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 30v3 feet from the entrance of the lift, find the speed of the lift which is descending.

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Answer 1

$\large\color{lime}\boxed{\colorbox{black}{Answer : - }}$

Solution:

Heigjt of tbe building AC = 90ft

AB = h. ft

BC = (90-h) ft

CD = 30√3

Time = 2min

From ∆ BCD :

tan 30° = BC / CD

$\frac{1}{ \sqrt{3} } = \frac{90 - h}{30 \sqrt{3} }$

$30 = 90 - h$

$h = 90 - 30$

$\color{blue}h \: = 60ft$

$Speed = \: \frac{distance}{time}$

$= \frac{60}{2}$

$\color{hotpink}Speed = 30 ft/ min$