Correct option is: B) 86.55 m
Given that
The height of the ship mast head is BC = 150 m.
Angle of depression is \(\angle\)CAB = \(60^\circ\)
In right \(\triangle\)ABC,
tan A = \(\frac {BC}{AB}\)
= AB = \(\frac {BC}{tan \, A}\) = \(\frac {BC}{tan \,60^\circ}\) (\(\because\) \(\angle\)CAB = \(60^\circ\))
= \(\frac {150}{\sqrt3} = 50\sqrt3 m\) (\(\because\) BC = 150 m)
= 50 \(\times\) 1.732 (\(\because\) \(\sqrt3\) = 1.732)
= 86.6 m
\(\approx\) 86.55 m
Hence, the distance of boat from the ship is 86.55 m.