Respuesta :
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,
[tex]a^{2}=b^{2}+c^{2}-2 b c \cos A[/tex]
Substitute the given values in the formula,
[tex]\mathrm{R}^{2}=700^{2}+80^{2}-(2 \times 700 \times 80 \times \cos 45^\circ)[/tex]
[tex]\mathrm{R}^{2}=490000+6400-(112000 \times\frac{1}{\sqrt{2} } )[/tex]
[tex]\mathrm{R}^{2}=496400-79195.95[/tex]
[tex]\mathrm{R}^{2}=417204.049[/tex]
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.
[tex]$\frac{\sin C}{c}=\frac{\sin A}{a}[/tex]
[tex]$\frac{\sin \theta}{80}=\frac{\sin 45}{645.91}[/tex]
Do cross multiplication, we get
[tex]${\sin \theta}}=\frac{\sin 45}{645.91}\times 80[/tex]
[tex]${\sin \theta}}=\frac{\frac{1}{\sqrt{2} } }{645.91}\times 80[/tex]
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. Â
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.
