Step 1

We are given the forces ${{\bf{F}}_{\bf{1}}}$, ${{\bf{F}}_{\bf{2}}}$ and ${{\bf{F}}_{\bf{3}}}$ as $400\;{\rm{N}}$, $200\;{\rm{N}}$ and $300\;{\rm{N}}$ respectively.

We are asked to determine the magnitude and direction of the resultant force $\left( {{F_R}} \right)$ measured counter-clockwise from the +x-axis.

 
Step 2

By using parallelogram law of addition, draw the free-body diagram of the resultant force $\left( {F'} \right)$ of the two forces ${F_2}$ and ${F_3}$ acting in the system as shown in the figure below,

To find the magnitude of the resultant force $\left( {F'} \right)$, we need to apply the parallelogram law of addition as follows:

\[\begin{array}{c} F' = \sqrt {{{\left( {{F_2}} \right)}^2} + {{\left( {{F_3}} \right)}^2} - 2{F_2}{F_3}\cos \left( {30^\circ } \right)} \\ F' = \sqrt {{{\left( {200\;{\rm{N}}} \right)}^2} + {{\left( {300\;{\rm{N}}} \right)}^2} - 2 \times \left( {200\;{\rm{N}}} \right) \times \left( {300\;{\rm{N}}} \right) \times 0.866} \\ F' = \sqrt {26.08 \times {{10}^3}} \;{\rm{N}}\\ F' = 161.49\;{\rm{N}} \end{array}\]
 
Step 3

To find the direction of the resultant force $\left( {F'} \right)$ of the two forces ${F_2}$ and ${F_3}$, the free-body diagram can be drawn as:

By applying the sine rule, we have,

\[\begin{array}{c} \frac{{\sin 30^\circ }}{{F'}} = \frac{{\sin \theta '}}{{{F_2}}}\\ \frac{{0.5}}{{161.49\;{\rm{N}}}} = \frac{{\sin \theta '}}{{200\;{\rm{N}}}}\\ \sin \theta ' = 0.619\\ \theta ' = {\sin ^{ - 1}}\left( {0.619} \right)\\ \theta ' = 38.24^\circ \end{array}\]
 
Step 4

By using parallelogram law of addition, draw the free-body diagram of the resultant force $\left( {{F_R}} \right)$ of the two forces $F'$ and ${F_1}$ acting in the system as shown in the figure below,

From the geometry, we have:

\[\begin{array}{c} \theta '' = 90^\circ - 30^\circ - 38.24^\circ \\ \theta '' = 21.76^\circ \end{array}\]

To find the magnitude of the resultant force $\left( {{F_R}} \right)$, we need to apply the parallelogram law of addition as follows:

\[\begin{array}{c} {F_R} = \sqrt {{{\left( {F'} \right)}^2} + {{\left( {{F_1}} \right)}^2} - 2F'{F_1}\cos \theta ''} \\ {F_R} = \sqrt {{{\left( {161.49\;{\rm{N}}} \right)}^2} + {{\left( {400\;{\rm{N}}} \right)}^2} - 2 \times \left( {161.49\;{\rm{N}}} \right) \times \left( {400\;{\rm{N}}} \right) \times \cos \left( {21.76^\circ } \right)} \\ {F_R} = 257.07\;{\rm{N}} \approx {\rm{257}}\;{\rm{N}}\\ {F_R} = 257\;{\rm{N}} \end{array}\]
 
Step 5

To find the direction of the resultant force $\left( {{F_R}} \right)$ of the two forces $F'$ and ${F_1}$, the free-body diagram can be drawn as:

By applying the sine rule, we have,

\[\begin{array}{c} \frac{{\sin \theta ''}}{{{F_R}}} = \frac{{\sin \theta }}{{F'}}\\ \frac{{\sin \left( {21.76^\circ } \right)}}{{257\;{\rm{N}}}} = \frac{{\sin \theta }}{{161.49\;{\rm{N}}}}\\ \sin \theta = 0.23\\ \theta = {\sin ^{ - 1}}\left( {0.23} \right)\\ \theta = 13.29^\circ \end{array}\]

As we have to measured counter-clockwise direction of the resultant force from the positive x-axis then,

\[\begin{array}{c} \phi = 90^\circ + 60^\circ + \theta \\ \phi = 90^\circ + 60^\circ + 13.29^\circ \\ \phi = 163.29^\circ \end{array}\]