Two points A and B are present on opposite sides of a tower of height 100√3 m. The angles of elevation from the two points are 30° and 60° respectively. If a car crosses these two points in 40 seconds, what is the speed of the car?
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Answers
Given : Two points A and B are present on opposite sides of a tower of height 100√3 m.
The angles of elevation from the two points are 30° and 60° respectively.
A car crosses these two points in 40 seconds,
To find : the speed of the car
Solution:
Let say O is the bottom/base of the tower
Hence Distance between A and B
= OA + OB as points are on opposites sides
Tan 30° = 100√3/OA
=> 1/√3 = 100√3/OA
=> OA = 300 m
Tan 60° = 100√3/OB
=> 3 = 100√3/OB
=> OB = 100 m
AB = OA + OB = 300 + 100 = 400 m
Distance covered = 400 m
Time taken = 40 sec
Speed of the car = 400/40
= 10 m/s
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