Question

From the top of a 9 m high building, the angle of elevation of the top of a cable tower is 60° and angle of depression of its foot is 45°. Determine the height of the tower and distance between building and tower. `("Use" sqrt3 = 1.732)`

Answer 1

Given,

Height of building = 9 m

Let AC = hm and BD = xm

In ΔBDE,

tan 45° = `(ED)/(BD)`

1 = `9/x`

x = 9m

In ΔACE

tan 60° = `(AC)/(CE)`

`sqrt3 = h/x`

h = `xsqrt3`

= 9 × 1.732

= 15.588 m

∴ Height of the tower (AB) = AC + CB

= (15.588 + 9)m

= 24.588 m