With the 6N force acting in the direction 090°, the diagram is redrawn as :

Resultant = \(\begin{pmatrix} 4 \cos 65° \\ 4 \sin 65° \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 8 \cos 145° \\ 8 \sin 145° \end{pmatrix} + \begin{pmatrix} 4 \cos 245° \\ 4 \sin 245° \end{pmatrix} + \begin{pmatrix} 5 \cos 295° \\ 5 \sin 295° \end{pmatrix}\)
= \(\begin{pmatrix} 4 \cos 65 \\ 4 \sin 65 \end{pmatrix} + \begin{pmatrix} 6 \cos 90 \\ 6 \sin 90 \end{pmatrix} + \begin{pmatrix} -8 \cos 35 \\ 8 \sin 35 \end{pmatrix} + \begin{pmatrix} -4 \cos 65 \\ -4 \sin 65 \end{pmatrix} + \begin{pmatrix} 5 \cos 65 \\ -5 \sin 65 \end{pmatrix}\)
= \(\begin{pmatrix} 4 \times 0.4226 \\ 4 \times 0.9063 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix} + \begin{pmatrix} -8 \times 0.8192 \\ 8 \times 0.5736 \end{pmatrix} + \begin{pmatrix} -4 \times 0.4226 \\ -4 \times 0.9063 \end{pmatrix} + \begin{pmatrix} 5 \times 0.4226 \\ -5 \times 0.9063 \end{pmatrix}\)
= \(\begin{pmatrix} 1.6904 \\ 3.6252 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix} + \begin{pmatrix} -6.5536 \\ 4.5888 \end{pmatrix} + \begin{pmatrix} -1.6904 \\ -3.6252 \end{pmatrix} + \begin{pmatrix} 2.113 \\ -4.5315 \end{pmatrix}\)
= \(\begin{pmatrix} -4.4406 \\ 6.0573 \end{pmatrix}\)
\(|Resultant| = \sqrt{(-4.4406)^{2} + (6.0573)^{2}} = \sqrt{56.41}\)
= 7.5107N.
The magnitude of the resultant = 7.5N.
(b) Let \(\theta\) be the direction of the resultant.
\(\tan \theta = \frac{6.0573}{-4.4406} = - 1.364\)
\(\theta = -53.75° \)
= \(180° - 53.75° = 126.25°\) with the positive x- axis.