Question

In the given figure, from the top of a building AB = 60 m hight, the angle 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 that AB is a building that is 60 m, high.

Let BC = DE = x and CD = BE = y

`=>` AE = AB – BE = 60 – y 

i. In right ΔAED, 

`tan 30^circ = (AE)/(DE)`

`=> 1/sqrt(3) = (60 - y)/(x)`

`=> x = 60sqrt(3) - ysqrt(3)`   ...(1)

In right ΔABC, 

`=> tan 60^circ = (AB)/(BC)`

`=> sqrt(3) = 60/x`

`=> x = 60/sqrt(3)`

`=> x = 60/sqrt(3) xx sqrt(3)/sqrt(3)`

`=> x = (60sqrt(3))/3`

`=> x = 20sqrt(3)`

`=>` x = 20 × 1.732

`=>` x = 34.64 m

Thus, the horizontal distance between AB and CD is 34.64 m.

ii. From (1), we get the height of the lamp post = CD = y

`x = 60sqrt(3) - ysqrt(3)`

`=> 20sqrt(3) = 60sqrt(3) - ysqrt(3)`

`=>` 20 = 60 – y

`=>` y = 60 – 20

`=>` y = 40 m

Thus, the height of the lamp post is 40 m.