Question

The upper part of a tree, broken by wind in two parts, makes an angle of ${30}^{\circ }$ with the ground. The top of the tree touches the ground at a distance of $20\text{m}$ from the foot of the tree. Find the height of the tree before it was broken

Answer 1

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

Given that:

$BC=20\text{m}$ and $\angle ACB={30}^{\circ }$

Solution:

In $△ABC$

$\tan \left(\angle ACB\right)=\frac{AB}{BC}$

or, $\tan {30}^{\circ }=\frac{AB}{20\text{m}}$

or, $AB=20\text{m}\times \frac{1}{\sqrt{3}}=\frac{20}{\sqrt{3}}\text{m}$

Similarly,

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

or, $AC=\frac{20\text{m}}{\cos {30}^{\circ }}$

or, $AC=\frac{20\text{m}\times 2}{\sqrt{3}}$

or, $AC=\frac{40}{\sqrt{3}}\text{m}$

Total length of the tree $=AB+AC=\frac{20}{\sqrt{3}}\text{m}+\frac{40}{\sqrt{3}}\text{m}=\frac{60}{\sqrt{3}}\text{m}=20\sqrt{3}\text{m}$

Therefore, total length of the tree$=20\sqrt{3}\text{m}=34.64\text{m}$