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°. Assuming that there is no slack in the string, the length of the string is

• A. \(40\sqrt3 \,\,m\)
• B. \(60\sqrt3 \,\,m\)
• C. \(120\sqrt3 \,\,m\)
• D. \(30\sqrt3 \,\,m\)

Answer 1

The Correct Option is A

Step 1: Represent the problem using trigonometry.

Let the length of the string be \( L \). The height of the kite (60 m) is the opposite side, and the string forms the hypotenuse of a right triangle. Using the sine function:

\[ \sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{60}{L}. \]

Step 2: Substitute the value of \( \sin 60^\circ \).

The value of \( \sin 60^\circ \) is \( \frac{\sqrt{3}}{2} \). Substituting this into the equation:

\[ \frac{\sqrt{3}}{2} = \frac{60}{L}. \]

Step 3: Solve for \( L \).

Rearrange the equation to isolate \( L \):

\[ L = \frac{60}{\frac{\sqrt{3}}{2}} = \frac{60 \cdot 2}{\sqrt{3}} = \frac{120}{\sqrt{3}}. \]

Rationalize the denominator:

\[ L = \frac{120}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3}. \]

Final Answer: The length of the string is \( \mathbf{40\sqrt{3} \, \text{m}} \), which corresponds to option \( \mathbf{(1)} \).