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 the flag on the mast of the boat?

(D) Velocity of the boat,$v_{b} = 51 \; km/h$ (towards North).
Velocity of the wind,$v_{w} = 72 \; km/h$ (towards North-East).
The flag flutters in the direction of the relative velocity of the wind with respect to the boat,given by $\vec{v}_{wb} = \vec{v}_{w} - \vec{v}_{b} = \vec{v}_{w} + (-\vec{v}_{b})$.
Here,$\vec{v}_{w}$ makes an angle of $45^{\circ}$ with the North (or East) direction. The vector $(-\vec{v}_{b})$ is directed towards the South.
The angle between $\vec{v}_{w}$ and $(-\vec{v}_{b})$ is $90^{\circ} + 45^{\circ} = 135^{\circ}$.
Using the formula for the direction $\beta$ of the resultant vector with respect to the East direction:
$\tan \beta = \frac{v_{w} \sin(135^{\circ}) - v_{b} \sin(90^{\circ})}{v_{w} \cos(135^{\circ}) + v_{b} \cos(90^{\circ})}$ is not the standard approach. Let's use the vector components:
$\vec{v}_{w} = 72 \cos(45^{\circ}) \hat{i} + 72 \sin(45^{\circ}) \hat{j} = 50.91 \hat{i} + 50.91 \hat{j}$
$\vec{v}_{b} = 51 \hat{j}$
$\vec{v}_{wb} = \vec{v}_{w} - \vec{v}_{b} = 50.91 \hat{i} + (50.91 - 51) \hat{j} = 50.91 \hat{i} - 0.09 \hat{j}$
The angle $\theta$ with the East direction is $\tan \theta = \frac{|v_{wb, y}|}{|v_{wb, x}|} = \frac{0.09}{50.91} \approx 0.00177$.
$\theta \approx \tan^{-1}(0.00177) \approx 0.1^{\circ}$ South of East.
Thus,the flag flutters almost due East.