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,

  1. the wind speed,
A north arrow with the direction of Rome to the right and the direction of the wind to the left with given angles and distances labelled

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, \(w\).

A right-angled triangle showing the speed of the plane in still air as the hypotenuse, the resultant speed and the unknown wind speed
\[\begin{align*} 380^2&=350^2+w^2 \\ \implies\quad w&=\quantity{148}{km/h} \end{align*}\]
  1. the direction the pilot should set for the flight.
A north arrow is added along with the 45 degree angle we already know, angle alpha is marked opposite the resultant wind speed of 350km/h

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.

A congruent version of the triangle we just drew is drawn and angle beta is marked opposite the side of the wind speed. At the angle a north arrow is drawn and the 315 degree angle we already knew is marked on

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 \(338^\circ\).