Question

The Top of a 25 meter high tower makes an angle of elevation of 45ⁿ with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pole.

Answer 1

Answer:

height=10m

Step-by-step explanation:

Let AB be the tower and CD be the electric pole

Then ∠ACB=60

0

,∠EDB=30

0

and AB=15 m

Let CD=h. Then BE=(AB−AE)

=(AB−CD)=(15−h)

We have

AC

AB

=tan60

0

=

3

⇒AC=

3

AB

=

3

15

And

DE

BE

=tan30

0

=

3

1

⇒DE=(BE×

3

) =

3

(15−h)

⇒3h=(45−15)

⇒h=10 m

Answer 2

Let AB be the tower and CD be the electric pole.

∠ACB = 45°

∠EDB = 30°

AB=25 Let's consider CD = h

then BE = ( AB- AE )

             =  AB-CD    

              = 25-h

we have AB/AC = tan 45° =1

                          =AC =AB/1 = 25/1

and BE/DE = tan 30° =1/√3

DE = BE × √3

     =( 25-h) × √3

   3h =( 75-25)

3h = 50

h=  50/3m

  = 16.6 m

Ans :- the height of the pole will be 16.6 meter