From Mechanics

From the previous part we have that

\displaystyle \mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}

Thus the magnitude \displaystyle Q is given by

\displaystyle Q{}^{2}={{\left( -292 \right)}^{2}}+{{\left( -62 \right)}^{2}}=85300+3840=89140

or

\displaystyle Q=299 \text{ N}

and

\displaystyle \tan \alpha =\frac{62\textrm{.}9}{292}=0\textrm{.}215

giving

\displaystyle \alpha ={{12\textrm{.}2}^{\circ }}