Question

The angles of depression and elevation of the top of a 12m high building from the top and the bottom of a tower are 60° and 30° respectively. Find the height of the tower, and its distance from the building. 

Answer 1

Let AB be the building and CD be the tower. 

In ΔABD

`"AB"/"BD" = tan 30^circ`

`12/"BD" = 1/sqrt(3)`

`"BD" = 12sqrt(3)`

In ΔACE

`"AE" = "BD" = 12sqrt(3)`

`"CE"/"AE" = tan 60^circ`

`"CE"/(12sqrt(3)) = sqrt(3)`

`"CE" = 12sqrt(3) xx sqrt(3) = 12 xx 3 = 36`

`"CD" = "CE" + "ED" = 36 + 12 = 48`

So, height of the tower is 48 m and its distance from the building is `12sqrt(3)` m = `12 xx 1.732` m = 20.78 m(approximately)