Question

The upper part of tree broken by wind makes an angle of 45° with the ground and horizontal distance from the foot of the tree to the point where the top of tree touches the ground is 12 m, find the height of tree before it was broken ?

Answer 1

Answer:

AC is the broken part of the tree and AB is the remaining part of the tree.

We get a 45°−45°−90° triangle.

We get AB=BC=12m

AC²=AB² +BC²

⇒AC²=12²+12²

=2×12²

⇒AC=12²√2m

Height of the tree=broken part+ remaining part

AC+AB=12+12√2=12(1+ √2)

Answer 2

Answer:

AC is the broken part of the tree and AB is the remaining part of the tree.

We get a 45 ∘ −45 ∘ −90 ∘ triangle.

We get AB=BC=12m

AC ^2 =AB^2 +BC^2

⇒AC^2 =12 ^2 +12 ^2 =2×12^2

⇒AC=12√2m

Height of the tree=broken part+ remaining part

AC+AB=12+12 ^2 =12(1+2)