A moving boat is observed from the top of a $ 150 \mathrm{~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 $ \mathrm{m} / \mathrm{h} $.
Given:
A moving boat is observed from the top of a \( 150 \mathrm{~m} \) high cliff moving away from the cliff. The angle of depression of the boat changes from \( 60^{\circ} \) to \( 45^{\circ} \) in 2 minutes.
To do:
We have to find the speed of the boat in \( \mathrm{m} / \mathrm{h} \).
Solution:
Let $AB$ be the high cliff and $C, D$ be the positions of the boat initially and 2 min later respectively.
From the figure,
$\mathrm{AB}=150 \mathrm{~m}, \angle \mathrm{BCA}=60^{\circ}, \angle \mathrm{BDA}=45^{\circ}$
Let the initial distance between the boat and the cliff be $AC=x\ m$ and the distance travelled by it in 2 min is $CD=y\ m$.
$\tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}}$
$=\frac{150}{x}$
$\Rightarrow \sqrt{3}=\frac{150}{x}$
$x=\frac{150}{\sqrt{3}} \mathrm{~m}$............(i)
Similarly,
$\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{AD}}$
$=\frac{150}{x+y}$
$\Rightarrow 1=\frac{150}{x+y}$
$\Rightarrow x+y=150$
$\Rightarrow y=150-x$
$\Rightarrow y=150-\frac{150}{\sqrt{3}}$ [From (i)]
$=\frac{150(\sqrt{3}-1)}{\sqrt{3}} \mathrm{~m}$
Time taken to move from point $\mathrm{C}$ to point $\mathrm{D}$ is $2 \mathrm{~min}=\frac{2}{60} \mathrm{~h}=\frac{1}{30} \mathrm{~h}$
We know that,
Speed $=\frac{\text { Distance }}{\text { Time }}$
$=\frac{y}{\frac{1}{30}}$
$=\frac{\frac{150(\sqrt{3}-1)}{\sqrt{3}}}{\frac{1}{30}}$
$=1500 \sqrt{3}(\sqrt{3}-1) \mathrm{m} / \mathrm{h}$
$=1500 \times 1.732(1.732-1) \mathrm{m} / \mathrm{h}$
$=2598 \times 0.732=1902 \mathrm{~m} / \mathrm{h}$
Therefore, the speed of the boat in \( \mathrm{m} / \mathrm{h} \) is $1902 \mathrm{~m} / \mathrm{h}$.
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