Question

In the adjoining figure, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp-post CD are observed to be 30° and 60° respectively. Find:

1. the horizontal distance between AB and CD. 
2. the height of the lamp-post.

Answer 1

Given:

AB = 60 m height of building.

From A, angle of depression to top of lamp-post point D = 30°.

From A, angle of depression to bottom of lamp-post point C = 60°.

Let horizontal distance between the verticals AB and CD be x and lamp-post height CD = h.

Step-wise calculation:

1. Triangle A–C (A at height 60, C on ground):

`tan 60^circ = "Vertical drop AC"/("Horizontal"  x)` 

= `60/x` 

⇒ `sqrt(3) = 60/x`

⇒ `x = 60/sqrt(3)`

= `20sqrt(3)`

= 34.64 m

2. Triangle A–D (D at height h):

Vertical drop AD = 60 – h

Using angle 30°:

`tan 30^circ = (60 - h)/x`

= `1/sqrt(3)`

Substitute `x = 20sqrt(3)`:

`(60 - h)/(20sqrt(3)) = 1/sqrt(3)` 

⇒ 60 – h = 20

⇒ h = 40 m

Horizontal distance between AB and CD = `20sqrt(3)` ≈ 34.64 m.

Height of the lamp-post CD = 40 m.