Question

from the top of a 60 Meter height building the angle of depression to the top and the foot of a cell tower is 30 degree and 60 degree respectively find the height of the cell tower ​

Answer 1

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

Againnnnnnnn,,,,

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

Thanks

Mark branliest

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Answer 2

The height of the cell tower=40 m

Step-by-step explanation:

Let AE=x

In triangle AEC

$\theta=30^{\circ}$

$tan\theta=\frac{perpendicular\;side}{base}$

Substitute the values

$tan30=\frac{x}{EC}$

$\frac{1}{\sqrt 3}=\frac{x}{EC}$

By using $tan30^{\circ}=\frac{1}{\sqrt 3}$

$EC=x\sqrt 3$...(1)

In triangle ABD

$AB=60 m$

EBDC is a rectangle

Therefore, BD=EC

EB=DC

$tan60^{\circ}=\frac{AB}{BD}=\frac{60}{CE}=\frac{60}{x\sqrt 3}$

$\sqrt 3=\frac{60}{x\sqrt 3}$

By using $tan60^{\circ}=\sqrt 3$

$3x=60$

$x=\frac{60}{3}=20 m$

$EB=DC=AB-AE=60-20=40 m$

Hence, the height of the cell tower=40 m

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https://brainly.in/question/14817038:Answered by Kumarilimbu