Question

A tree is broken over by the wind forms a right angled triangle with the ground. If the broken part makes an angle of ${60}^o$ with the ground and the top of the tree is now $20m$ from its base, how tall was the tree?

Answer 1

Let $△ABC$ be a right angled triangle where $\angle B={90}^0$ and $\angle C={60}^0$ as shown in the above figure.

Let $x$ be the length of remaining part of the tree and $y$ be the length of part of tree which is broken.

We know that $\tan \theta =\frac{Oppositeside}{Adjacentside}=\frac{AB}{BC}$

Here, $\theta ={60}^0$, $BC=20$ m and $AB=x$ m, therefore,

$\tan \theta =\frac{AB}{BC}\Rightarrow \tan {60}^0=\frac{x}{20}\Rightarrow \sqrt{3}=\frac{x}{20}(\because \tan {60}^0=\sqrt{3})\Rightarrow x=20\sqrt{3}...........\left(1\right)$

We also know that $\cos \theta =\frac{Adjacentside}{Hypotenuse}=\frac{BC}{AC}$

Here, $\theta ={60}^0$, $BC=20$ m and $AC=y$ m, therefore,

$\cos \theta =\frac{BC}{AC}\Rightarrow \cos {60}^0=\frac{20}{y}\Rightarrow \frac{1}{2}=\frac{20}{y}(\because \cos {60}^0=\frac{1}{2})\Rightarrow y=20\times 2\Rightarrow y=40...........\left(2\right)$

Now, total length of the tree is $x+y=20\sqrt{3}+40=20(\sqrt{3}+2)$ m.

Hence, the length of the tree is $20(\sqrt{3}+2)$ m.