Review question

# In a $\quantity{60}{km/h}$ wind, what bearing should this plane take? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6009

## Solution

An aeroplane can fly at $\quantity{300}{km/h}$ in still air. If a steady $\quantity{60}{km/h}$ wind is blowing from the direction $030^\circ$, determine

1. the direction in which the aeroplane must head in order that it actually flies due North,

We can draw a diagram of the velocity vectors for the wind and the aeroplane.

We can now use the sine rule to find the angle $\alpha$. We have

\begin{align*} \frac{\sin \alpha}{60} &= \frac{\sin 150^\circ}{300}\\ \alpha &= \sin^{-1} \frac{\sin 150^\circ}{5} = 5.739...^\circ. \end{align*}

The aeroplane needs to take a bearing of $006^\circ$ (to the nearest degree) to counteract the wind and fly due north.

1. the aeroplane’s resultant velocity.

#### Using the sine rule

The angle between the velocity vectors of wind and aeroplane is $30^\circ - \alpha = 24.261...^\circ$. Hence we find

\begin{align*} \frac{\sin 150^\circ}{300} &= \frac{\sin 24.261^\circ}{x}\\ x&=300 \frac{\sin 24.261^\circ}{\sin 150^\circ}. \end{align*}

This gives a final velocity of the aeroplane of $v = 246.54...\approx \quantity{247}{km/h}$.

Alternatively…

#### Calculating individual components

The aeroplane’s component of velocity northward is $v_1=300 \cos \alpha.$

The wind subtracts from this a velocity component of $v_2 = 60 \cos 30^\circ.$

Hence we obtain $v= v_1-v_2 =300 \cos 5.739 - 60 \cos 30 = 246.53... \approx \quantity{247}{km/h}$, as before.

Would a similar method also work for the first part of the problem?