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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.
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.
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