Question

from the top of a 60 meter height
building, the angles of depression to the
top and foot of a cell tower are 30degree
and 60degree respectively. find the height of
the cell tower ?​

Answer 1

Answer:

Let the building be AB,cell tower be CD,and o is the point on AB

Now

In triangle ABD

AB=P=60m

BD=b=?

Theta=60

Tan theta=p/b

Tan60=60/BD

√3=60/BD

BD=60/√3

Again

In triangle AOC

AO=AB-OB

=60-OB=p

OC=BD=60/√3=b

Tan theta=p/b

Tan 30=60-OB/60/√3

1/√3=60-OB/34.64

34.64/√3=60-OB

20=60-OB

OB=60-20

OB=40m

BD=OB=40 m

Height of cell tower is 40 m

Step-by-step explanation:

Answer 2

Answer:

give diagram of this question