Two men on either side of the cliff $ 80 \mathrm{~m} $ high observes the angles of elevation of the top of the cliff to be $ 30^{\circ} $ and $ 60^{\circ} $ respectively. Find the distance between the two men.

Given:

Two men on either side of the cliff \( 80 \mathrm{~m} \) high observes the angles of elevation of the top of the cliff to be \( 30^{\circ} \) and \( 60^{\circ} \) respectively.

To do:

We have to find the distance between the two men.

Solution:  

Let $AB$ be the cliff and $C, D$ are two men on either side of the cliff.

From the figure,

$\mathrm{AB}=80 \mathrm{~m}, \angle \mathrm{BCA}=30^{\circ}, \angle \mathrm{BDA}=60^{\circ}$.

Let the distance between the point $A$ and point $C$ be $\mathrm{CA}=x \mathrm{~m}$ and the distance between the point $A$ and point $D$ be $\mathrm{AD}=y \mathrm{~m}$.

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{AD}$

$\Rightarrow \tan 60^{\circ}=\frac{80}{y}$

$\Rightarrow \sqrt3=\frac{80}{y}$

$\Rightarrow y=\frac{80}{\sqrt3} \mathrm{~m}$.........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{AC}$

$\Rightarrow \tan 30^{\circ}=\frac{80}{x}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{80}{x}$           

$\Rightarrow x=80\sqrt3 \mathrm{~m}$..........(ii)

From (i) and (ii), we get,

$\Rightarrow x+y=\frac{80}{\sqrt3}+80\sqrt3 \mathrm{~m}$     

$=\frac{80+80\sqrt3(\sqrt3)}{\sqrt3} \mathrm{~m}$

$=\frac{80+240}{\sqrt3} \mathrm{~m}$

$=\frac{320}{1.732} \mathrm{~m}$

$=184.8 \mathrm{~m}$

Therefore, the distance between the two men is $184.8 \mathrm{~m}$.

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Updated on: 10-Oct-2022

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