Solution 4.6a
From Mechanics
\displaystyle \begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}
\displaystyle F1=40\ \text{N}
and
\displaystyle \alpha 1=20{}^\circ
this gives
\displaystyle \begin{align}
& \mathbf{F}1=40\cos 20{}^\circ
\mathbf{i}+40\sin 20{}^\circ
\mathbf{j}=40\times 0\textrm{.}94\mathbf{i}+40\times 0\textrm{.}342\mathbf{j} \\
& =37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\
\end{align}
\displaystyle F2=50\ \text{N} and \displaystyle \alpha 2=-30{}^\circ this gives
\displaystyle \begin{align}
& \mathbf{F}2=50\cos \left( -30
\right) {}^\circ
\mathbf{i}+50\sin \left( -30 \right) {}^\circ
\mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\
& =43\textrm{.}3\mathbf{i}-25\mathbf{j} \ \text{N}\\
\end{align}
