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