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A tower is 50 metre high. Its shadow is x metres shorter when the sun’s altitude is 45° than when it is 30°. The value of x in metre is :
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- 50 √3
- 50 (√3 - 1)
- 50 (√3 + 1)
- 50
Correct Option: B
AB = Height of tower = 50 metre
∠ ACB = 30°; ∠ADB = 45°
CD = x metre (let)
In ∆ ABD,
| tan45° = | BD |
| ⇒ 1 = | ⇒ BD = 50 metre | BD |
In ∆ ABC,
| tan30° = | ⇒ | = | BC | √3 | 50 + x |
⇒ 50 + x = 50 √3
⇒ x = 50 √3 - 50
= 50 (√3 - 1) metre
