The angles of elevation of the top of a tower from the top and bottom of a tree of height 15 m are 30° and 60° respectively. Find the height of the tower?

Question: The angles of elevation of the top of a tower from the top and bottom of a tree of height 15 m are 30° and 60° respectively. Find the height of the tower?Solution:Let the CE be h meter.Height of tree be AD = 15mBE is the height of tower = BC + CE = 15 + hAB = CD, let it is = X mFrom ΔCDE, tan30° = EC/CD⇒ 1/√3 = h/X∴ X = √3h ..........(1)From ΔABE, tan60° = EB/AB⇒ √3 = (h + 15)/X∴ X = (h + 15)/√3 ..............(2)From (1) and (2) we get,√3h = (h + 15)/√3⇒ 3h = h + 15⇒ 2h = 15∴ h = 7.5∴ Height of tower = 15 + 7.5 = 22.5 m