Question

Angle of elevation of top of a tower from a 7 m high building is 60° and angle of depression of its foot is 45°. Find the height of the tower.

Answer 1

Let AB is a tower and CD is a building of height 7 m.

The angle of elevation and angle of depression are 60° and 45° respectively.

i.e., ∠ACE = 60°

and ∠ECB = 45°

BD || CE, CD || BE

∴ CD = BE = 7 m

From right angled ∆CBD

tan 45° = CD/DB

⇒ 1 = 7/DB

∴ DB = 7 m

CE = DB = 7 m

Again from right angled ∆AEC

tan 60° = AE/CE

√3 = AE/7

AE = 7√ m

Hence, height of tower AB = AE + EB

= 7√3 + 7

= 7(√3 + 1) m

Hence, height of tower AB = 7(√3 + 1) m