The angle of elevation of the top of a tower from point on the ground is60degree. From a point 10 m vertically above the first the angle of elevation is 30.find the height of tower
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In ∆ AEC,
tan30° = AE/CE
1/√3 = x/CE
CE = √3x -------1)
Now In ∆ ABD,
tan60° = AB/BD
√3 = x+10/BD
CE = x+10/√3 (CE=BD) -------2)
From eq 1) and 2)
x+10/√3=√3x
x+10 = 3x
10 = 3x-x
10 = 2x
x = 5
Height of the tower = x+10
= 5+10
= 15m Ans.
tan30° = AE/CE
1/√3 = x/CE
CE = √3x -------1)
Now In ∆ ABD,
tan60° = AB/BD
√3 = x+10/BD
CE = x+10/√3 (CE=BD) -------2)
From eq 1) and 2)
x+10/√3=√3x
x+10 = 3x
10 = 3x-x
10 = 2x
x = 5
Height of the tower = x+10
= 5+10
= 15m Ans.
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