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Bearings of a two-leg boat trip: distance, an angle between the legs, the direct course and its travel time

Bearings of a two-leg boat trip: distance, an angle between the legs, the direct course and its travel time

16 marks

A boat leaves a harbour at \(H\) and sails on a bearing of \(062^{\circ}\) at a constant speed of \(18\text{ km h}^{-1}\). After \(40\) minutes it reaches a buoy at \(P\).

  1. Calculate the distance \(\mathrm{HP}\).[2]

From \(P\), the boat sails \(9.5\text{ km}\) on a bearing of \(157^{\circ}\) to a lighthouse at \(L\), as shown in the diagram.

    1. Show that \(\mathrm{H\hat{P}L}=85^{\circ}\).
    2. Calculate the distance \(\mathrm{HL}\).[5]
  2. Calculate \(\mathrm{P\hat{L}H}\).[3]

A second boat is to travel in a straight line from the harbour to the lighthouse.

  1. Find the bearing of \(L\) from \(H\).[3]

The second boat travels at a constant speed of \(20\text{ km h}^{-1}\).

  1. Find the time that the second boat takes to reach the lighthouse, giving your answer to the nearest minute.[3]

Solution:

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Answer:(a) \(\mathrm{HP}=12\text{ km}\) (b)(i) \(\mathrm{H\hat{P}L}=85^{\circ}\) (b)(ii) \(\mathrm{HL}=14.6\text{ km}\) (c) \(\mathrm{P\hat{L}H}=54.7^{\circ}\) (d) \(102^{\circ}\) (e) \(44\) minutes

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