A person observed the angle of elevation of the top of a tower as $ 30^{\circ} $. He walked $ 50 \mathrm{~m} $ towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as $ 60^{\circ} $. Find the height of the tower.

Given:

A person observed the angle of elevation of the top of a tower as \( 30^{\circ} \). He walked \( 50 \mathrm{~m} \) towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as \( 60^{\circ} \). 

To do:

We have to find the height of the tower.

Solution:  

Let $AB$ be the tower and $CD$ be the distance walked by the person starting from $C$.

From the figure,

$\mathrm{CD}=50 \mathrm{~m}, \angle \mathrm{ACB}=30^{\circ}, \angle \mathrm{ADB}=60^{\circ}$

Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$ and the distance between the point $C$ and the foot of the tower be $\mathrm{BC}=x \mathrm{~m}$.

This implies,

$\mathrm{DB}=x-50 \mathrm{~m}$

We know that,

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

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

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

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

$\Rightarrow x=h(\sqrt3) \mathrm{~m}$

$\Rightarrow x=\sqrt3 h \mathrm{~m}$...........(i)

Similarly,

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

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

$\Rightarrow \tan 60^{\circ}=\frac{h}{x-50}$

$\Rightarrow \sqrt3=\frac{h}{x-50}$

$\Rightarrow (x-50)\sqrt3=h \mathrm{~m}$

$\Rightarrow (\sqrt3 h-50)\sqrt3=h \mathrm{~m}$           [From (i)]

$\Rightarrow 3h-50\sqrt3=h \mathrm{~m}$

$\Rightarrow 3h-h=50\sqrt3 \mathrm{~m}$

$\Rightarrow h=\frac{50\times1.73}{2}=43.25 \mathrm{~m}$

Therefore, the height of the tower is $43.25 \mathrm{~m}$.      

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Updated on: 10-Oct-2022

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