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The shadow of a tower standing on a level plane is found to be 40 m longer when the sun’s altitude is 45°, than when it is 60°. The height of the tower is
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- 30 (3 + √3) metre
- 40 (3 + √3) metre
- 20 (3 + √3) metre
- 10 (3 + √3) metre
Correct Option: C
∠ACB = 60°; BC = x metre
CD = 40 metre, AB = Tower = h metre
From ∆ABC,
| tan 60° = | BC |
| ⇒ √3 = | x |
⇒ h = √3 x ...............(i)
From ∆ABD,
| tan 45° = | BD |
| ⇒ 1 = | x + 40 |
| ⇒ h = x + 40 = | + 40 | √3 |
| ⇒ h - | = 40 | √3 |
| ⇒ | = 40 | √3 |
⇒ ( √3 – 1)h = 40 √3
| ⇒ = | √3 - 1 |
| ⇒ = | (√3 - 1)(√3 + 1) |
= 20 (3 + √3 ) metre
