Question

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.

Answer 1

Let AC was the original tree. Due to storm, it was broken into two parts. The broken part A'B is making 30° with the ground.
In ∆A'BC,
$\frac{BC}{ A'C} = tan 30°$

$\frac{Bc} 8 = \frac{1 }{ \sqrt3}$

$BC = (\frac{8}{\sqrt3})m$

$\frac{A'C}{ A'B} = cos 30°$

$\frac{8}{A'B} = \frac{\sqrt3}2$

$A'B = (\frac{16}{\sqrt3})m$

Height of the tree = $A’B + BC$
= $(\frac{16}{\sqrt3} + \frac{8}{ \sqrt3})m = \frac{24}{ \sqrt3 }m$

= $8\sqrt3m$

Hence, the height of the tree is $8\sqrt3m$.