Question

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making ${30}^{\circ }$ angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is $6m$. Find the height of the tree before falling down.

Answer 1

Let $AC$ is the broken part of the tree and $AB$ is standing part.

Given that,

$BC=6m$; $\angle ACB={30}^{\circ }$

Solution:

In $△ABC$

$\tan \angle ACB=\frac{AB}{BC}$

or, $\tan {30}^{\circ }=\frac{AB}{6m}$

or, $AB=6m\times \frac{1}{\sqrt{3}}=2\times \sqrt{3}m$

Similarly,

$\cos {30}^{\circ }=\frac{BC}{AC}$

or, $AC=\frac{6m}{\cos {30}^{\circ }}$

or, $AC=\frac{6m\times 2}{\sqrt{3}}$

or, $AC=4\times \sqrt{3}m$

Total length of the tree $=AB+AC=2\times \sqrt{3}m+4\times \sqrt{3}m=6\sqrt{3}m$

Therefore, total length of the tree $=6\sqrt{3}m=10.4m$