Question

From a point on the ground the angles of elevation of the bottom and top of a communication tower fixed on the top of a 20-m-high building are 45° and 60° respectively. Find the height of the tower. [Take `sqrt(3) = 1.732`.]

Answer 1

Let AB be the building and BC be the tower and let O be the point of observation. 

Then, AB = 20 m, ∠AOB = 45° and ∠AOC = 60°.

Let BC = h metres.

From right ΔOAB, we have

`(OA)/(AB) = cot 45^circ = 1`

⇒ OA = AB = 20 m.

From right ΔOAC, we have

`(AC)/(OA) = tan 60^circ = sqrt(3)`

⇒ `(20 + h)/20 = sqrt(3)`

∴ `h = 20(sqrt(3) - 1)`

= 20(1.732 – 1)

= (20 × 0.732)

= 14.64

∴ Height of the tower = 14.64 m.