sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances

it is given that The horizontal distance between two towers is 140 m.

The angle of elevation of the top of the first tower when seen from the top of the second tower is 30.

If the height of the second tower is 60 m,

find the height of the first tower

question no 25 , heights and distances , ICSE board

Consider the height of the first tower TR = x

It is given that

Height of the second tower PQ = 60 m

Distance between the two towers QR = 140 m

Construct PL parallel to QR

LR = PQ = 60 m

PL = QR = 140 m

So we get

TL = (x – 60) m

In right triangle TPL

tan θ = TL/PL

Substituting the values

tan 30

^{0}= (x – 60)/ 140So we get

1/√3 = (x – 60)/ 140

By further calculation

x – 60 = 140/√3

Multiply and divide by √3

x – 60 = 140/√3 × √3/√3 = 140√3/3

We get

x = 140√3/3 + 60

x = (140 × 1.732)/ 3 + 60

x = 80.83 + 60

x = 140.83

Hence, the height of first tower is 140.83 m.