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Question

An aircraft flies at 400 km/h in still air. A wind of 2002 km/h is blowing from the south. The pilot wishes to travel from A to a point B north-east of A. The direction he must steer is


A
75 from North towards East.
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B
75 from East towards North.
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C
45 from North towards East.
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D
45 from East towards North.
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Solution

The correct option is A 75 from North towards East.

Given, velocity of wind vw=2002 km/h
and velocity of aircraft w.r.t wind vaw=400 km/h
va should be along AB or in north-east direction.
Thus, the direction of vaw should be such that the resultant of vw and vaw is along AB or in north-east direction.

Let vaw makes an angle α with AB as shown in the figure.
Applying sine law in triangle ABC, we get
ACsin45=BCsinα
or, sinα=(BCAC)sin45=(2002400)12=12
α=30
Therefore, the pilot should steer in a direction at an angle of (45+α)=75 from north towards east.

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