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The shadow of a vertical tower becomes 30 metres longer when the altitude of the sun changes from 60° to 45°. Find the height of the tower.
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- 15√3 metres
- 15 (3 + √3) metres
- 15 (3 - √3) metres
- 12 (3 + √3) metres
- 15√3 metres
Correct Option: B
AB = Tower = h metre
DC = 30 metre
BD = x metre
From ∆ABC,
| tan 45° = | ⇒ 1 = | ⇒ h = x + 30 ......(i) | ||
| BC | x + 30 |
From ∆ABD,
| tan 60° = | ⇒ √3 = | ||
| BD | x |
h = √3x
| ⇒ x = | .........(ii) | |
| √3 |
∴ h = x + 30
| ⇒ h = | + 30 | |
| √3 |
⇒ (√3 + 1)h = 30√3
| ⇒ h = | ||
| √3 - 1 |
| = | ||
| (√3 - 1)(√3 + 1) |
= 15(3 + √3) metre.
