Question

In a harbour, wind is blowing at the speed of $72km.h$ and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. It 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

The angle between $v_{w}and(-v_b)=90+45$
$\tan \beta =\frac{51\sin 90+45}{72+51\cos 90+45}$
Substituting and solving we get-
$\tan \beta =\frac{51}{50.80}=1.0038$
$\beta ={\tan }^{-1}1.0038={45.11}^{\circ }$
Angle with respect to the east direction $={45.11}^{\circ }-{45}^{\circ }={0.11}^{\circ }$
Hence the flag will flutter almost due east.