Question

An observer 1.5 metres tall is 20.5 m away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer 1

Let AB be an observer of height 1.5 m and CE be a tower of height 22 m.
`:. DE=AB=1.5m`
and `CD=CE-DE=(22-1.5)m=20.5m`
Now `BD=AE=20.5m`
Let the angle of elevation be `theta`.
In`DeltaBCD`,
`tantheta=(CD)/(BD) rArr tantheta =(20.5)/(20.5)=1`
`rArr theta=45^(@)`
`:."Angle of elevation is "45^(@)`.