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The shadow of the tower becomes 60 metres longer when the altitude of the sun changes from 45° to 30°. Then the height of the tower is
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- 20 (√3 + 1) m
- 24 (√3 + 1) m
- 30 (√3 + 1) m
- 30 (√3 - 1) m
Correct Option: C
AB = Tower = h metre
∠ADB = 30°
∠ACB = 45°
CD = 60 metre
BC = x metre
From ∆ABC,
| tan45° = | BC |
| ⇒ 1 = | ⇒ h = x | x |
From ∆ABD,
| tan30° = | BD |
| ⇒ | = | |||
| √3 | x + 60 |
| ⇒ | = | |||
| √3 | h + 60 |
⇒ √3 h = h + 60
⇒ √3 h – h = 60
⇒ h(√3 - 1) = 60
| ⇒ h = | = | |||
| √3 - 1 | (√3 - 1)(√3 + 1) |
= 30(√3 + 1) metre
