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An Observer , 1.7 M Tall , is 20 √ 3 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. - Mathematics

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. 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.

 
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 12 Trigonometry
Q 10 | Page 41
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