A man is standing on the banks of a river, at distance of 36 m from a boat crossing the river. The angle of elevation from the position of the man to the boat changes from 30°`to 60° in 8 seconds . The speed (in m/s ) of the boat is -

5.196

in triangle ABC

tan 30° = \(\frac{BC}{AB}\)

\(\frac{1}{\sqrt{3}}\) =  \(\frac{BC}{AB}\)

As , AB = 36 m . so , BC =  \(\frac{36}{\sqrt{3}}\)  = 12 \(\sqrt{3}\)

in triangle ABD

tan 60° =  \(\frac{BD}{AB}\)

\(\sqrt{3}\)  = \(\frac{BD}{AB}\)

AB = 36 m . So, BD = 36\(\sqrt{3}\) m

Distance covered by the boat = DC = DB - BC =   36\(\sqrt{3}\)  - 12 \(\sqrt{3}\) = 24\(\sqrt{3}\) = 41.568

Distance  = Speed × Time

Speed of the boat = \(\frac{41.568}{8}\) = 5.196 m/s