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Question

# An aircraft flies at 400 km/h in still air. A wind of 200√2 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=200√2 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∘=(200√2400)1√2=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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