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Q2:
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30°$ with it. 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.

Solution :

Given:

  • A tree breaks at a certain point, and the top touches the ground.
  • The angle of elevation of the top of the tree from the point where it touches the ground is $\theta = 30^\circ$.
  • The distance between the foot of the tree and the point where the top touches the ground is $BC = 8$ m.

To Find:

The total height of the tree ($H = AB + AC$, where $A$ is the top, $B$ is the foot, and $C$ is the point where the top touches the ground, with the break occurring at point $D$).

B C D 8 m 30°

Step 1: Define the variables and model the triangle.

Let the original tree be $AD$. Let it break at point $D$. The part $AD$ bends such that the top $A$ touches the ground at point $C$. Let $B$ be the foot of the tree. Thus, $BD$ is the vertical part remaining standing, and $DC$ is the broken part that touches the ground. The total height of the tree is $H = BD + DC$.

In the right-angled triangle $\triangle DBC$ (right-angled at $B$):

  • Base $BC = 8$ m
  • Angle $\angle DCB = 30^\circ$

Step 2: Calculate the height of the standing part ($BD$).

Using the trigonometric ratio for tangent: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(30^\circ) = \frac{BD}{BC}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ [Standard trigonometric value]:

$\frac{1}{\sqrt{3}} = \frac{BD}{8}$

$BD = \frac{8}{\sqrt{3}}$ m

Step 3: Calculate the length of the broken part ($DC$).

Using the trigonometric ratio for cosine: $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

$\cos(30^\circ) = \frac{BC}{DC}$

Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ [Standard trigonometric value]:

$\frac{\sqrt{3}}{2} = \frac{8}{DC}$

$DC = \frac{16}{\sqrt{3}}$ m

Step 4: Calculate the total height of the tree ($H$).

$H = BD + DC$

$H = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}}$

$H = \frac{24}{\sqrt{3}}$

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

$H = \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3}$ m

Using $\sqrt{3} \approx 1.732$:

$H = 8 \times 1.732 = 13.856$ m

Final Answer: The height of the tree is $8\sqrt{3}$ m (or approximately $13.86$ m).


More Questions from Class 10 Mathematics Applications of Trigonometry EXERCISE 9.1


CBSE Solutions for Class 10 Mathematics Applications of Trigonometry


Chapters in CBSE - Class 10 Mathematics


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