The upper part of a tree is broken over by the strong wind makes an angle of 30° with the ground. The top of the broken tree meets the ground at a distance of 25 m from the foot of the tree. Find the original height of the tree.
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Let length of tree before windstorm is BD.
After windstorm the upper part of tree C falls from point C to point A on the ground
. Now, let CD = AC = h2 m AB = 10 m
Broken part makes an angle 60° from the ground.
So, ∠CAB = 60°
From right angled ∆ABC tan 60° = BC/AB ⇒ √3 = h1/10 ⇒ h1 = 10√3 and cos 60° = AB/AC ⇒ 1/2 = 10/h2 ⇒ h2 = 10 × 2 = 20 m
Hence, total length of tree BD = BC + CD = h1 + h1
= 10√3 + 20
= 10 × 1.732 + 20
= 17.32 + 20
= 37.32 m
Hence, height of the tree 37.32 m.
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