An observer $1.6$ metres tall is $10.5$ 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 $1.6$ metres tall is $10.5$ metres away from a tower $22$ metres high.

To do: To find the angle of elevation of the top of the tower from the eye of the observer.

Solution:

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



As given, the height of the observer is $CD=1.5\ m$

Height of the tower $AB=22\ m$

$AE=AB-BE=22-1.5=20.5$                          [$\because BE=CD=1.5\ m$]

In $\vartriangle AEC$,

$tan\theta=\frac{AE}{CE}$

$tan\theta=\frac{20.5}{20.5}$                      [$\because CE=BD=20.5$]

$tan\theta=1=tan45^o$

$\Rightarrow \theta=45^o$

Thus, the angle of elevation of the top of the tower from the eye of the observer is $45^o$.

Updated on: 10-Oct-2022

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