Question

A wire of length 18 m had been tied with electric pole at an angle of elevation 30º with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60º with the ground. How much length of the wire was cut?​

Answer 1

Answer:

I hope the answer is 36 .

Step-by-step explanation:

a wire length - 18 m

Answer 2

Answer:

$\tt{\implies AD = 18m}$

$\tt{\implies In\:∆ABD,}$

$\tt{\implies ∠ABD = 30°}$

$\tt{\implies ∴ sin30° = \dfrac{AB}{AD} implies AB = ADsin30° = 9m}$

$\tt{\implies In\:∆ABC,}$

$\tt{\implies ∠ABC = 90°\:&∠ACB = 60°}$

$\tt{\implies sin60° = \dfrac{AB}{AC} = \dfrac{9}{AC} }$

$\tt{\implies \dfrac{\dsqrt{3}}{2} = \drac{9}{AC}\:» AC = \dfrac{9×2}{\sqrt{3}\:6\:\sqrt{3}m}$

$\tt{\implies ∴ wire\:length\:cut\:down = AB-AC = (18-6\sqrt{3})m = 7.6 m}$