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The angle of elevation of the top of a tower of height 100 √3 metre from a point at a distance of 100 metre from the foot of the tower on a horizontal plane is
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- 45°
- 60°
- 30°
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22 1 ° 2
Correct Option: B
AB = Tower = 100 √3 metre
BC = 100 metre
From ∆ ABC,
| tan θ = | BC |
| ⇒ tan θ = | = √3 | 100 |
⇒ tanθ = tan 60° ⇒ θ = 60°
