Arun
Last Activity: 6 Years ago
Dear Abhishek
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.