Question

A moving boat is observed from the top of a $150m$ high cliff moving away from the cliff. The angle of depression of the boat changes from ${60}^0$ to ${45}^0$ in $2$ minutes. Find the speed of the boat in $m/h$.

Answer 1

REF.Image

In $\Delta ABC$

$tan{60}^{\circ }=\frac{AB}{BC}=\frac{150}{x}$

$\Rightarrow \sqrt{3}=\frac{150}{x}\Rightarrow x=\frac{150}{\sqrt{3}}m$

In $\Delta ABD$

$tan{45}^{\circ }=\frac{AB}{BD}=\frac{150}{x+y}$

$1=\frac{150}{x+y}\Rightarrow y=150-x$

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

$S=\frac{s}{t}=\frac{\vartheta }{1/30}=\frac{\frac{150(\sqrt{3}-1)}{\sqrt{3}}}{1/30}=1500\sqrt{3}(\sqrt{3}-1)m/h$