(v) What is the angle of depression from top of tower to point O ?
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1
According to me :-
Let AB=height of the building=100m
DE=AC=AB+BC=h= height of the tower
∠ADC=60
∘
∠BDC=60
∘
Now
tan30
∘
=
CD
CB
=
d
x
3
1
=
d
x
(1)
tan60
∘
=
CD
CA
=
d
x+100
3
=
d
x+100
(2)
From (1)
d=
3
x (3)
Substituting (3) in (2)
3
=
3
x
(x+100)
3x=x+100
2x=100
x=50
Height of the tower=DE=h=AB+CB=100m+x=100m+50m=150m
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