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in Trigonometry by (41.9k points)
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A wire of length 24 m had been tied with electric pole at an angle of elevation 30° with the ground. As it is covering a long distance, it was cut and tied at angle of elevation 60° with the ground. How much length of the wire was cut?

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Let AD be length of wire before cut = 24 m 

Let AC be length of wire after cut (AC) = x m 

Height of the electric pole = AB

Angle of elevation 

∠BDA = 30° 

∠BCA = 60° 

In right Δ ABD 

sin 30° = \(\frac{AB}{AD}\)

\(\frac{1}{2}=\frac{AB}{24}\) ⇒ 2AB = 24 ⇒ AB = 12m 

In right Δ ABC 

sin 60° = \(\frac{AB}{AC}\) ⇒ \(\frac{\sqrt{3}}{2}=\frac{12}{AC}\)

= √3 AC = 24 ⇒ AC = \(\frac{24}{\sqrt{3}}\)

= 8√3m = 8 × 1.732 = 13.856 m 

Length of the wire was cut 

= 24 – 13.856 = 10.144 m

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