Question

A moving boat is observed from the top of a cliff of 150 m height. The angle of depression of the boat changes from 60${}^o$ to 45${}^o$ in 2 minutes. What is the speed of the boat in metres per hour?

• A. $\frac{4500}{\sqrt{3}}$
• B. $\frac{4500(\sqrt{3}-1)}{\sqrt{3}}$
• C. $4500\sqrt{3}$
• D. $\frac{4500(\sqrt{3}+1)}{\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{4500(\sqrt{3}-1)}{\sqrt{3}}$

As the Angle changes from ${60}^o$ to ${45}^o$ The $Boat$ must Traveled a distance of $150-x$ where x can be determined as

$\Rightarrow$ $tan{60}^o$ $=\frac{150}{x}$

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

$\Rightarrow x=50\sqrt{3}$

Now, Speed of $Boat$ is

$\Rightarrow \frac{150-50\sqrt{3}}{2/60}$

$\Rightarrow 30\left(50\right)(3-\sqrt{3})$

$\Rightarrow 1500\times 3\times$ $(\frac{\sqrt{3}-1}{\sqrt{3}})$

$\Rightarrow 4500(\frac{\sqrt{3}-1}{\sqrt{3}})$