Question

From the top 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 45°. Determine the height of the tower.

Answer 1

Let the distance between the foots of building and cable tower is (m). 

The height of cable tower = AB = AE+EB 

⇒ (h+7)m. 

In ∆AED, 

tan 60° = \(\frac{AE}{DE}\)

√3 = \(\frac{h}{x}\)

h = √3 x---------(1) 

The height of cable tower = AB = AE+EB 

⇒ (h+7)m. 

In ∆DEB, 

tan 45° = \(\frac{BE}{DE}\)

1 = \(\frac{7}{x}\)

x = 7.........(2)

On substituting value of x in eqn. (1) 

h = 7√3

Height of cable tower is (h+7)m. 

⇒ 7√3+7 

⇒ 7(√3+1)m. 

Therefore height of cable tower is 7(√3+1)m.