Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide, From a point P between them on the road, the angle of elevation of the top of one pole is 60 and the angle of depression from the top of another pole at P is 30 . Find the height of each pole and distance of the point P from the poles.

#### Solution

Let AB and CD be the equal poles; and BD be the width of the road.

We have,

∠AOB = 60° and ∠COD = 60°

In ΔAOB,

`tan 60° = (AB)/(BO)`

`⇒ sqrt(3) = (AB)/ (BO)`

`⇒ BO = (AB) / sqrt(3)`

Also, in ΔCOD,

`tan 30° = (CD)/(DO)`

`⇒ 1/ sqrt(3) = (CD)/(DO)`

`⇒ DO = sqrt(3)CD`

As , BD = 80

⇒ BO +DO = 80

`⇒ (AB) /sqrt(3) + sqrt(3) CD = 80`

`⇒ (AB) /sqrt(3) + sqrt(3) AB = 80` (Given : AB - CD)

`⇒AB (1/sqrt(3) + sqrt(3) ) = 80`

`⇒ AB ((1+3)/sqrt(3) )=80`

`⇒ AB (4/ sqrt(3) ) =80`

`⇒ AB = (80+sqrt(3))/4`

`⇒ AB = 20 sqrt(3) m`

`Also , BO = (AB)/ sqrt(3) = (20sqrt(3))/sqrt(3) = 20m`

So, DO = 80 - 20 = 60m

Hence, the height of each pole is 20 `sqrt(3)`m and point P is at a distance of 20 m from left pole ad 60 m from right pole.