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

Given:

A man in a boat rowing away from a light house \( 100 \mathrm{~m} \) high takes 2 minutes to change the angle of elevation of the top of the light house from \( 60^{\circ} \) to \( 30^{\circ} \).

To do:

We have to find the speed of the boat.

Solution:

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

Let the speed of the boat be $x\ metres/minute$ and $CD$ be the distance travelled by the man in 2 minutes.

Distance$=$speed$\times$ Time

Therefore,

$CD=x\times2=2x$

In $\vartriangle ABC$,

$tan\ 60^{o}=\frac{AB}{BC}$

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

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

In $\vartriangle ABD$,

$tan\ 30^{o}=\frac{AB}{BD}$

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

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

From the figure,

$CD=BD-BC$

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

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

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

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

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

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

Therefore, the speed of the boat is $57.80$ metres per minute. 

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Updated on: 10-Oct-2022

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