Question

The horizontal distance between two towers is 140m. The angle of depression of the top of the
first tower when seen from the top of the second tower is 30 degrees . If the height of the first tower is 60m,
find the height of the second tower.

Answer 1

Answer:

Let AB be the height of second tower and CD be the height of first tower.

Given, BD = AE = 140 m

& AB = DE = 60 m

In ΔAEC,

tan 30°= Perpendicular /Base= CE/AE

tan 30 ° = CE/140

1/√3= CE /140

CE= 140 / √3

CE= 140 ×√3/ (√3×√3)

[ Rationalising the denominator]

CE= 140√3 /3

Height of the first tower CD= CE+DE

= 140√3/3 + 60

=( 140 × 1.73) /3 + 60

[ √3=1.73]

= 242.2/3 + 60

= 80.73 + 60

= 140.73 m

Height of the first tower (CD)=140.73 m

Thank you

I hope my answer is helpful for you