From Mechanics

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

\displaystyle F1=100\ \text{N} and \displaystyle \alpha 1=30{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}1=100\cos 30{}^\circ \mathbf{i}+100\sin 30{}^\circ \mathbf{j}=100\times 0\textrm{.}866\mathbf{i}+100\times 0\textrm{.}5\mathbf{j} \\ & =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\ \end{align}

\displaystyle F2=90\ \text{N} and \displaystyle \alpha 2=180{}^\circ-70{}^\circ=110{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}2=90\cos 110 {}^\circ \mathbf{i}+90\sin 110 {}^\circ \mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\ & =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\ \end{align}

\displaystyle F3=80\ \text{N} and \displaystyle \alpha 3=-90{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}3=80\cos \left(-90{}^\circ \right) \mathbf{i}+80\sin \left(-90{}^\circ \right) \mathbf{j}=80\times 0\mathbf{i}+80\times \left(-1\right)\mathbf{j} \\ & =-80\mathbf{j}\ \text{N}\\ \end{align}