An aircraft flies at 400km/h in still air. A wind of 200√2km/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 A75∘from North towards East.
Given, velocity of wind vw=200√2km/h and velocity of aircraft w.r.t wind vaw=400km/h →va should be along AB or in north-east direction. Thus, the direction of −−→vaw should be such that the resultant of −→vwand−−→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.