more calculation.
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