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 300 = (x – 60)/ 140
So 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.