Expand each vector into their component forms:
[tex]\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N[/tex]
Similarly,
[tex]\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N[/tex]
[tex]\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N[/tex]
Then assuming the resultant vector [tex]\vec R[/tex] is the sum of these three vectors, we have
[tex]\vec R=\vec A+\vec B+\vec C[/tex]
[tex]\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N[/tex]
and so [tex]\vec R[/tex] has magnitude
[tex]\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N[/tex]
and direction [tex]\theta_R[/tex] such that
[tex]\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ[/tex]
