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The elevation of the top of a tower from a point on the ground is 45°. On travelling 60 m from the point towards the tower, the elevation of the top becomes 60°. The height of the tower (in metres) is
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- 30
- 30(3 - √3)
- 30 (3 + √3)
- 30 √3
Correct Option: C
AB = tower = h metre
∠ACB = 45°, ∠ADB = 60°
CD = 60 metre] BD = x metre
From ∆ABC,
tan 45° = | BC |
⇒ 1 = | x + 60 |
⇒ h = x + 60 ....................(i)
From ∆ABD
tan 60° = | BD |
⇒ √3 = | x |
⇒ h = √3x
⇒ h = √3(h - 60)
⇒ √3h - h= 60√3
⇒ h(√3 - 1) = 60√3
⇒ h = | = | |||
√3 - 1 | (√3 - 1)(√3 + 1) |
= 30√3(√3 + 1)
= 30(3 + √3) metre