Question

A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.

Answer 1

Let seg AB represent the tree before the storm.

Due to storm, tree gets broken at C.

Broken part AC takes the position CD.

∴ AC = CD   ...(1)

BD = 20 m and ∠CDB = 60°

In right angled ΔCBD,

`tan 60^circ = "CB"/"BD"`   ...(By definition)

∴ `sqrt(3) = "CB"/20`   ...`(tan 60^circ = sqrt(3))`

∴ CB = `20sqrt(3)` m

Also `sin 60^circ = "CB"/"CD"`   ...(By definition)

∴ `sqrt(3)/2 = (20sqrt(3))/"CD"`   ...`(sin 60^circ = sqrt(3)/2)`

∴ CD = `(2 xx 20sqrt(3))/sqrt(3)`

∴ CD = 40 m

AC = CD = 40 m   ...[From (1)]

AB = AC + CB   ...(A – C – B)

∴ AB = `(40 + 20sqrt(3))` m

The height of the tree is `(40 + 20sqrt(3))` m.