Question

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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
​

Answer 1

Speed of Lift = 30 ft/min

Step-by-step explanation:

Tan 60°  = Initial Height of Lift / Distance of Fountain from foot of Lift

=> √3  = Initial Height of Lift / 30 √3

=>  Initial Height of Lift = 90 feet

Tan 30°  = Height of Lift after 2 minutes / Distance of Fountain from foot of Lift

=> 1/√3  = Height of Lift after 2 minutes / 30 √3

=> Height of Lift after 2 minutes= 30 feet

Distance covered by lift in 2 Minutes = 90 - 30 = 60 feet

Speed of Lift = 60/2 = 30 ft/min

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Answer 2

The speed of the lift is 30 feet/min.

Step-by-step explanation:

Consider the provided information.

The Initial height of lift is 90 feet.

At the top of the building, the angle of depression to a fountain  in the garden is 60°.

The  fountain is $30\sqrt{3}$ feet from the entrance of the lift.

Two minutes later, the angle of depression reduces to 30°.

$\tan30^0=\frac{opp}{30\sqrt{30}}$

$\frac{1}{\sqrt{3}}=\frac{opp}{30\sqrt{3}}$

$opp=\frac{30\sqrt{3}}{\sqrt{3}}$

$opp=30$

The initial height was 90 feet and after 2 min the height decreases to 30 feet. That means the height decrease 60 feet in 2 minutes.

$Speed=\frac{Distance}{Time}$

Substitute the respective values in the above formula.

$Speed=\frac{60}{2}=30$

Hence, the speed of the lift is 30 feet/min.

#Learn more

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