A a person from the top of the building of height 15 m observe the the top of the the bottom of a cell tower with the angle of elevation as 60° and the the angle of of depression as 45° respectively. Then find the height of the cell tower.
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Answer:
Let the Height of the Tower be AD=h
and let the height of Building be CE=15
In △ADE
⇒tan60=
DE
AD
=
DE
h
⇒DE=
3
h
and we can see that BCDE is a Rectangle
⇒BC=DE=
3
h
and BD=CE
so, in △ABC
⇒tan30
o
=
BC
AB
=
h/
3
AB
⇒h=3AB⇒AB=
3
h
Now, it is given that CE=15⇒AD−AB=15⇒h−
3
h
=
3
2h
=15⇒h=22.5 m
and DE=
3
h
=
3
22.5
=13 m (Take
3
=1.73)
Answered by
0
you go now
Step-by-step explanation:
Let the Height of the Tower be AD=h
and let the height of Building be CE=15
In △ADE
⇒tan60=
DE
AD
=
DE
h
⇒DE=
3
h
and we can see that BCDE is a Rectangle
⇒BC=DE=
3
h
and BD=CE
so, in △ABC
⇒tan30
o
=
BC
AB
=
h/
3
AB
⇒h=3AB⇒AB=
3
h
Now, it is given that CE=15⇒AD−AB=15⇒h−
3
h
=
3
2h
=15⇒h=22.5 m
and DE=
3
h
=
3
22.5
=13 m (Take
3
=1.73)
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