Respuesta :
so.. checking the picture below
we can say y = y, or just do a substitution and end up with
[tex]\bf tan(72^o)=tan(30^o)(30+x)\implies 3.08x=\cfrac{30}{\sqrt{3}}+\cfrac{1}{\sqrt{3}}x \\\\\\ 3.08x-\cfrac{1}{\sqrt{3}}x=\cfrac{30}{\sqrt{3}}\implies 2.5x\approx 17.32\implies x\approx \cfrac{17.32}{2.5}[/tex]
once you know, how long is "x", then you can simply use the cosine of 72° to get "r"
thus [tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(72^o)=\cfrac{x}{r}\implies r=\cfrac{x}{cos(72^o)}[/tex]
we can say y = y, or just do a substitution and end up with
[tex]\bf tan(72^o)=tan(30^o)(30+x)\implies 3.08x=\cfrac{30}{\sqrt{3}}+\cfrac{1}{\sqrt{3}}x \\\\\\ 3.08x-\cfrac{1}{\sqrt{3}}x=\cfrac{30}{\sqrt{3}}\implies 2.5x\approx 17.32\implies x\approx \cfrac{17.32}{2.5}[/tex]
once you know, how long is "x", then you can simply use the cosine of 72° to get "r"
thus [tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(72^o)=\cfrac{x}{r}\implies r=\cfrac{x}{cos(72^o)}[/tex]

Answer:
[tex]r=21.45{\text{feet}[/tex]
Step-by-step explanation:
To find: The length of the wire.
Solution:
From the figure, using trigonometry, we have
[tex]tan72^{\circ}=\frac{y}{x}[/tex]
⇒[tex]y=xtan72^{\circ}[/tex] (1)
And, [tex]tan30^{\circ}=\frac{y}{30+x}[/tex]
⇒[tex]y=(30+x)tan30^{\circ}[/tex] (2)
Thus, from equation (1) and (2), we get
[tex]xtan72^{\circ}=(30+x)tan30^{\circ}[/tex]
⇒[tex]\frac{x}{30+x}=\frac{tan30^{\circ}}{tan72^{\circ}}[/tex]
⇒[tex]\frac{x}{30+x}=\frac{0.577}{3.077}[/tex]
⇒[tex]\frac{x}{30+x}=0.181[/tex]
⇒[tex]x=5.43+0.181x[/tex]
⇒[tex]x=6.63{\text{feet}[/tex]
Also, [tex]\frac{x}{r}=cos72^{\circ}[/tex]
⇒[tex]\frac{x}{cos72^{\circ}}=r[/tex]
⇒[tex]\frac{6.63}{0.309}=r[/tex]
⇒[tex]r=21.45{\text{feet}[/tex]
Therefore, the length of the wire is 21.45 feet.
