A plane flies in a straight line from London to Rome, a distance of \(\quantity{1400}{km}\) on a bearing of \(135^\circ\). Given that the plane’s speed in still air is \(\quantity{380}{km/h}\), that the wind direction is \(225^\circ\) and that the journey takes \(4\) hours, determine,

- the wind speed,

The diagram shows the direction of the wind as well as the distance from London to Rome. We can use this distance along with the time it takes to fly to calculate the resultant speed. \[\frac{1400}{4}=\quantity{350}{km/h}\]

We can now construct a triangle to show the resultant speed as the vector sum of the wind speed and the plane’s speed. It is a right-angled triangle so we can use Pythagoras’ theorem to calculate the wind speed, \(w\).

\[\begin{align*} 380^2&=350^2+w^2 \\ \implies\quad w&=\quantity{148}{km/h} \end{align*}\]- the direction the pilot should set for the flight.

This diagram shows us the triangle we just used to calculate \(w\) with an arrow pointing towards North. We can calculate the angle \(\alpha\) using trigonometry and then get the direction of the flight by adding the \(45^\circ\) angle that we already know. \[\alpha=sin^{-1}\frac{350}{380}\approx67.08^\circ\] So the bearing to set for the flight is \[\alpha+ 45^\circ\approx112^\circ.\]

Find also the direction the pilot should set for the return flight, assuming that the speed and direction of the wind remain unchanged.

The triangle in this diagram is congruent to the previous one. We can calculate the angle \(\beta\) as \[90^\circ-\alpha \approx 22.9^\circ.\] To get the bearing for the return flight we add on the bearing of the resultant journey to London. \[22.9^\circ+315^\circ=337.9^\circ\] So the pilot should set the controls to a bearing of \(338^\circ\).