An observer $ 1.5 $ metres tall is $ 20.5 $ metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Given:
An observer \( 1.5 \) metres tall is \( 20.5 \) metres away from a tower 22 metres high.
To do:
We have to determine the angle of elevation of the top of the tower from his eye.
Solution:
Let $AB$ be the tower and $CD$ be the observer who is $20.5\ m$ away from the tower.
From the figure,
$AB = 22\ m, CD=1.5\ m, AC=20.5\ m$
This implies,
$AE=CD=1.5\ m, DE=AC=20.5\ m$ and $BE=22-1.5=20.5\ m$
Let \( \theta \) be the angle of elevation of the top of the tower from the eye of the observer.
In $\Delta \mathrm{BDE}$,
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow \tan \theta=\frac{\mathrm{BE}}{\mathrm{DE}}$
$=\frac{20.5}{20.5}$
$=1$
$=\tan 45^{\circ}$ [Since $\tan 45^{\circ}=1$]
$\Rightarrow \theta=45^{\circ}$
Therefore, the angle of elevation of the top of the tower from his eye is $45^{\circ}$.
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