the angle of elevation of the top of the building at the distance of 80m from its foot on a horizontal plane is found to be 60°, find the height of the building
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Answer:
Given height of tower CD=50m
Let the height of the building, AB=h
In right angled △BDC,
⇒ tan60
o
=
BD
CD
⇒
3
=
BD
50
⇒ BD=
3
50
m
In right angled △ABD,
⇒ tan30
o
=
BD
AB
⇒
3
1
=
3
50
h
∴
3
1
=
50
3
h
∴ h=
3
50
=16.66m
∴ The height of the building is 16.66m.
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