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The length of the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45° to 30°. Then the height of the tower is
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- 5 ( √3 + 1) metres
- 5 ( √3 – 1) metres
- 5 √3 metres
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5 metres √3
Correct Option: A
AB = Tower = h metre
BD = Shadow = x metre
∠ ADB = 45°
∠ ACB = 30°
In ∆ ABD,
| tan 45° = | BD |
| ⇒ 1 = | ⇒ h = x ..........(i) | x |
In ∆ ABC
| tan 30° = | BC |
| ⇒ | = | |||
| √3 | x + 10 |
| ⇒ | = | |||
| √3 | h + 10 |
⇒ √3 h = h + 10
⇒ √3 h – h = 10
⇒ h ( √3 –1) = 10
| ⇒ h = | √3 - 1 |
| = | × | |||
| √3 - 1 | √3 + 1 |
| = | = 5(√3 + 1) metre | 3 - 1 |
