Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use `sqrt(3)` = 1.73)

Answer 1

Let the height of the tower AB be h m and the horizontal distance between the tower and the building BC be x m.

So,

AE = (h – 50) m

In ∆AED,

tan 45° = `(AE)/(ED)`

 ⇒ 1 = `(h - 50)/x`

 ⇒ x = h – 50   ...(1)

 In ∆ABC,

tan 60° = `(AB)/(BC)`

⇒ `sqrt(3) = h/x`

⇒ x = `sqrt(3)` = h  ....(2)

Using (1) and (2), we get

x = `sqrt(3)x - 50`

⇒ x = `(sqrt(3) - 1) = 50`

⇒ `x = (50(sqrt(3) + 1))/2`

⇒ x = 25 × 2.73

⇒ x = 68.25 m

Substituting the value of x in (1), we get

68.25 = h – 50

⇒ h = 68.25 + 50

⇒ h = 118.25 m

Hence, the height of tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.