Question

The horizontal distance between two towers is 60 metres. The angle of depression 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 90 metres, find the height of the first tower. [Use `sqrt(3) = 1.732`.]

Answer 1

Let DE be the first tower and AB be the second tower.
Now, AB = 90 m and AD = 60 m such that CE = 60 m and ∠BEC = 30°.
Let DE = h m such that AC = h m and BC = 90°.

In the right ΔBCE,we have:

`(BC)/(CE) = tan 30° = 1/ sqrt(3)`

`⇒ ((90-h )) /60 = 1/ sqrt(3) `

`⇒ (90-h ) sqrt(3) = 60`

`⇒ h sqrt(3) = 90 sqrt( 3 ) -60`

`⇒ h = 90- 60/ sqrt(3) = 90- 34.64 = 55.36 m`

∴ Height of the first tower = DE = h = 55.36m