A boat is being rowed away from a cliff 150 m high.  At the top of the cliff the angle of depression of the boat changes from 60° to 45° in 2 min.  Then the speed of the boat  $\left(\text{in\hspace{0.17em}}\frac{m}{n}\right)$ is

Let  $PQ$ be the cliff  $\Rightarrow PQ=150m$

Let the two position of the boat be  $A\text{\hspace{0.17em}and\hspace{0.17em}}B$,

$\begin{array}{cc}From\Delta APQ& From\Delta BPQ\\ \tan 60°=\frac{PQ}{AP}& \tan 45°=\frac{PQ}{BP}\\ \sqrt{3}=\frac{150}{AP}& 1=\frac{150}{BP}\\ AP=\frac{150}{\sqrt{3}}& BP=150\end{array}$

Now, $AB=BP-AP$

$=150-\frac{150}{\sqrt{3}}$

$=\frac{150}{\sqrt{3}}(\sqrt{3}-1)$

The speed of the boat $=\frac{D}{T}=\frac{AB}{2}=\frac{1}{2}\cdot \frac{150}{\sqrt{3}}(\sqrt{3}-1)$

$=\frac{75}{\sqrt{3}}(\sqrt{3}-1)\times 60m/n$

$=\frac{4500}{\sqrt{3}}(\sqrt{3}-1)m/n$