Question

A man in a boat rowing away from a lighthouse 180 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° and 30°. Find the speed of the boat. 

Answer 1

Let AB be the lighthouse. 
Initial position of boat is C, which changes to D after 2 minutes. 
In ΔADB

${\displaystyle \frac{\text{AB}}{\text{DB}}={\tan 60}^{\circ }}$

${\displaystyle \frac{180}{\text{x}}=\sqrt{3}}$

${\displaystyle \text{x}=\frac{180}{\sqrt{3}}}$

In ΔABC
${\displaystyle \frac{\text{AB}}{\text{BC}}={\tan 30}^{\circ }}$

${\displaystyle \frac{180}{\text{x + y}}=\frac{1}{\sqrt{3}}}$

${\displaystyle 180\sqrt{3}=\text{x + y}}$

${\displaystyle 180\sqrt{3}=\frac{180}{\sqrt{3}}+\text{y}}$

${\displaystyle \text{y}=180(\sqrt{3}-\frac{1}{\sqrt{3}})=180\left(\frac{2}{\sqrt{3}}\right)=\frac{360}{\sqrt{3}}}$

Time taken by car to travel DC distance ${\displaystyle (\text{i.e}.\frac{360}{\sqrt{3}})}$ = 2 minutes = 120 seconds

Speed of the boat = ${\displaystyle \frac{\text{Distance}}{\text{Time}}=\frac{\frac{360}{\sqrt{3}}}{120}=\frac{3}{\sqrt{3}}=\frac{3}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}=1.732}$

Thus , the speed of the boat is 1.732 m/sec.