Question

HELP PLEASE
An airplane is heading north at an airspeed of 700 km/hr, but there is a wind blowing from the southwest at 80 km/hr.

The plane will end up flying ___?___ degrees off course
The plane's speed relative to the ground will be ___?__ km/hr

Answer 1

The plane will end up flying 5.02°.

The plan's speed relative to the ground will be 645.91 km/hr.

Solution:

Use the cosine formula,

$a^{2}=b^{2}+c^{2}-2 b c \cos A$

Substitute the given values in the formula,

$\mathrm{R}^{2}=700^{2}+80^{2}-(2 \times 700 \times 80 \times \cos 45^\circ)$

$\mathrm{R}^{2}=490000+6400-(112000 \times\frac{1}{\sqrt{2} } )$

$\mathrm{R}^{2}=496400-79195.95$

$\mathrm{R}^{2}=417204.049$

Taking square root on both sides, we get

R = 645.91 km/hr

This is the grouped speed of the aircraft.

To find θ use sine rule.

$\frac{\sin C}{c}=\frac{\sin A}{a}$

$\frac{\sin \theta}{80}=\frac{\sin 45}{645.91}$

Do cross multiplication, we get

${\sin \theta}}=\frac{\sin 45}{645.91}\times 80$

${\sin \theta}}=\frac{\frac{1}{\sqrt{2} } }{645.91}\times 80$

sin θ = 0.0875

θ = 5.02°

This is known as the drift angle and is the correction the pilot should apply to remain on course.  

The heading is the direction the aircraft's nose is pointing which is

The track is the actual direction over the ground which is θ = 5.02°

An alternative method to this would be to separate each vector into vertical and horizontal components and add.

The resultant can be found using Pythagoras.