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Question
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is `30sqrt(3)` feet from the entrance of the lift, find the speed of the lift which is descending.
Solution
Let the speed of the lift is “x” feet/minute
Distance AB = 2 × feet ...(speed × time)
BC = (90 – 2x)
In the right ∆BCD,
tan 30° = `"BC"/"DC"`
`1/sqrt(3) = (90 - 2x)/(30 sqrt(3))`
`sqrt(3) (90 - 2x) = 30sqrt(3)`
(90 – 2x) = `(30sqrt(3))/sqrt(3)`
⇒ (90 – 2x) = 30
2x = 60
x = `60/2` = 30
x = 30 feet/minute
Speed of the lift = 30 feet/minute
or
`[30/60 "second"]` 0.5 feet/second
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