 Krishna
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Step 1: Draw a figure from the given information

Given that

Wind speed along the north-east direction   v_w = 72 km/h

Boat velocity along the north v_b = 51 km/h Since the wind blows from north to east, the angle between NE and E, \angle NEOE = 45\degree

Angle between S and E = 90\degree

Step 2: Calculating the direction of the flag

The wind's velocities are along the North-East axis. When the boat begins to move, the flag will flutter in the direction of the relative velocity of the wind with respect to the boat.

Angle between the velocity of the wind and boat ( v_w \text{ and } - v_b ), \theta = (90\degree + 45\degree)

Components of resultant vector(flag direction),

v_w = v_{w} \cos \theta

v_b = v_b + v_{w} \sin \theta

Direction of the flag, \tan \beta = \frac{v_{b} \sin \theta}{v_w + v_{b} \cos \theta}

\tan\theta=\frac{51\sin(90+45)\degree}{72+51(\cos(90+45)\degree)}

\tan\theta=\frac{51\cos45\degree}{72+51(-\sin45\degree)}

\tan\theta=\frac{51*\frac{1}{\sqrt{2}}}{72-51*\frac{1}{\sqrt{2}}}

\tan \theta = \frac{51}{72*\sqrt{2} - 51}

\theta = \tan^{-1} (1.003)

\theta = 1.1 \degree

Hence, direction of the flag \theta = 1.1 \degree almost due east