SOLUTION
From the question, we are given
[tex]\begin{gathered} F_1=(11,000,5000) \\ F_2=(14,500,-8,000) \end{gathered}[/tex]
Now, to find the angle between the ropes, we will take the dot product of two vectors. This is given as
[tex]\begin{gathered} a.b=|a||b|cos\theta \\ so,\text{ we have } \\ cos\theta=\frac{a.b}{|a||b|} \end{gathered}[/tex]
But writing the forces in vector form, we have
[tex]\begin{gathered} F_1=11,000i+5000j \\ F_2=14,500i-8,000j \end{gathered}[/tex]
Applying the formula above, we have
[tex]\begin{gathered} cos\theta=\frac{a.b}{|a||b|} \\ cos\theta=\frac{(11,000)(14,500)+(5000)(-8000)}{\sqrt{11,000^2+5000^2)\sqrt{14500^2+(-8000)^2}}} \\ cos\theta=\frac{159,500,000-40,000,000}{\sqrt{121,000,000+25,000,000}\sqrt{210,250,000+64,000,000}} \\ cos\theta=\frac{119,500,000}{\sqrt{146,000,000}\sqrt{274,250,000}} \\ cos\theta=\frac{119,500,000}{12,083.04579\times16560.4952} \\ cos\theta=\frac{119,500,000}{200,101,221.806675} \\ cos\theta=0.5971977 \end{gathered}[/tex]
Now, theta becomes
[tex]\begin{gathered} \theta=cos^{-1}0.5971977 \\ \theta=53.3305 \end{gathered}[/tex]
hence the answer is option B, 53 degrees
