Question

question from some applications of trigonometry​

Answer 1

Let AB=10= Height of Pole

and AD be the length of the wire

From △ABD,

$sin \: 45 = \frac{ab}{ad} = > \frac{1}{ \sqrt{2} } = \frac{10}{ad } \\ ad = 10 \sqrt{2 } = > 10 \times 1.414 \\ = \geqslant 14.14metres \: \leqslant =$

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