Discussion
A kite is flying at a height of 75 m above the ground, attached to a string inclined at \( 60^{\circ} \) to the horizontal. The length of the string is:
Answer: B
The string is the hypotenuse and the kite's height is opposite the \( 60^{\circ} \) inclination, so sine applies.
\[ \sin 60^{\circ} = \frac{75}{L} \quad\Longrightarrow\quad \frac{\sqrt{3}}{2} = \frac{75}{L} \]
\[ L = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3} \approx 86.6 \text{ m} \]
Rationalising \( \tfrac{150}{\sqrt{3}} \) into \( 50\sqrt{3} \) is what matches it to the listed form. The string must be longer than the 75 m height, which rules out the two divided options immediately.
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