An observer, $ 1.5 \mathrm{~m} $ tall, is $ 28.5 \mathrm{~m} $ away from a tower $ 30 \mathrm{~m} $ high. Determine the angle of elevation of the top of the tower from his eye.

Given:

An observer, \( 1.5 \mathrm{~m} \) tall, is \( 28.5 \mathrm{~m} \) away from a tower \( 30 \mathrm{~m} \) 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 $28.5\ m$ away from the tower.

From the figure,

$AB = 30\ m, CD=1.5\ m, AC=28.5\ m$

This implies,

$AE=CD=1.5\ m, DE=AC=28.5\ m$ and $BE=30-1.5=28.5\ m$

Let \( \theta \) be the angle of elevation of the top of the tower from the eye of the observer.

In $\Delta \mathrm{BDE}$,

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$\Rightarrow \tan \theta=\frac{\mathrm{BE}}{\mathrm{DE}}$

$=\frac{28.5}{28.5}$

$=1$

$=\tan 45^{\circ}$ [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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