Question

The horizontal distance between two towers is 150 m. The angle of depression of the top of one tower as observed from the top of other tower, which is 120 m in
height, is 30°. Find the height of the first tower.
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Answer 1

see the given figure

Horizontal distance between two towers = 150m

let the height of first tower be x m

ED = 120 m

but ED = BC = 120

AC = 120 m

AB = AC - BC

AB = (x - 120)m

Now , in ∆ABE

tanA = perpendicular/base = AB/BE

tan30° = (x - 120)/150

1/√3 = (x - 120)/150

(x - 120)√3 = 150

x√3 - 120√3 = 150

x√3 = 120√3 + 150

x√3 = 30(4√3 + 5)

x = {30(4√3 + 5)}/√3

rationalise √3

x = {30(4√3 + 5)√3} / √3.√3

x = {30 × √3(4√3 + 5)} / 3

x = 10√3(4√3 + 5)

x = (120 + 50√3)

or

x = 10(12 + 5√3)

Hence, the height of tower is 10(12 + 5√3)m

Answer 2

Ur answer is 33.4 m!!

Mark me as brainliest if you think my answer is helpful