The shadow of a tower, when the angle of elevation of the sun is \( 45^{\circ} \), is found to be \( 10 \mathrm{~m} \). longer than when it was \( 60^{\circ} \). Find the height of the tower.

Given:

The shadow of a tower, when the angle of elevation of the sun is \( 45^{\circ} \), is found to be \( 10 \mathrm{~m} \). longer than when it was \( 60^{\circ} \).

To do:

We have to find the height of the tower.

Solution:  

Let $AB$ be the tower and $CB$ be the shadow when the angle of elevation of the sun is \( 45^{\circ} \) and $DB$ be the shadow when the angle of elevation of the sun is \( 60^{\circ} \).

From the figure,

$\mathrm{CD}=10 \mathrm{~m}, \angle \mathrm{ACB}=45^{\circ}, \angle \mathrm{ADB}=60^{\circ}$

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,

$\mathrm{DB}=x-10 \mathrm{~m}$

We know that,

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

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

$\Rightarrow \tan 45^{\circ}=\frac{h}{x}$

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

$\Rightarrow h=x(1) \mathrm{~m}$

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

Similarly,

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

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

$\Rightarrow \tan 60^{\circ}=\frac{h}{x-10}$

$\Rightarrow \sqrt3=\frac{h}{x-10}$

$\Rightarrow (x-10)\sqrt3=h \mathrm{~m}$

$\Rightarrow (h-10)\sqrt3=h \mathrm{~m}$           [From (i)]

$\Rightarrow \sqrt3h-10\sqrt3=h \mathrm{~m}$

$\Rightarrow h(\sqrt3-1)=10\sqrt3 \mathrm{~m}$

$\Rightarrow h=\frac{10\times1.732}{1.732-1} \mathrm{~m}$

$\Rightarrow h=\frac{17.32}{0.732}=23.66 \mathrm{~m}$

Therefore, the height of the tower is $23.66 \mathrm{~m}$.      

Tutorialspoint

Updated on: 10-Oct-2022

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