Question

From the top of a 60m high building the angles of depression of the top and bottom of a lamp post are 30° and 60° respectively. Find the distance on the ground between the building and the lamp post and the difference in their heights. 

Answer 1

Let AB be the building. Then, AB = 60 m. 
Let the height of the lamp post (CD) be h. 
Let the distance between the building and the lamp post be x. 
In ΔACB, 

`tan60^circ = "AB"/"BC"`

∴ `sqrt(3) = 60/X`

∴ `X = 60/sqrt(3) = 20sqrt(3) = 20 × 1.732 = 34.64`  ...(1)

Thus, the distance between the building and the lamp post is 34. 64 m 
In ΔADE, 

`tan30^circ = "AE"/"DE"`

∴ `1/sqrt(3) = (60 - h)/X`

∴ `X = sqrt(3)(60 - h)`  ..(2)

From (1) and (2): 

`sqrt(3)(60 - h) = 20sqrt(3)`

60 - h = 20

h = 40