A man in a boat rowing away from a light house $100\ m$ high takes $2$ minutes to change the angle of elevation of the top of the light house from $60^{o}$ to $30^{o}$. Find the speed of the boat in meter per minute. $( Use\ \sqrt{3}=1.732)$

Given: Height of the light house$=100\ m$, Time taken to change the angle of elevation$=2\ minutes$,  Change in angle of elevation is from $60^{o}$ to $30^{o}$.

To do: To find the speed of the boat.

Solution:

AB is a lighthouse of height $100\ m$. 

Let the speed of boat be $x\ meter/minute$.

And $CD$ is the distance which man travelled to change the angle of elevation.

$\because Distance=speed\times time$

Therefore, $CD=x\times2=2x$

In $\vartriangle ABC$,

$tan60^{o}=\frac{AB}{BC}$

$\Rightarrow \sqrt{3}=\frac{100}{BC}$ 

$\Rightarrow BC=\frac{100}{\sqrt{100}}$

In $\vartriangle ABD$,

$tan30^{o}=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BD}$

$\Rightarrow BD=100\sqrt{3}$

We know, $CD=BD-BC$

$\Rightarrow 2x=100\sqrt{3}-\frac{100}{\sqrt{3}}$

$\Rightarrow 2x=\frac{300-100}{\sqrt{3}}$

$\Rightarrow 2x=\frac{200}{\sqrt{3}}$

$\Rightarrow x=\frac{1}{2}.\frac{200}{\sqrt{3}}$

$\Rightarrow x=\frac{100}{\sqrt{3}}$

Using $\sqrt{3}=1.73$

$x=\frac{100}{3.173}=57.80\ meter/minute$ 

Hence, The speed of the boat is 57.80 meters per minute.
 

Updated on: 10-Oct-2022

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