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

Given:

The shadow of a tower standing on a level ground is found to be \( 40 \mathrm{~m} \) longer when Sun's altitude is \( 30^{\circ} \) than when it was \( 60^{\circ} \). 

To do:

We have to find the height of the tower.

Solution:  

Let $AB$ be the height of the tower, $CB$ be the shadow when Sun's altitude is \( 60^{\circ} \) and $DB$ be the shadow when Sun's altitude is \( 30^{\circ} \).

From the figure,

$\mathrm{CD}=40 \mathrm{~m}, \angle \mathrm{ADB}=30^{\circ}, \angle \mathrm{ACB}=60^{\circ}$

Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$ and the length of the shadow when Sun's altitude is \( 60^{\circ} \) be $\mathrm{CB}=x \mathrm{~m}$.

This implies,

$\mathrm{DB}=40+x \mathrm{~m}$

We know that,

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

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

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

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

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

Similarly,

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

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

$\Rightarrow \tan 30^{\circ}=\frac{h}{x+40}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{h}{x+40}$

$\Rightarrow x+40=h \sqrt3 \mathrm{~m}$

$\Rightarrow x=(x\sqrt3)\sqrt3-40 \mathrm{~m}$            [From (i)]

$\Rightarrow 3x-x=40 \mathrm{~m}$

$\Rightarrow x=\frac{40}{2} \mathrm{~m}$

$\Rightarrow x=20 \mathrm{~m}$

$\Rightarrow h=20\sqrt3 \mathrm{~m}$

Therefore, the height of the tower is $20\sqrt3 \mathrm{~m}$.      

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Updated on: 10-Oct-2022

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