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The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to 60°. The height of the tower is
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- √3 m
- 5 √3 m
- 10 √3 m
- 20 √3 m
Correct Option: C
Let PQ = h metre and BQ = x metre.
From ∆ APQ,
| tan30° = | x + 20 |
| ⇒ | = | |||
| √3 | x + 20 |
⇒ √3h = x + 20 ...........(i)
From ∆ PQB,
| tan 60° = | = | |||
| BQ | x |
| ⇒ √3 = | ⇒ h = √3x | x |
| ⇒ x = | h .......(ii) | √3 |
| ∴ √3h = | h + 20 | √3 |
[From equation (i) and (ii)]
⇒ 3h – h = 20 √3
⇒ 2h = 20 √3
∴ h = 10 √3 metre
