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An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 30° and 60° respectively. The distance between the two planes at that instant is
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- 6520 m
- 6000 m
- 5000 m
- 6250 m
Correct Option: D
A and C ⇒ position of planes
BC = 3125m
Let AC = x metre
In ∆ABD
| tan60° = | BD |
| ⇒ √3 = | BD |
| ⇒ BD = | √3 |
In ∆BCD,
| tan30° = | BD |
| ⇒ | = | |||||
| √3 | ||||||
| √3 |
⇒ 3 (3125) = 3125 + x
⇒ 9375 = 3125 + x
⇒ x = 9375 – 3125
x = 6250 metre
