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for Solution 4.6a

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<math>\begin{align}
& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}  
\end{align}</math>


<math>F1=40\ \text{N}</math>
and 
<math>\alpha 1=20{}^\circ
</math>
this gives 


<math>\begin{align}
& \mathbf{F}1=40\cos 20{}^\circ
\mathbf{i}+40\sin 20{}^\circ
\mathbf{j}=40\times 0.94\mathbf{i}+40\times 0.342\mathbf{j} \\ 
& =37.6\mathbf{i}+13.7\mathbf{j} \\ 
\end{align}</math>



<math>F2=50\ \text{N}</math>
and 
<math>\alpha 2=-30{}^\circ
</math>
this gives 


<math>\begin{align}
& \mathbf{F}2=50\cos \left( -30
\right) {}^\circ
\mathbf{i}+50\sin \left( -30 \right) {}^\circ
\mathbf{j}=50\times 0.866\mathbf{i}-50\times 0.50\mathbf{j} \\ 
& =43.3\mathbf{i}+25\mathbf{j} \\ 
\end{align}</math>

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