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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