Question

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of \[{{30}^{{}^\circ }}\] with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer 1

Hint: In this question, the broken part of the tree forms a right-angled triangle with the ground and remaining part of the tree. So, we should have knowledge of trigonometric ratios to find the height of a tree.

Complete step by step answer:
Let us assume that the tree is positioned at A and after the tree breaks down from point C it touches the ground at point B, which forms a right-angled triangle, as shown below in figure.

Now, to find the height of the tree, we need to find the length of the side’s AC and BC, so that if we add them up, we will get the height of the tree.
Now, we are given that the top of the tree touches the ground and makes an angle of \[{{30}^{\circ }}\], with the ground.
\[\Rightarrow \angle ABC={{30}^{\circ }}\]
So, here in \[\Delta ABC\], we will apply trigonometric ratios to find the length of side AC and side BC. We know that \[\tan \theta \] is always equal to the ratio of perpendicular to the base, mathematically it can be written as \[\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}\]
\[\Rightarrow \tan \theta =\dfrac{AC}{AB}\]
We know that \[\angle ABC=\theta ={{30}^{\circ }}\] and \[\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}\], therefore, \[\tan \theta =\dfrac{AC}{AB}\]
\[\Rightarrow \tan {{30}^{\circ }}=\dfrac{AC}{AB}\]
We have been given that the top of the tree is at a distance of 8 m from the foot of tree, so, we get,
\[\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{AC}{8}\]
\[\Rightarrow \dfrac{8}{\sqrt{3}}=AC\], or we can say,
\[\Rightarrow AC=\dfrac{8}{\sqrt{3}}......\left( i \right)\]
In the same manner, we will find the length of BC by using cosine function, we will get,
\[\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}}\]
\[\Rightarrow \cos \theta =\dfrac{AB}{BC}\]
\[\Rightarrow \cos {{30}^{\circ }}=\dfrac{8}{BC}\]
\[\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{8}{BC}\]
\[\Rightarrow BC=\dfrac{16}{\sqrt{3}}......\left( ii \right)\]
Now, we know that the height of the tree is the sum of lengths AC and BC, which we can write as
\[\text{height of tree = }AC+BC......\left( iii \right)\]
Now, we will put the values of AC and BC from equation (i) and (ii) to equation (iii), so, we get,
\[\text{height of tree = }\dfrac{8}{\sqrt{3}}+\dfrac{16}{\sqrt{3}}\]
\[\Rightarrow \text{height of tree = }\dfrac{24}{\sqrt{3}}\]
\[\Rightarrow \text{height of tree = 8}\sqrt{3}\]
Therefore, the height of the tree is \[\text{8}\sqrt{3}\]m.

Note: In this question, one can possibly mistake while writing the answer is considering AC as the length of the tree, which is not. Also, in a hurry, we might take wrong ratios which will give us wrong answers.