In a violent storm, a tree got bent by the wind. The top of the tree meets the ground at an angle of \[{{30}^{o}},\] at a distance of 30m from the root. At what height from the bottom did the tree get bent? What was the original height of the tree? \[\left[ \text{Take }\sqrt{3}=1.73 \right]\]

Hint: In order to solve this question, we should have a good visualization and a basic knowledge of the trigonometric ratios. So, by the visualization power, we will imagine the situation and then draw the figure accordingly. So, we will have a triangle formed with base as ground, perpendicular as part of tree standing and hypotenuse as bent part of tree meeting ground at \[{{30}^{o}},\]. Then by the help of trigonometric ratios, we will be able to find the required values.

Complete step-by-step answer:
In this question, we have been given that in a violent storm, a tree got bent by the wind such that the top of the tree meets the ground at an angle of \[{{30}^{o}},\] 30 m away from the foot of the tree and we have been asked a couple of things to find.
So, to solve this question, we will first visualize the given situation and draw the diagram accordingly. So, we can say the tree will look something like this,

Here, we have represented A’B as the total height of the tree such that after the storm, it got bent at point O so that the point A’ touches the ground A at an angle of \[{{30}^{o}},\] 30 m away from the foot of the tree. So, we can say OA’ = OA. And therefore, we can say that the total height of the tree is AO + OB.
Now, we know that the tangent ratio of any angle is a right angle triangle and is given by \[\dfrac{\text{perpendicular}}{\text{base}}.\] So, we can say, in triangle OAB, right-angled at B, we get,
\[\tan \angle OAB=\dfrac{OB}{BA}\]
Now, we have been given that, \[\angle OAB={{30}^{o}}\] and BA = 30 m. So, we can say,
\[\tan {{30}^{o}}=\dfrac{OB}{30}\]
Now, we know that \[\tan {{30}^{o}}=\dfrac{1}{\sqrt{3}}.\] So, we get,
\[\dfrac{1}{\sqrt{3}}=\dfrac{OB}{30}\]
And therefore, we can say,
\[OB=\dfrac{30}{\sqrt{3}}\]
And we have been given that \[\sqrt{3}=1.73.\] So, we get,
\[OB=\dfrac{30}{1.73}\]
Hence, we can say the height at which the tree got bent from the bottom is OB = 17. 34 m.
Now, we have been asked to find the total length of the tree, that is, A’B or we can say AO + OB. Now, we know that the cosine ratio of any angle is right angle triangle is given by \[\dfrac{\text{base}}{\text{hypotenuse}}.\]
So, we can say,
\[\cos \angle OAB=\dfrac{AB}{OA}\]
Now, we know that \[\angle OAB={{30}^{o}}\] and AB = 30m. So, we get,
\[\cos {{30}^{o}}=\dfrac{30}{OA}\]
Now, we know that, \[\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}.\] So, we can say,
\[\dfrac{\sqrt{3}}{2}=\dfrac{30}{OA}\]
\[\Rightarrow OA=\dfrac{30\times 2}{\sqrt{3}}\]
\[\Rightarrow OA=\dfrac{30\times 2}{1.73}\]
\[\Rightarrow OA=34.68m\]
Hence, we get the value of OA = 34.68 m. Therefore, we can say the total height of the tree is OA + OB = 34.68 + 17.321 = 52.02m.

Note: While solving this question, we need to remember that the tangent ratio is given by \[\dfrac{\text{perpendicular}}{\text{base}}\] and cosine ratio is given by \[\dfrac{\text{base}}{\text{hypotenuse}}.\] Also, a few standard trigonometric ratios like \[\tan {{30}^{o}}=\dfrac{1}{\sqrt{3}},\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}\] and so on. Because if we forget these values, we may get the wrong answer.