Question

An aeroplane has to go from a point A to another point B , 500 km away due 30 degrees east of north. Wind is blowing with a speed of 20m/s . The air speed of the plane is 150m/s . Find the direction in which the pilot should head the plane to reach point B.

Answer 1

Given :
Wind travels at a Speed =u=20m/s

Aeroplane travels at a Speed  =v=150m/s

To reach the point B, The pilot should travel along AB.
Let us assume that θ be the angle at which the plane heads towards east of line AB.

so angle between two velocities is θ+30°


tanθ°=usin(θ+30)/(v+ucos(θ+30))

sinθ°/cosθ°=usin(θ+30)/(v+ucos(θ+30))
By cross multiplying we get

sinθ°(v+ucos(θ+30))=cosθusin(θ+30)

sinθ° [v+ucosθcos30° - usinθsin30)]=ucosθ°sinθ°cos30°+ucosθ°cosθ°sin30

Since Cos(A+B)=cosAcosB-SinASinB]
Sin(A+B)=sinACosB+cosAsin B

Vsinθ°+ucosθ°sinθ°√3/2-usin²θ1/2=ucosθ°sinθ°√3/2+ucos²θ°1/2

on solving the above equation,
we get
vsinθ°-u/2sin²θ=u/2cos²θ

Vsinθ°=(u/2)[cos²θ+sin²θ]

V sinθ°=u/2 [ ∵cos²θ+sin²θ=1]

sinθ=u/2v
=20/2x150
=20/300
=1/15

θ=sin⁻¹[1/15]
=sin⁻¹[0.066]

∴ in order to reach point B, the plane should move with an angle of sin⁻¹[1/15]  east of line AB.