Question

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 30°. Determine the height of the tower.

Answer 1

Given, height of building = 7 m

Let AC = h m and BD = x m

In ΔBDE,

tan 30° = `(ED)/(BD)`

⇒ `1/sqrt(3) = 7/x`

⇒ x = `7sqrt (3)` m

In ΔACE,

tan 60° = `(AC)/(CE)`

⇒ `sqrt(3) = h/x`  ...[∵ CE = BD]

⇒ h = `xsqrt(3)`

= `7sqrt(3) xx sqrt(3)`

= 7 × 3

= 21 m

∴ Height of the tower = AB = AC + CB

= 21 + 7

= 28 m.