Question

From the top of the tower 60 m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post (tan 38° = 0.7813, `sqrt(3)` = 1.732)

Answer 1

Let the height of the lamp post be h

The height of the tower (BC) = 60 m

∴ EC = 60 – h

Let AB be x

In the right ∆ABC,

tan 60° = `"BC"/"AB"`

`sqrt(3) = 60/x`

x = `60/sqrt(3)`  ...(1)

In the right ∆DEC, tan 38° = `"EC"/"DE"`

0.7813 = `(60 - "h")/x`

x = `(60 - "h")/(0.7813)`  ...(2)

From (1) and (2) we get

`60/sqrt(3) = (60 - "h")/(0.7813)`

60 × 0.7813 = `60 sqrt(3) = sqrt(3)`h

`sqrt(3)"h" = 60 sqrt(3) - 46.88`

= 60 × 1.732 – 46.88

= 103.92 – 46.88

1.732h = 57.04

⇒ h = `(57.04)/(1.732)`

h = `(570440)/(1732)`

= 32.93 m

∴ Height of the lamp post = 32.93 m