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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

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प्रश्न

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.

योग

उत्तर

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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अध्याय 12: Trigonometry - Exercise 12.2 [पृष्ठ ४१]

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आरडी शर्मा Mathematics [English] Class 10
अध्याय 12 Trigonometry
Exercise 12.2 | Q 10 | पृष्ठ ४१

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