An aeroplane is headed due north with a speed of 240 km/hr. If there is a wind speed of 100 km/hr from west to east, a) what is the direction of the airplane relative to the earth?

The aeroplane's velocity vector relative to the air, $\mathbf{v}_{pA}$, is due north with a magnitude of

240 km/hr.

The wind's velocity vector relative to the earth, $\mathbf{v}_{aE}$, is from west to east (due east)

with a magnitude of 100 km/hr.

The aeroplane's velocity relative to the earth, $\mathbf{v}_{pE}$, is the vector sum of these two

velocities:

$\mathbf{v}_{pE} = \mathbf{v}_{pA} + \mathbf{v}_{aE}$

The two velocity vectors are perpendicular to each other. The resultant velocity vector

$\mathbf{v}_{pE}$ points to the northeast. The angle $\theta$ relative to the north direction can be found

using trigonometry:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{|\mathbf{v}_{aE}|}{|\mathbf{v}_{pA}|} = \frac{100 \text{ km/hr}}{240 \text{ km/hr}}$

$\theta = \arctan\left(\frac{100}{240}\right)$

$\theta \approx 22.62^\circ$

The direction is $22.62^\circ$ east of north (or $67.38^\circ$ north of east).

The direction of the airplane relative to the earth is $22.62^\circ$ east of north (or $67.38^\circ$

north of east).