Question

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

Answer 1

Let the distance BC be x m and CD be y m.


$$\begin{gathered} \mathrm{In}∆\mathrm{ABC}, \\ \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{150}{x} \\ \Rightarrow \sqrt{3}=\frac{150}{x} \\ \Rightarrow x=\frac{150}{\sqrt{3}}\mathrm{m}.....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In}∆\mathrm{ABD}, \\ \tan 45°=\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{150}{x+y} \\ \Rightarrow 1=\frac{150}{x+y} \\ \Rightarrow x+y=150 \\ \Rightarrow y=150-x \end{gathered}$$

Using (1), we get

$\Rightarrow y=150-\frac{150}{\sqrt{3}}=\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}\mathrm{m}$
Time taken to move from point C to point D is 2 min = $\frac{2}{60}\mathrm{h}=\frac{1}{30}\mathrm{h}$.
Now,

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{y}{{\displaystyle \frac{1}{30}}}$
$=\frac{\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}}{{\displaystyle \frac{1}{30}}}=1500\sqrt{3}\left(\sqrt{3}-1\right)\mathrm{m}/\mathrm{h}$