Question

From the top of a 7 meter high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45° . Determine the height of the tower.

Answer 1

Let AB be the 7-m high building and CD be the cable tower,
We have,
AB = 7m, ∠CAE = 60°,∠DAE = ∠ADB = 45°
Also, DE = AB = 7m
In  ΔABD,

` tan 45° = (AB)/(BD)`

`⇒ 1= 7/(BD) `

⇒  BD = 7 m

So, AE = BD = 7m
Also, in ΔACE,

` tan 60° = (CF)/(AE)`

`⇒  sqrt(3) = (CE) /7`

`⇒ CE = 7 sqrt(3) m`

Now, CD = CE  +DE

`= 7 sqrt(3) +7`

`= 7 ( sqrt(3) +1) m`

= 7(1.732 +1) 

= 7(2.732) 

= 19.124

`~~ 19.12m`

So, the height of the tower is 19.12m.