Question

The upper part of a tree is broken by windstorm  and it makes an angle of 60° with the ground. The distance from the bottom of the tree to the point where the top touches the ground is 10 m. Find the original height of the tree. (√3 = 1.732)

Answer 1

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