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Q14:

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 $30°$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Solution :

Given:

  • Height of the girl ($h_g$) = $1.2$ m.
  • Height of the balloon from the ground ($H$) = $88.2$ m.
  • Initial angle of elevation ($\theta_1$) = $60^\circ$.
  • Final angle of elevation ($\theta_2$) = $30^\circ$.

To Find:

The distance travelled by the balloon during the interval, let this be $d$.

A B C D 60° 30°

Step 1: Determine the effective height of the balloon from the girl's eye level.

Since the girl is $1.2$ m tall, the height of the balloon from her eye level ($h$) is the total height minus the girl's height.

$h = 88.2\text{ m} - 1.2\text{ m} = 87\text{ m}$.

Step 2: Analyze the first position of the balloon (Triangle 1).

Let the first position of the balloon be $C$ and the girl's eye position be $E$. Let the point on the ground directly below the balloon be $F$.

In the right-angled triangle formed by the eye level, the balloon, and the vertical line:

$\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{x_1}$

$\sqrt{3} = \frac{87}{x_1}$

$x_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3}\text{ m}$.

Step 3: Analyze the second position of the balloon (Triangle 2).

Let the second position of the balloon be $D$.

In the right-angled triangle formed by the eye level, the new balloon position, and the vertical line:

$\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{x_2}$

$\frac{1}{\sqrt{3}} = \frac{87}{x_2}$

$x_2 = 87\sqrt{3}\text{ m}$.

Step 4: Calculate the distance travelled by the balloon.

The distance travelled ($d$) is the difference between the two horizontal distances $x_2$ and $x_1$.

$d = x_2 - x_1$

$d = 87\sqrt{3} - 29\sqrt{3}$

$d = (87 - 29)\sqrt{3}$

$d = 58\sqrt{3}\text{ m}$.

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

$d = 58 \times 1.732 = 100.456\text{ m}$.

Final Answer: The distance travelled by the balloon is $58\sqrt{3}$ m (or approximately $100.46$ 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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