Question

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

Answer 1

$\Rightarrow$ Here, let $DC$ be an observer and $\theta$ is the angle of elevation of the top of the tower from his eye.

$\Rightarrow$ $AB$ is height of tower and $CB$ is distance between tower and observer.

$\Rightarrow$ $CB=DE=28.5m$

$\Rightarrow$ So in $△ADE$, $\tan \theta =\frac{AE}{DE}$

$\Rightarrow$ $\tan \theta =\frac{30-1.5}{28.5}$

$\Rightarrow$ $\tan \theta =\frac{28.5}{28.5}$

$\Rightarrow$ $\tan \theta =1$

$\therefore$ $\theta ={\tan }^{-1}\left(1\right)$

$\therefore$ $\theta ={45}^o$