Question

From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.

Answer 1

Let AB be the cliff and CD be the tower. 

Here AB = 60 m, ∠ADE = 30° and ∠ACB = 60°

In ΔABC, 

`(AB)/(BC) = tan 60^circ = sqrt(3)`

`=> BC = (60)/(sqrt(3))`

In ΔADE, 

`(AE)/(DE) = tan 30^circ`

`=>` AE = DE tan 30°

= `60/(sqrt(3)) xx 1/sqrt(3)`    ...[∵ DE = BC]

= 20 m

∴ CD = EB

= AB – AE

= (60 – 20)

= 40 m

Hence, height of the tower is 40 m.