A) $ 40(3-\sqrt{3})m $
B) $ 40(3\sqrt{3}-1)m $
C) $ 120(\sqrt{3}-1)m $
D) $ 12(3-\sqrt{3})m $
Correct Answer: A
$ \Rightarrow $ $ BD=120m $
In $ \Delta ACB, $ $ \tan 60{}^\circ =\frac{AB}{BC} $
$ \Rightarrow $ $ \sqrt{3}=\frac{120}{BC} $
$ \Rightarrow $ $ BC=\frac{120}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=40\sqrt{3}m $ Thus, distance between both the boats, $ CD=BD-BC $ $ =120-40\sqrt{3}=40(3-\sqrt{3})m $