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Some Applications Of Trigonometry
EXERCISE 9.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Step-by-Step Solution

Key Concept: Use the definition of sine in a right‑angled triangle: \(\sin \theta = \dfrac{\text{opposite side}}{\text{hypotenuse}}\). Here the height of the kite is the side opposite the given angle, and the string is the hypotenuse.
1. Draw a right‑angled triangle:
- The vertical side (opposite the angle) = height of the kite = $60\,\text{m}$.
- The angle between the string and the ground = $\theta = 60^{\circ}$.
- The string is the hypotenuse (let its length be $\ell$).

2. Apply the sine definition:
$$\sin 60^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{60}{\ell}$$

3. Evaluate $\sin 60^{\circ}$ using the known value $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$:
$$\frac{\sqrt{3}}{2} = \frac{60}{\ell}$$

4. Solve for $\ell$:
$$\ell = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}}$$
Rationalise the denominator:
$$\ell = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{120\sqrt{3}}{3}=40\sqrt{3}\,\text{m}$$

5. Numerical value (optional):
$$40\sqrt{3} \approx 40 \times 1.732 = 69.28\,\text{m}$$

Thus, the length of the string is $40\sqrt{3}\,\text{m}$ (approximately $69.3\,\text{m}$).

Correct Answer: Length of the string $= 40\sqrt{3}\ \text{m} \approx 69.3\ \text{m}$
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