Question

An observer , 1.7 m tall , is` 20 sqrt3`  m away from a tower . The angle of elevation from the eye of an observer to the top of tower is 30⁰ . Find the height of the tower.

Answer 1

Let AB be the height of the observer and EC be the height of the tower.

Given: 

`AB=1.7 m ⇒ CD= 1.7 m` 

`BC=20 sqrt3 m`

Let ED be h m. 

In ∆ADE,

`tan 30° = (ED)/(AD)`  

`⇒ 1/sqrt3= h/(20sqrt3)`

`⇒ h=20 m`

`∴ EC=ED+DC=(h+1.7)m=21.7 m`

Hence, the height of the tower is 21.7 m.