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Question

# An aircraft flies at a speed of 400 km/h in still air. A wind is blowing with a speed of 200√2 km/h from South. The pilot wishes to travel from point A to a point B, where point B is North-East of A. Find the direction he must head if AB=1000 km.

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

## The correct option is C 75∘ from North towards East Given that speed of wind w.r.t ground vw=200√2 km/h and speed of airplane w.r.t wind vaw=400 km/h and Velocity of airplane w.r.t ground →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 make an angle α with AB as shown in vector triangle ABC: Applying sine law in triangle ABC, we get: ACsin45∘=BCsinα ⇒sinα=(BCAC)sin45∘=(200√2400)1√2=12 ∴α=30∘ Therefore, the pilot should head in a direction at an angle of (45∘+α) or 75∘ from north towards east.

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