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 metres from the foot of the tree. Find the height of the tree before it is broken.
Last updated date: 17th Mar 2023
•
Total views: 303.9k
•
Views today: 6.82k
Answer
303.9k+ views
Hint: Here when the tree is broken, a triangle is formed and using the triangle we can find the height of the tree.
Complete step-by-step answer:
First with the given data we have to draw the figure.
In this figure BD is the tree, due to wind it is broken from the point C where $CD=CA$. The broken tree makes an angle of ${{30}^{\circ }}$ with the ground. i.e. $\angle BAC={{30}^{\circ }}$
Here, the distance between the top of the tree that touches the ground and foot of the tree is 20 m. i.e. $AB=20\text{ m}$.
Here, we have to calculate the height of the tree, i.e. $BD$.
From the figure, we can say that,
$\begin{align}
& BD=BC+CD \\
& BD=BC+CA \\
\end{align}$
For that, first consider $\vartriangle ABC$, the tangent of A is given by,
$\begin{align}
& \tan {{30}^{\circ }}=\dfrac{oposite\text{ }side}{adjacent\text{ }side} \\
& \tan {{30}^{\circ }}=\dfrac{BC}{AB} \\
& \tan {{30}^{\circ }}=\dfrac{BC}{20}\text{ }....\text{ (1)} \\
\end{align}$
We know that $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$, by substituting this value in equation (1) we get:
$\dfrac{1}{\sqrt{3}}=\dfrac{BC}{20}$
By cross multiplication we get,
$\begin{align}
& 20=\sqrt{3}BC \\
& BC=\dfrac{20}{\sqrt{3}} \\
\end{align}$
To find $CA$, consider the cosine of A, we obtain:
$\begin{align}
& \cos {{30}^{\circ }}=\dfrac{\text{adjacent }side}{hypotenuse} \\
& \cos {{30}^{\circ }}=\dfrac{AB}{CA} \\
& \cos {{30}^{\circ }}=\dfrac{20}{CA}\text{ }.....\text{ (2)} \\
\end{align}$
We also know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ our equation (2) becomes:
$\dfrac{\sqrt{3}}{2}=\dfrac{20}{CA}$
By cross multiplication we get,
$\begin{align}
& \sqrt{3}CA=2\times 20 \\
& \sqrt{3}CA=40 \\
& CA=\dfrac{40}{\sqrt{3}} \\
\end{align}$
Next, we have to calculate $BD$ where.
$\begin{align}
& BD=BC+CA \\
& BD=\dfrac{20}{\sqrt{3}}+\dfrac{40}{\sqrt{3}} \\
\end{align}$
By taking the LCM we get:
$\begin{align}
& BD=\dfrac{20+40}{\sqrt{3}} \\
& BD=\dfrac{60}{\sqrt{3}} \\
& BD=\dfrac{60}{1.73}\text{ }....\text{(}\sqrt{3}=1.73) \\
& BD=34.682 \\
\end{align}$
Hence, we obtain the height of the tree is $34.682$.
Note: Here, you may get confused regarding the tree and the broken part of the tree, from the figure we can say that CA is the broken part of the tree and CA=CD.
Complete step-by-step answer:
First with the given data we have to draw the figure.

In this figure BD is the tree, due to wind it is broken from the point C where $CD=CA$. The broken tree makes an angle of ${{30}^{\circ }}$ with the ground. i.e. $\angle BAC={{30}^{\circ }}$
Here, the distance between the top of the tree that touches the ground and foot of the tree is 20 m. i.e. $AB=20\text{ m}$.
Here, we have to calculate the height of the tree, i.e. $BD$.
From the figure, we can say that,
$\begin{align}
& BD=BC+CD \\
& BD=BC+CA \\
\end{align}$
For that, first consider $\vartriangle ABC$, the tangent of A is given by,
$\begin{align}
& \tan {{30}^{\circ }}=\dfrac{oposite\text{ }side}{adjacent\text{ }side} \\
& \tan {{30}^{\circ }}=\dfrac{BC}{AB} \\
& \tan {{30}^{\circ }}=\dfrac{BC}{20}\text{ }....\text{ (1)} \\
\end{align}$
We know that $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$, by substituting this value in equation (1) we get:
$\dfrac{1}{\sqrt{3}}=\dfrac{BC}{20}$
By cross multiplication we get,
$\begin{align}
& 20=\sqrt{3}BC \\
& BC=\dfrac{20}{\sqrt{3}} \\
\end{align}$
To find $CA$, consider the cosine of A, we obtain:
$\begin{align}
& \cos {{30}^{\circ }}=\dfrac{\text{adjacent }side}{hypotenuse} \\
& \cos {{30}^{\circ }}=\dfrac{AB}{CA} \\
& \cos {{30}^{\circ }}=\dfrac{20}{CA}\text{ }.....\text{ (2)} \\
\end{align}$
We also know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ our equation (2) becomes:
$\dfrac{\sqrt{3}}{2}=\dfrac{20}{CA}$
By cross multiplication we get,
$\begin{align}
& \sqrt{3}CA=2\times 20 \\
& \sqrt{3}CA=40 \\
& CA=\dfrac{40}{\sqrt{3}} \\
\end{align}$
Next, we have to calculate $BD$ where.
$\begin{align}
& BD=BC+CA \\
& BD=\dfrac{20}{\sqrt{3}}+\dfrac{40}{\sqrt{3}} \\
\end{align}$
By taking the LCM we get:
$\begin{align}
& BD=\dfrac{20+40}{\sqrt{3}} \\
& BD=\dfrac{60}{\sqrt{3}} \\
& BD=\dfrac{60}{1.73}\text{ }....\text{(}\sqrt{3}=1.73) \\
& BD=34.682 \\
\end{align}$
Hence, we obtain the height of the tree is $34.682$.
Note: Here, you may get confused regarding the tree and the broken part of the tree, from the figure we can say that CA is the broken part of the tree and CA=CD.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
