Question

From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer 1

Let the height of the tower be BD

In ΔPAB,

`tan 45^circ = "AB"/"AP"`

⇒ `1 = 20/"AP"`

⇒ AP = 20 m

In ΔPAD,

`tan 60^circ = "AD"/"AP"`

= `(20 + BD)/20`

⇒ `sqrt(3) = (20 + BD)/20`

⇒ `20 + BD = 20sqrt(3)`

⇒ `BD = 20sqrt(3) - 20`

= `20 (sqrt(3) - 1)`

= 20 (1.732 – 1) 

= 20 × 0.732

= 14.64 cm