NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 14

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Given:

  • Height of the girl = 1.2 m
  • Height of the balloon from the ground = 88.2 m
  • Height of the balloon above the girl's eyes = 88.2 - 1.2 = 87 m
  • Initial angle of elevation (\( \theta_1 \)) = \( 60^\circ \)
  • Final angle of elevation (\( \theta_2 \)) = \( 30^\circ \)

To Find:

Distance travelled by the balloon during the interval.

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 initial horizontal distance from the girl to the balloon be \( x_1 \), and the final distance be \( x_2 \). The vertical height from the girl's eyes to the balloon is 87 m.

Step 2: Using the tangent formula for the initial position (\( \theta_1 = 60^\circ \)):

$$ \tan 60^\circ = \frac{87}{x_1} $$ $$ \sqrt{3} = \frac{87}{x_1} $$ $$ x_1 = \frac{87}{\sqrt{3}} $$

Step 3: Calculate \( x_1 \):

$$ x_1 = \frac{87}{1.732} \approx 50.23 \ \text{m} $$

Step 4: Using the tangent formula for the final position (\( \theta_2 = 30^\circ \)):

$$ \tan 30^\circ = \frac{87}{x_2} $$ $$ \frac{1}{\sqrt{3}} = \frac{87}{x_2} $$ $$ x_2 = 87 \times \sqrt{3} $$

Step 5: Calculate \( x_2 \):

$$ x_2 = 87 \times 1.732 \approx 150.68 \ \text{m} $$

Step 6: Find the distance travelled by the balloon:

$$ \text{Distance} = x_2 - x_1 = 150.68 - 50.23 = 100.45 \ \text{m} $$

Result:

The distance travelled by the balloon during the interval is approximately 100.45 m.

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