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A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is
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- 80 m
- 100 m
- 160 m
- 200 m
Correct Option: A
AB = Tower = h metre
CD = 100 metre; BC = 160 metre
∠ACB = ∴ ∠ADB = 2θ
In ∆ ABC,
| tanθ = | BC |
| ⇒ tanθ = | .............(i) | 160 |
In ∆ ABD,
| tan2θ = | BD |
| ⇒ tan2θ = | 60 |
| ⇒ | = | |||
| 1 - tan2θ | 60 |
| = | 2 × | |||||
| 160 | ||||||
| 1 - | ||||||
| 160 × 160 | ||||||
| 60 |
| ⇒ | 1 | = | |||||
| 80 |  | 1 - |  | 60 | |||
| 160 × 160 | |||||||
| ⇒ 4 | | 1 - | | = 3 | 160 × 160 |
| ⇒ | = 1 - | = | ||||
| 160 × 160 | 4 | 4 |
⇒ h2 = 6400
⇒ h = √6400 = 80 metre
