An aircraft flies at 400km/h in still air. A wind of `300sqrt(2)` km/h is blowing from the south. The pilot wishes to travel from point A to a point B north east of A. find the direction he must steer and time of his journey if AB=1000km.

Explanation

Given that `v_w=200sqrt2km//h`
`v_(aw)=400km//h ` and `vecV_a` should be along AB or in north-east direction. Thus, the direction of `vecv_(aw)`
should be such as the resultant of `vecv_(w)` and `vecv_(aw)` si along AB or in north-east direction
Let `vecv_(aw)` makes an angle `prop` with AB as shown in figure.
Applying sine law in triangle ABC, we get
`(AC)/(sin45^(@))=(BC)/(sinprop)`

or `sin prop=((BC)/(AC))sin45^(@)=((200sqrt2)/(400))1/sqrt2=1/2`
`:. prop=30^(@)`
Therefore, the pilot should steer in a direction at an angle of `(45^(@)+prop)` or `75^(@)` from north towards east.
Further `(|vecV_a|)/(sin180^(@)-45^(@)-30^(@))=(400)/(sin 45^(@))`
or `|vecv_a|(sin105^(@))/(sin45^(@))xx(400)(km)/(h)`
`=(cos15^(@))/(sin45^(@))(400)(km)/h=((0.9659)/(0.707))(400)(km)/h`
=546.47 km/h
`:.` The time of journey from A to B is
`t=(AB)/(|vecv_a|)=(1000)/(546.47)himplies1.83h`