Question

A man rowing a boat away from a lighthouse which is 100m 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 100m. So, AB = 100m.

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

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

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

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