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If the Angle of Elevation of a Tower from a Distance of 100 Metres from Its Foot is 60°, Then the Height of the Tower is - Mathematics

MCQ

If the angle of elevation of a tower from a distance of 100 metres from its foot is 60°, then the height of the tower is

Options

  •  100\[\sqrt{3}\]

  • \[\frac{100}{\sqrt{3}} m\]

  • \[50 \sqrt{3}\]

  • \[\frac{200}{\sqrt{3}} m\]

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Solution

Let AB be the height of tower is h meters

Given that: angle of elevation is `60°` from tower of foot and distance `BC=100` meters.

Here, we have to find the height of tower. 

So we use trigonometric ratios. 

In a triangle ABC 

`⇒ tan C= (AB)/(BC)`  

`⇒tan 60°= (AB)/(BC)`  

`⇒ sqrt3= h/100` 

`⇒ h= 100sqrt3` 

 

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 12 Trigonometry
Q 2 | Page 41
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