Question

a tree due to storm and broken parts bends so that the top of tree touches the ground with angle 30' distance between the two is 8m find height if tree

Answer 1

let BD be the length of tree before storm After Storm AD=AC=length of Broken part the Angle of Elevation Is 30' various arrangement are Shown in figure.
In right angled ABC (diagram is attached with photo)
$\frac{ab}{bc} = \tan(30) \\ \frac{h1}{8} = \frac{1}{ \sqrt{3}} \\h1 = \frac{8}{ \sqrt{3}}\times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \frac{8}{3} \sqrt{3m} \\ also \: \frac{bc}{ac} = \cos(30) \\ \frac{8}{h2} = \frac{ \sqrt{3} }{2} \\ h2 = \frac{8 \times 2}{ \sqrt{3} } = \frac{16}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ h2 = \frac{16}{3} \sqrt{3}$
Total height of tree
$\frac{8}{3} \sqrt{3} + \frac{16}{3} \sqrt{3} \\ \binom{8 + 6}{3} \sqrt{3} \\ \frac{24}{3} \sqrt{3} \\ 8 \sqrt{3m}$

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Answer 2

let BD be the length of tree before storm After Storm AD=AC=length of Broken part the Angle of Elevation Is 30' various arrangement are Shown in figure.
In right angled ABC (diagram is attached with photo)

Total height of tree