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A tower stands vertically on the ground. From a point on the ground, $ 20 \mathrm{~m} $ away from the foot of the tower, the angle of elevation of the top of the tower is $ 60^{\circ} $. What is the height of the tower?
Given:
A tower stands vertically on the ground. From a point on the ground, \( 20 \mathrm{~m} \) away from the foot of the tower, the angle of elevation of the top of the tower is \( 60^{\circ} \).
To do:
We have to find the height of the tower.
Solution:
Let $AB$ be the tower and $C$ is the point \( 20 \mathrm{~m} \) away from the foot of the tower.
From the figure,
$\mathrm{BC}=20 \mathrm{~m}, \angle \mathrm{ACB}=60^{\circ}$
Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\text { AB }}{BC}$
$\Rightarrow \tan 60^{\circ}=\frac{h}{20}$
$\Rightarrow \sqrt{3}=\frac{h}{20}$
$\Rightarrow h=20 \sqrt{3} \mathrm{~m}$
Therefore, the height of the tower is $20 \sqrt{3} \mathrm{~m}$.
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