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Question

An airplane has to go from a point A to another point B due 45 east of north. Wind is blowing due north at speed of 2002 km/h. The steering-speed of the plane is 400 km/h. Find the direction in which the pilot should head the plane so as to reach the point B.


A
75east of north.
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B
30east of north.
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C
60east of north.
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D
15east of north.
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Solution

The correct option is A 75east of north.
Given that,

Velocity of wind, Vw=2002 km/h,

The steering-speed of plane,Vaw=400 km/h

and velocity of plane Va should be along AB or in north-east direction.

The direction of Vaw should be such as the resultant of Vw and Vawis along AB or in north -east direction.

Let, Vaw makes an angle α with AB as shown in figure.


Applying sine law in triangle ABC, we get

ACsin45=BCsinα

sinα=(BCAC)sin45

sinα=(VwVaw)sin45

sinα=(2002400)×12

sinα=12

α=30

From the figure, the direction in which the pilot should head the plane is

=α+45

=30+45

=75 east of the north.

Hence, option (a) is the correct answer.

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