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A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is \( 2 \mathrm{~m} \) away from the wall and the ladder is making an angle of \( 60^{\circ} \) with the level of the ground. Determine the height of the wall.
Given:
A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is \( 2 \mathrm{~m} \) away from the wall and the ladder is making an angle of \( 60^{\circ} \) with the level of the ground.
To do:
We have to determine the height of the wall.
Solution:
Let $AB$ be the wall and $AC$ be the ladder.
The foot of the ladder(point $C$) is \( 2 \mathrm{~m} \) away from the wall.
From the figure,
$\mathrm{BC}=2 \mathrm{~m}, \angle \mathrm{ACB}=60^{\circ}$
Let the height of the wall be $\mathrm{AB}=h \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\text { AB }}{BC}$
$\Rightarrow \tan 60^{\circ}=\frac{h}{2}$
$\Rightarrow \sqrt3=\frac{h}{2}$
$\Rightarrow h=2 \times \sqrt3 \mathrm{~m}$
$\Rightarrow h=2\sqrt3 \mathrm{~m}$
Therefore, the height of the wall is $2\sqrt3 \mathrm{~m}$.
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