In a harbour, wind is blowing at the speed of ` 72 km //h` and the flag on the mast of a boat anchored in the harbour flutters along the N-E` direction`. If the boat starts moving at a speed of ` 51 km// h` to the North, what is the direction of flag on the mast of the bat ?

Explanation

When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the north-east direction. When the boat starts moving, the flug will flutter along the direction of relative velocity of wind w.r.t boat. Let `V_(wb)` be the realtive velocity of wind w.r.t boat and P be the angle between `V_(wb)` and vw

Now, `vecv_(wb) = vecv_(w) + (-vecv_(b))`
Here, `|vecv_(w)|=72km/h`
`|-vecv_(b)|= 51km/h`
Angle between `vecv_(w)` and `-vec_(b)` is `135^(@)`. Then
`tanbeta=(51sin135^(@))/(72+51cos135^(@))= (51sin45^(@))/(72+51(-cos45^(@))`
`=(51xx(1//sqrt(2))/(72-51(1//sqrt(2)) = 1.0039`
`beta = tan^(-)(1.0039) = 45.1^(@)`
Angle w.r.t. east direction = `45.1^(@)- 45^(@)= 0.1^(@)`.
It means the flag will further almost due east.