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Some Applications Of Trigonometry
EXERCISE 9.1
CBSE_NCERT_TEXTBOOK
Grade 10
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° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.
Step-by-Step Solution
Key Concept: The vertical distance between the girl’s eyes and the balloon remains constant. Using the definition of tangent (tan θ = opposite/adjacent) for the two positions of the balloon, compute the horizontal distances from the girl to the balloon at the two instants and subtract them.
1. Determine the constant vertical distance\
Height of balloon = 88.2 m\
Height of girl = 1.2 m\
Therefore, vertical distance from the girl's eyes to the balloon \(h = 88.2 - 1.2 = 87.0\) m.
2. Let the horizontal distances from the girl to the balloon at the two instants be \(x_1\) (when the angle of elevation is 60°) and \(x_2\) (when the angle becomes 30°).
3. Apply the tangent definition for each position:
- For \(\theta_1 = 60^{\circ}\):
$$\tan 60^{\circ} = \frac{h}{x_1} \quad\Rightarrow\quad \sqrt{3} = \frac{87}{x_1}$$
Hence \(x_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3}\) m.
- For \(\theta_2 = 30^{\circ}\):
$$\tan 30^{\circ} = \frac{h}{x_2} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{87}{x_2}$$
Hence \(x_2 = 87\sqrt{3}\) m.
4. Find the horizontal distance travelled by the balloon:
$$\text{Distance} = x_2 - x_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\ \text{m}.$$
5. Numerical value (optional):
$$58\sqrt{3} \approx 58 \times 1.732 = 100.5\ \text{m}.$$
Thus the balloon travels a horizontal distance of \(58\sqrt{3}\) metres (approximately 100.5 m) during the interval.
Height of balloon = 88.2 m\
Height of girl = 1.2 m\
Therefore, vertical distance from the girl's eyes to the balloon \(h = 88.2 - 1.2 = 87.0\) m.
2. Let the horizontal distances from the girl to the balloon at the two instants be \(x_1\) (when the angle of elevation is 60°) and \(x_2\) (when the angle becomes 30°).
3. Apply the tangent definition for each position:
- For \(\theta_1 = 60^{\circ}\):
$$\tan 60^{\circ} = \frac{h}{x_1} \quad\Rightarrow\quad \sqrt{3} = \frac{87}{x_1}$$
Hence \(x_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3}\) m.
- For \(\theta_2 = 30^{\circ}\):
$$\tan 30^{\circ} = \frac{h}{x_2} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{87}{x_2}$$
Hence \(x_2 = 87\sqrt{3}\) m.
4. Find the horizontal distance travelled by the balloon:
$$\text{Distance} = x_2 - x_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\ \text{m}.$$
5. Numerical value (optional):
$$58\sqrt{3} \approx 58 \times 1.732 = 100.5\ \text{m}.$$
Thus the balloon travels a horizontal distance of \(58\sqrt{3}\) metres (approximately 100.5 m) during the interval.
Correct Answer: 58√3 m \;(≈ 100.5 m)
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