Question

From the top of a 60m high building, the angles ol depression of the top and the bottom of a tower are 45º and 60º respectively. Find the height of the tower.[Take √3 =1.73]

Answer 1

\(\tan 60° = \frac{60}{CD}\)

⇒ \(CD = \frac{60}{\sqrt 3}\)

\(\therefore CD = 34.64 m\)

\(\tan 45° = \frac{AB}{BE}\)

⸫ AB = 34.64 m [⸪ BE = CD]

Height of tower = ED = BC 

= 60 – AB 

= 60 – 34.64 

= 25.36 m

Answer 2

Let the height of the building be BD=60m. 

The height of the tower is 'h'. The distance between the base of the building and the tower be 'd'.