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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.
Solution
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.
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