A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of `60^@` with the horizontal. Find the height of the tree.

Explanation

Let seg AB represent the tree before the storm.
Due to storm, tree gets broken at C.
Broken part AC takes and position CD.
`therefore` AC = CD …(1)
BD = 20 m and
`angleCDB = 60^@`
In right angled `triangle`CBD,
`tan 60^@ = (CB)/(BD)` ....(By definition)
`therefore sqrt(3) = (CB)/(20) " "......(tan 60^@ = sqrt(3) )`
`therefore CB = 20sqrt(3)` m
also `sin 60^@ = (CB)/(CD)` ...(By definition)
`therefore sqrt3/2 = (20sqrt(3))/(CD)` ... (`because 60 = (sqrt(3))/(2)`)
`therefore CD = (2 xx 20sqrt(3))/(sqrt(3))`
`therefore` CD = 40 m ....[From (1) ]
AC = CD + CB ....(A-C-B)
`therefore AB = (40 + 20 sqrt(3)` m
The height of the tree is `(40 + 20sqrt(3))`m.