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Question
The top of a palm tree having been broken by the wind struck the ground at an angle of 60° at a distance of 9m from the foot of the tree. Find the original height of the palm tree.
Solution
Let AC was original tree. It was broken into two parts. The broken part A' B is making 60° with ground.
In ΔA'BC
`"BC"/"A'C" = tan 60^circ`
`"BC"/9 = sqrt(3)`
`BC = 9sqrt(3)`
`"A'C"/"A'B" = cos 60^circ`
`9/"A'B" = 1/2`
`A'B = 18`
Height of tree = A'B + BC
= `9sqrt(3) + 18 = 9 × 1.732 + 18 = 15.588 + 18 = 33.588`
Hence , the height of tree was 33.588 m = 33.6 m (approximately).
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