Assuming point K is held in equilibrium, by Newton's second law we have
• net horizontal force
[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]
• net vertical force
[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]
where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies
[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]
Solve the first equation for [tex]F_C[/tex].
[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)
Substitute this into the second equation and solve for [tex]F_B[/tex].
[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)
[tex]F_B = \dfrac57 F_A[/tex]
[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]
