Let the broken part of tree be AC
It is given that, distance between foot of the tree B and point C=8m.So, BC=8m
Also, broken parts of tree makes an angle 30∘ with ground.
So, ∠C=30∘
We need to find height of the tree
Height of the tree=Height of broken part+Height of remaining tree
=AB+AC
Since, tree was vertical to ground. So, ∠ABC=90∘
In rightangled △ABC,
cosC=BCAC
⇒cos30∘=8AC
⇒√32=8AC
⇒AC=8×2√3=16√3
In rightangled △ABC,
sinC=ABAC
⇒sin30∘=AB16√3
⇒12=√316×AB
⇒AB=162√3=8√3
So, height of the tree=AC+AB=16√3+8√3=24√3
24√3×√3√3=24√33=8√3m
Hence, height of tree is 8√3m