Question

A kite is at a height of 30 m from the earth and its string makes an angle 60\(^\circ\) with the earth. Then the length of the string is

• A. \(30\sqrt{2}\) m
• B. \(35\sqrt{3}\) m
• C. \(20\sqrt{3}\) m
• D. \(45\sqrt{2}\) m

Hint:

It's good practice to rationalize the denominator in your final answer. This involves removing any square roots from the bottom of the fraction.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem can also be modeled with a right-angled triangle. The height of the kite is the opposite side, the length of the string is the hypotenuse, and the angle the string makes with the ground is the angle of elevation.

Step 2: Key Formula or Approach:
In a right-angled triangle, the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse.
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \]

Step 3: Detailed Explanation:
The height of the kite (opposite side) = 30 m.
Let the length of the string be \(l\) (hypotenuse).
The angle with the earth, \(\theta\), is \(60^\circ\).
Using the sine ratio:
\[ \sin 60^\circ = \frac{30}{l} \] We know the standard value \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).
\[ \frac{\sqrt{3}}{2} = \frac{30}{l} \] To solve for \(l\), we can cross-multiply:
\[ l \times \sqrt{3} = 30 \times 2 \] \[ l\sqrt{3} = 60 \] \[ l = \frac{60}{\sqrt{3}} \] To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{3}\):
\[ l = \frac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{60\sqrt{3}}{3} \] \[ l = 20\sqrt{3} \text{ m} \]

Step 4: Final Answer:
The length of the string is \(20\sqrt{3}\) m. This matches option (C).