A) \( \Large 8\sqrt{3} \) m |
B) 8m |
C) \( \Large 6\sqrt{3} \) m |
D) 6m |
A) \( \Large 8\sqrt{3} \) m |
So, in \( \Large \triangle ADE\)
\( \Large \sin 60 ^{\circ} = \frac{DE}{AD} \)
\( \Large \left[ AD = length\ of\ wire \right] \)
=> \( \Large AD = \frac{DE}{\sin 60 ^{\circ} } = \frac{12}{\frac{\sqrt{3}}{2}} \)
= \( \Large \frac{12 \times 2}{\sqrt{3}} \)
\( \Large \left[ on\ multiply\ and\ divide\ by\ \sqrt{3} \right] \)
= \( \Large \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3} } = \frac{24\sqrt{3}}{3} = 8\sqrt{3} m \)