The shadow of a tower standing on a level ground is found to be $40\ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $60^{\circ}$. Find the height of the tower.

Given: The shadow of a tower standing on a level ground is found to be $40\ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $60^{\circ}$.

To do: To find the height of the tower.

Solution:




When the sun’s altitude is the angle of elevation of the top of the tower from the tip of the shadow. 

Let $AB$ be $h\ m$ and $BC$ be $x\ m$. 

From the question, $DB$ is $40\ m$ longer than $BC$. 

So, $BD=( 40+x)\ m$ And two right triangles $ABC$ and $ABD$ are formed. 

In $\vartriangle ABC$,

$tan\ 60^o=\frac{AB}{BC}$ 

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

$\Rightarrow x=\frac{h}{\sqrt{3}}$ ........ $( i)$ 

In $\vartriangle ABD$,

$\Rightarrow tan\ 30^o=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{( x+40)}$

$\Rightarrow x+40=\sqrt{3}h$ 

$\Rightarrow \frac{h}{\sqrt{3}+40}=\sqrt{3}h$    [using $( i)$] 

$\Rightarrow h+40\sqrt{3}=3h $

$\Rightarrow 2h=40\sqrt{3}$

$\Rightarrow h=\frac{40\sqrt{3}}{2}$

$\Rightarrow h=20\sqrt{3} $

Therefore, the height of the tower is $20\sqrt{3}\ m$.

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Updated on: 10-Oct-2022

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