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`

Concept: Heights and Distances

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