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Answer :
Let OA be the horizontal ground, and let K be the position of the kite at height 60 m above the ground. Let the length of the string OK be x metres. It is given \( ∠ \ KOA \ = \ 60º \)
In \( ∆ \ AOK \), we have
\( \frac{AK}{OK} \ = \ sin60º \ => \ \frac{60}{x} \ = \ \frac{ \sqrt{3}}{2} \)
\( x \ = \ \frac{120}{ \sqrt{3}} \ = \ 40 \sqrt{3} \)
\(\therefore \) the length of the string is \( 40 \sqrt{3} \) m