from the top of a building 30 M high the angle of depression of the top and bottom of a tower are observed to be 30 degree and 60 degree respectively find the height of the tower
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Let AC be the tower and BE be the building .
Let height of the tower be h m. It is given that the that the angles of depression of the top C and bottom A of the tower , observed from the top of the building be 30° and 60° respectively.
Now , Refer the Attachment ! ^^^^^^
In Right triangle CDE we have ....
tan 30° = DE/CD
=> 1/√3 = 60-h / CD
=> CD = √3(60-h) .... (i)
In right triangle ABE , we have ....
tan 60° = BE/AB
=> √3 = 60/CD (AB=CD)
=> CD = 60/√3 ........... (ii)
Comparing (i) and (ii) , we get .....
√3(60-h) = 60/√3
=> 3 ( 60-h ) = 60
=> 180-3h = 60
=> 3h = 120
=> h = 40
Hence , the height of the tower is 40 m.
I THINK IT HELPED YOU ^
Let AC be the tower and BE be the building .
Let height of the tower be h m. It is given that the that the angles of depression of the top C and bottom A of the tower , observed from the top of the building be 30° and 60° respectively.
Now , Refer the Attachment ! ^^^^^^
In Right triangle CDE we have ....
tan 30° = DE/CD
=> 1/√3 = 60-h / CD
=> CD = √3(60-h) .... (i)
In right triangle ABE , we have ....
tan 60° = BE/AB
=> √3 = 60/CD (AB=CD)
=> CD = 60/√3 ........... (ii)
Comparing (i) and (ii) , we get .....
√3(60-h) = 60/√3
=> 3 ( 60-h ) = 60
=> 180-3h = 60
=> 3h = 120
=> h = 40
Hence , the height of the tower is 40 m.
I THINK IT HELPED YOU ^
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