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A moving boat is observed from the top of a cliff of $150 \mathrm{~m}$ height. The angle of depression of the boat changes from $60^{\circ}$ to $45^{\circ}$ in 2 minutes. What is the speed of the boat in metres per hour?
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The correct answer is:
$\frac{4500(\sqrt{3}-1)}{\sqrt{3}}$
$\tan 60^{\circ}=\frac{150}{x} \Rightarrow x=\frac{150}{\sqrt{3}}$
Also, $\tan 45^{\circ}=\frac{150}{x+y}$
$\Rightarrow \quad x+y=150$
$\Rightarrow \quad y=150-x=150-\frac{150}{\sqrt{3}}$
$\Rightarrow \quad y=150\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=$ distance travelled
$\quad$ Speed $(\operatorname{in} \mathrm{m} / \mathrm{hr})=\frac{150(\sqrt{3}-1)}{\sqrt{3}} \times \frac{60}{2}=4500 \frac{(\sqrt{3}-1)}{\sqrt{3}}$
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