Question

A man rowing a boat away from a lighthouse which is 100 m high takes 2 mins to change the angle of elevation of the lighthouse from ${60}^{\circ }$ to ${45}^{\circ }$. The speed of the boat is:

• A. $\frac{50}{3}m/min$
• B. $\frac{50}{3}(3-\sqrt{3})m/min$
• C. $\frac{50-\sqrt{3}}{3}m/min$
• D. $\frac{50\sqrt{3}-1}{\sqrt{3}}m/min$

Answer 1

The correct option is B $\frac{50}{3}(3-\sqrt{3})m/min$

Let AB be the lighthouse and C, D be the positions of the man when the angle of elevation changes from ${60}^{\circ }to{45}^{\circ }$.
The man has covered a distance CD in 2 min. As per question, the height of the lighthouse is 100 m. So, AB = 100 m.

$Speed=\frac{Distance}{Time}$

In 2 mins, the man moves from point C to D.
$\therefore Speed=\frac{CD}{2}m/min$
In$\Delta ABC,wehave$
$\tan {60}^{\circ }=\frac{AB}{BC}$
$\Rightarrow \sqrt{3}=\frac{100}{BC}$
$\Rightarrow BC=\frac{100}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow BC=\frac{100\sqrt{3}}{3}$ .... (i)

In $\Delta ABD$, we have
$\tan {45}^{\circ }=\frac{AB}{BD}$
$\Rightarrow 1=\frac{100}{BD}$
$\Rightarrow BD=100$
Also $CD=DB-BC$
$=100-\frac{100\sqrt{3}}{3}\left[using\right(i\left)\right]$
$=100(1-\frac{\sqrt{3}}{3})=100\left(\frac{3-\sqrt{3}}{3}\right)$
$\therefore Speed=\frac{CD}{2}=\frac{100\left(\frac{3-\sqrt{3}}{3}\right)}{2}$
$=50\left(\frac{3-\sqrt{3}}{3}\right)m/min$

Hence, the required speed of boat is $\frac{50}{3}(3-\sqrt{3})m/min$