\(P = (4N, 030°) ; R = (10N, 300°)\)
Let \(F = (a N, \theta)\) be the force that will keep the particle in equilibrium.
Then P + R + F = 0.
\(\begin{pmatrix} 4 \cos 30 \\ 4 \sin 30 \end{pmatrix} + \begin{pmatrix} 10 \cos 300 \\ 10 \sin 300 \end{pmatrix} + \begin{a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).
\(\begin{pmatrix} 4(\frac{\sqrt{3}}{2}) \\ 4(\frac{1}{2}) \end{pmatrix} + \begin{pmatrix} 10(\frac{1}{2}) \\ 10(-\frac{\sqrt{3}}{2}) \end{pmatrix} + \begin{pmatrix} a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
\(\begin{pmatrix} 2\sqrt{3} + 5 \\ 2 - 5\sqrt{3} \end{pmatrix} + \begin{pmatrix} a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \begin{pmatrix}\)
\(\implies 2\sqrt{3} + 5 + a \cos \theta = 0 \)
\(2\sqrt{3} + 5 = -a \cos \theta\)
\(2 - 5\sqrt{3} = -a \sin \theta\)
\(\tan \theta = \frac{2 - 5\sqrt{3}}{2\sqrt{3} + 5} = \frac{-6.66}{8.486}\)
\(\tan \theta = -0.7848 \implies \theta = -38.13°\)
= \(141.87°\)
\(-a \cos \theta = 8.486\)
\(-a \cos (141.87) = -a (-0.7866) = 8.486\)
\(0.7866a = 8.486 \implies a = \frac{8.486}{0.7866} = 10.79N\)
\(\therefore F = (10.79N, 141.87°)\)