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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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