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Grade: 12th pass
        
An aeroplane has to go from a point a to another point b 500 km away due 30 degree East of north wind is blowing due north at the speed of 20 metre per second the steering speed of plane is 150 metre per second
 
Find the direction in which the pilot should head the pain to reach the point p
 
Find the time taken by the plane to go from a to b
 
Kindly use sine formula.
4 months ago

Answers : (2)

Arun
22993 Points
							
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]
 
4 months ago
Khimraj
3008 Points
							

The vector diagram for various velocities is shown.

Vector AC gives velocity of aero plane relative to quiet air.

Vector CB gives velocity of wind(air) relative to ground.

Vector AB gives velocity of plane relative to ground.

Using law of sines for triangle,

(150/sin 30 degree) = (20/sin theta).

Therefore, sin( theta)=(20/300)=(1/15). Then,

Theta=sin^-1(1/15)=3.8 degree.

Angle NAC=33.8 degree.

The pilot should head the plane in the direction 33.8 degree east of north.

3 months ago
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