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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 2002 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=2002 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=(2002400)12=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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