Question

The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find the distance between the lamp post and the apartment `(sqrt(3) = 1.732)`

Answer 1

Let the height of the lamp post AE be h m

DE = h – 66

Let AB be x

In the right ∆ABC, tan 30° = `"BC"/"AB"`

`1/sqrt(3) = 66/x`

x = `66sqrt(3)`  ...(1)

In the right ∆CDE, tan 60° = `"DE"/"DC"`

`sqrt(3) = ("h" - 66)/x`

⇒ `sqrt(3)x` = h – 66

x = `("h" - 66)/sqrt(3)`  ...(2)

From (1) and (2) we get

`("h" - 66)/sqrt(3) = 66sqrt(3)`

h – 66 = `66sqrt(3) xx sqrt(3)` = 66 × 3

h – 66 = 198 ⇒ h = 198 + 66

h = 264 m

Distance between the lamp post and the apartment 

= `66 sqrt(3)` m

= 66 × 1.732

= 114.31 m