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The height of broken part of tree from ground is 5.569m.
Justification:
Let BD is a tree of height 12 m.
<u>Suppose it got bent at a point C and let the part CD take the position CA, meeting the ground at A</u>.
i.e., CD = AC = h m
<u>Broken part makes 60° angle from ground</u>
So, ∠BAC = 60°
<u>Now, height of remaining part of tree</u> = (12 – h)m.
In right angled ∆ABC,
sin 60° = BC/AC
⇒ √3/2 = (12 - h)/h
⇒ √3h = 2(12 – h)
⇒ √3h = 24 – 2h
⇒ √3h + 2h = 24
⇒ h(√3 + 2) = 24
⇒ h(1.732 + 2) = 24
⇒ h(3.732) = 24
⇒ h = 24/3.732 = 6.4308 m
<u>Hence, height of broken tree from ground</u>
⇒ BC = 12 – h
⇒ 12 – 6.4308 = 5.569m
<u>Hence, tree is broken 5.569 m from ground</u>.
<u>Note</u>: See attached picture.
