Question

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.

Answer 1

 

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ΔABO,

`("AB")/("BO") = tan 60°`

`("AB")/("BO") = sqrt3`

`"BO" = ("AB")/sqrt3`

In ΔCDO,

`("CD")/("DO") = tan 30°`

`("CD")/(80-"BO") = 1/sqrt3`

`"CD"sqrt3  = 80 - "BO"`

`"CD"sqrt3 = 80 - ("AB")/sqrt3`

`"CD"sqrt3+ ("AB")/sqrt3 = 80`

Since the poles are of equal heights,

CD = AB

`"CD"[sqrt3 + 1/sqrt3] = 80`

`"CD"((3+1)/sqrt3) = 80`

`"CD" = 20sqrt3`

`"BO"= ("AB")/sqrt3 = ("CD")/sqrt3`

= `((20sqrt3)/sqrt3)`m

= 20 m

DO = BD – BO

= (80 – 20) m

= 60 m

Therefore, the height of poles is `20sqrt3` and the point is 20 m and 60 m far from these poles.

Answer 2

Let AB and CD be the two poles of equal height.

O be the point makes an angle of elevation from the top of the poles of 60° and 30°, respectively.

Let OA = 80 – x, OD = x.

And ∠BOA = 30°, ∠COD = 60°.

Here, we have to find the height of the poles and the distance of the points from the poles.

We have the corresponding figure as follows:.

So we use trigonometric ratios.

In a triangle COD,

⇒ tan 60° = `"CD"/"DO"`

⇒ `sqrt(3) = "h"/"x"`

⇒ `"x" = "h"/sqrt(3)`

Again in a triangle AOB,

⇒ tan 30° = `"AB"/"OA"`

⇒ `(1)/sqrt(3) = "h"/(80 -"x")`

⇒ `sqrt(3)"h" = 80 - "x"`

⇒ `sqrt(3)"h" = 80 - "h"/sqrt(3)`

⇒ `sqrt(3)"h" + "h"/sqrt(3) = 80`

⇒ `3"h" + "h" = 80sqrt(3)`

⇒ `4"h" = 80sqrt(3)`

⇒ `"h" = 20sqrt(3)`

⇒ `"x" = (20sqrt(3))/sqrt(3)`

⇒ x = 20

And

⇒ OA = 80 – x

⇒ OA = 80 – 20

⇒ OA = 60

Hence, the height of pole is `20sqrt(3)` and distances are 20 m, 60 m respectively.