Question

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

विकल्प

•  100\[\sqrt{3}\]

• \[\frac{100}{\sqrt{3}} m\]

• \[50 \sqrt{3}\]

• \[\frac{200}{\sqrt{3}} m\]

Answer 1

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`