Sum

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 30^{0 }. Find the height of the tower.

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#### Solution

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.

Concept: Heights and Distances

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