From Mechanics

\displaystyle \begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}

\displaystyle F1=7\ \text{N} and \displaystyle \alpha 1=35{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}1=7\cos 35{}^\circ \mathbf{i}+7\sin 35{}^\circ \mathbf{j}=7\times 0\textrm{.}819\mathbf{i}+7\times 0\textrm{.}574\mathbf{j} \\ & =5\textrm{.}73\mathbf{i}+4\textrm{.}02\mathbf{j}\ \text{N}\\ \end{align}

\displaystyle F2=90 \text{N} and \displaystyle \alpha 2=90{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}2=9\cos 90 {}^\circ \mathbf{i}+9\sin 90 {}^\circ \mathbf{j}=9\times 0\mathbf{i}+9\times 1\textrm{.}0\mathbf{j} \\ & =9\textrm{.}0\mathbf{j} \ \text{N}\\ \end{align}

\displaystyle F3=6\ \text{N} and \displaystyle \alpha 3=-180{}^\circ+72{}^\circ=-108{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}3=6\cos \left(-108{}^\circ \right) \mathbf{i}+6\sin \left(-108{}^\circ \right) \mathbf{j}=6\times \left(-0\textrm{.}309\right)\mathbf{i}+6\times \left(-0\textrm{.}951\right)\mathbf{j} \\ & =-1\textrm{.}85\mathbf{i}-5\textrm{.}71\mathbf{j} \ \text{N}\\ \end{align}

\displaystyle F4=8\ \text{N} and \displaystyle \alpha 3=-25{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}4=8\cos \left(-25{}^\circ \right) \mathbf{i}+8\sin \left(-25{}^\circ \right) \mathbf{j}=8\times 0\textrm{.}906\mathbf{i}+8\times \left(-0\textrm{.}423\right)\mathbf{j} \\ & =7\textrm{.}25\mathbf{i}-3\textrm{.}38\mathbf{j} \ \text{N}\\ \end{align}

The resultant \displaystyle \mathbf{R} is the sum of these four forces.

\displaystyle \begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}

Then if \displaystyle R is the magnitude of this resultant

\displaystyle \begin{align} & {{R\ }^{2}}={{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}} \\ & R=\sqrt{{{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}}}=11\textrm{.}8\ \text{N} \\ \end{align}