The angle of elevation of the top of a tower from a point on the ground which is at a distance of 70 m from the foot of the tower is 60° . Then the height of the tower is:?

LohaHridoy
  1. \(\frac{70}{\sqrt{3}}\)
  2. 80\(\sqrt{3}\)
  3. \(\frac{80}{\sqrt{3}}\)
  4. 70\(\sqrt{3}\)

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AhmedNazir

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Option 4 : 70\(\sqrt{3}\)

Given:

Distance of the point from the foot of the tower = 70 m

Angle of elevation = 60° 

Formula used:

tan θ = Perpendicular/Base

Calculation:

[ alt="F4 Arindam ghosh 25-5-2021 Swati D2" src="//storage.googleapis.com/tb-img/production/21/05/F4_Arindam%20ghosh_25-5-2021_Swati_D2.png" style="width: 163px; height: 174px;"> 

Let the height of the tower be h.

In ΔABC,

tanθ = BC/AB

⇒ tanθ = h/70

⇒ tan60° = h/70

⇒ √3 = h/70

⇒ h = 70√3

∴ The height of the tower is 70√3 m.

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