Question

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

Answer 1

As a first step, you could make a neat diagram representing the velocity vectors given in the question. Then you could use your basic knowledge in geometry to find certain quantities that would get you the answer. We used the standard formula for tangent of the angle and thus found the answer.

Complete step by step answer:
Since the flag flutters the boat in the NE direction when anchored in the harbor, we could conclude that the velocity of wind is in the NE direction. When the boat starts to sail due north from the position where it has been anchored, the flag would flutter along the direction of relative velocity of wind with respect to the boat.
Let us consider the relative velocity of the wind with respect to the boat to be $\overrightarrow{{{v}_{wb}}}$ and let $\beta$ be the angle between $\overrightarrow{{{v}_{wb}}}$ and$\overrightarrow{{{v}_{w}}}$

The angle between ${{v}_{w}}$ and ${{v}_{b}}$ would be $90+45=135{}^\circ$
We know that,
$\tan \theta =\dfrac{51\sin \left( 90+45 \right)}{72+51\cos \left( 90+45 \right)}$
$\Rightarrow \tan \theta =\dfrac{51}{50.80}=1.0038$
$\therefore \theta ={{\tan }^{-1}}\left( 1.0038 \right)=45.11{}^\circ$
The angle with respect to the east direction would be,
$45.11-45=0.11{}^\circ$
Therefore, we could conclude that the fluttering of the flag would be almost east.

Note: At the final step we have found the angle with respect to the east direction. This is done so as to make a conclusion regarding the direction of fluttering of the flag. So, in the solution by finding the direction of the fluttering with respect to the east direction we made a standard while defining the direction.