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From the top of a cliff 100 metre high, the angles of depression of the top and bottom of a tower are 45° and 60° respectively. The height of the tower is
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100 (3 - √3) metre 3 -
100 (√3 - 1) metre 3 -
100 (2√3 - 1) metre 3 -
100 √3 - √2 metre 3
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Correct Option: A
AB = Height of cliff = 100 metre.
CD = Height of tower = h metre.
∠ADE = 45°, ∠ACB = 60°
In ∆ABC,
| tan 60° = | BC |
| ⇒ √3 = | BC |
| ⇒ BC = | metre ....(1) | √3 |
In ∆ADE,
| tan 45° = | DE |
| ⇒ 1 = | = | BC | BC |
⇒ BC = 100 – h
| ∴ | = 100 - h | √3 |
| ⇒ h = 100 – | √3 |
| = | - 100 | √3 |
| = | = | √3 | 3 |
| = | metre | 3 |
