question_answer

A moving boat is observed from the top of a \[150\text{ }m\] high cliff moving away from the cliff. The angle of depression of the boat changes from \[60{}^\circ \] to \[45{}^\circ \] in 2 minutes. Find the speed of the boat in m/h.

Answer:

From \[\Delta \,ABC,\frac{AB}{BC}=\tan \,\,60{}^\circ \]

or                                 \[BC=\frac{AB}{\tan \,\,60{}^\circ }\]

\[BC=\frac{150}{\sqrt{3}}m\]

From \[\Delta \,ABD,\frac{AB}{BD}=\tan \,\,45{}^\circ \] or \[AB=BD[\because \,tan\,\,45{}^\circ =1]\]

\[\Rightarrow \]                           \[BD=150\,m\]

Distance covered in 2 min \[=BD-BC\]

\[=150-\frac{150}{\sqrt{3}}=\frac{150\sqrt{3}-150}{\sqrt{3}}\]

Distance covered in 1 hour \[=\frac{150\left( \sqrt{3}-1 \right)}{\sqrt{3}\times 2}\times 60\,\,m\]

\[speed=\frac{4500(\sqrt{3}-1)}{\sqrt{3}}\]

\[=4500-1500\sqrt{3}\]

\[=4500-2598=1902\text{ }m/hr.\]

Hence, the speed of boat is \[1902\text{ }m/hr\].