NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 2
NCERT Class X Chapter 9: Some Applications of Trigonometry
Question:
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 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.
Given:
- The tree breaks at a certain point and the broken part touches the ground, making an angle of 30° with the ground.
- The distance from the foot of the tree to the point where the top touches the ground is 8 m.
To Find:
The original height of the tree.
Formula:
The tangent of an angle in a right triangle is given by:
$$
\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}
$$
Solution:
Step 1: Let the tree break at a point such that the unbroken part is of height \( h \) meters and the broken part is of length \( x \) meters. The broken part touches the ground at a distance of 8 m from the foot of the tree, making an angle of 30° with the ground.
Step 2: In the right triangle formed, the broken part of the tree acts as the hypotenuse, the vertical unbroken part is one side, and the distance from the foot to the point where the top touches the ground is the base (8 m).
Step 3: Let the point where the tree breaks be \( A \), the foot be \( B \), and the point where the top touches the ground be \( C \). Then, \( AB = h \), \( BC = 8 \) m, and \( AC = x \).
Step 4: Since the broken part \( AC \) touches the ground at 30°, use trigonometry:
$$ \sin 30^\circ = \frac{AB}{AC} = \frac{h}{x} $$ $$ \cos 30^\circ = \frac{BC}{AC} = \frac{8}{x} $$Step 5: From the cosine relation, solve for \( x \):
$$ \cos 30^\circ = \frac{8}{x} $$ $$ \frac{\sqrt{3}}{2} = \frac{8}{x} $$ $$ x = \frac{8 \times 2}{\sqrt{3}} = \frac{16}{\sqrt{3}} $$Step 6: Now, from the sine relation, solve for \( h \):
$$ \sin 30^\circ = \frac{h}{x} $$ $$ \frac{1}{2} = \frac{h}{x} $$ $$ h = \frac{x}{2} = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}} $$Step 7: The total height of the tree is the sum of the unbroken and broken parts:
$$ \text{Total height} = h + x = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} $$Step 8: Rationalized form (optional):
$$ 8\sqrt{3} \approx 8 \times 1.732 = 13.856 \approx 13.86 \text{ m} $$Result:
The height of the tree was \( 8\sqrt{3} \) meters (approximately 13.86 m).
Comments
Post a Comment