An aircraft flies at `400 km//h` in still air. A wind of `200sqrt2 km//h` is blowing from the south towards north. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if `AB=1000 km.`

Explanation

Given that `v_(w)=200 sqrt(2)km//h`.
`v_(aw)=400 km//h` and `v_(a)` should be along AB or in north - east direction. Thus, the direction of `v_(aw)` should be such as the resultant of `v_(w)` and `v_(aw)` is along AB or in north - east direction.

Let `v_(aw)` makes an angle `alpha` with AB as shown in Fig. Applying sine law triangle ABC, we get
`(AC)/(sin 45^(@)) = (BC)/(sin alpha)`
or `sin alpha = ((BC)/(AC))sin 45^(@)`
`=((200sqrt(2))/(400))(1)/(sqrt(2))`
`=(1)/(2)`
`therefore alpha = 30^(@)`
Therefore, the pilot should steer in a direction at an angle of `(45^(@)+alpha)` or `75^(@)` from north towards east.
Further, `(|v_(a)|)/(sin(180^(@)-45^(@)-30^(@)))`
`=(400)/(sin 45^(@))`
or `|v_(a)| = (sin 105^(@))/(sin 45^(@))xx(400)km//h`
`=((cos 15^(@))/(sin 45^(@)))(400 km//h`
`=546.47 km//h`
`therefore` The time of journey from A to B is
`t=(AB)/(|v_(a)|)=(1000)/(546.47)h`
`t=1.83 h`