Question

A storm broke a tree and the tree top rested on ground 20 m away from the
base of the tree, making an angle of 60o with the ground. Find the height
of the tree.

Answer 1

AB = Height of the tree
Tree is broken at C
AC = CD                    ....... (1)
∠ CDB = 60°
BD = 20 m

In right angled Δ CBD,
`tan60° = (CB)/(BD)`
`sqrt3 = (CB)/20`
`CB = 20sqrt3  m.`
`sin60° = (CB)/(CD)`
CD =`sqrt3/2 = (20sqrt3)/(CD)`
`CD = (2 xx 20sqrt3)/sqrt3`
CD = 40 m.
∴ AC = CD = 40 m. ....... (From (1) )

AB = AC + CB
AB = `(40 + 20sqrt3)` m.
∴ height of the tree = `(40 + 20sqrt3)` m.