Step 1

We are given the force ${F_1} = 400\;{\rm{N}}$ and ${F_2} = 125\;{\rm{N}}$.

We are asked to determine the magnitude and coordinate direction angles of the resultant force and also to sketch the vector on the coordinate system.

 
Step 2

Resolve the forces ${F_1}$ and ${F_2}$ into their x and y directions as shown in the figure below,

 
Step 3

From the above figure, we can write the force ${F_1}$ in the Cartesian vector form as follows:

\[\begin{array}{c} {F_1} = \left( {{F_1}\cos 45^\circ } \right)\hat i + \left( {{F_1}\cos 60^\circ } \right)\hat j - \left( {{F_1}\cos 60^\circ } \right)\hat k\\ {F_1} = {F_1}\left( {\cos 45^\circ \hat i + \cos 60^\circ \hat j - \cos 60^\circ \hat k} \right)\\ {F_1} = \left( {400\;{\rm{N}}} \right) \times \left[ {\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{2}\hat j - \frac{1}{2}\hat k} \right]\\ {F_1} = \left( {282.84\hat i + 200\hat j - 200\hat k} \right)\;{\rm{N}} \end{array}\]

Similarly, we can consider the force ${F_2}$ in the Cartesian vector form as follows:

\[\begin{array}{c} {F_2} = {F_2}\left( {\frac{4}{5}\cos 20^\circ \hat i - \frac{4}{5}\sin 20^\circ \hat j + \frac{3}{5}\hat k} \right)\\ {F_2} = \left( {125\;{\rm{N}}} \right) \times \left( {0.75\hat i - 0.27\hat j + 0.6\hat k} \right)\\ {F_2} = \left( {93.75\hat i - 33.75\hat j + 75\hat k} \right)\;{\rm{N}} \end{array}\]
 
Step 4

The resultant force in the Cartesian vector form can be calculated as follows:

\[\begin{array}{c} {F_R} = {F_1} + {F_2}\\ {F_R} = \left( {282.84\hat i + 200\hat j - 200\hat k} \right)\;{\rm{N}} + \left( {93.75\hat i - 33.75\hat j + 75\hat k} \right)\;{\rm{N}}\\ {F_R} = \left( {282.84\hat i + 93.75\hat i} \right)\;{\rm{N}} + \left( {200\hat j - 33.75\hat j} \right)\;{\rm{N}} + \left( { - 200\hat k + 75\hat k} \right)\;{\rm{N}}\\ {F_R} = \left( {376.59\hat i + 166.25\hat j - 125\hat k} \right)\;{\rm{N}} \end{array}\]

Now, to find the magnitude of the resultant force, we have:

\[\begin{array}{c} {F_R} = \sqrt {\left( {{F_R}} \right)_x^2 + \left( {{F_R}} \right)_y^2 + \left( {{F_R}} \right)_z^2} \\ {F_R} = \sqrt {{{\left( {376.59\;{\rm{N}}} \right)}^2} + {{\left( {166.25\;{\rm{N}}} \right)}^2} + {{\left( { - 125\;{\rm{N}}} \right)}^2}} \\ {F_R} = \sqrt {185.08 \times {{10}^3}} \;{\rm{N}}\\ {F_R} = 430.20\;{\rm{N}} \end{array}\]

Now, to show the coordinate system acting on the resultant force, the free-body diagram can be drawn as:

From the above figure, the coordinate direction angle of the resultant force along the x-axis can be calculated as follows:

\[\begin{array}{c} \cos \alpha = \frac{{{{\left( {{F_R}} \right)}_x}}}{{{F_R}}}\\ \alpha = {\cos ^{ - 1}}\left( {\frac{{376.59\;{\rm{N}}}}{{430.20\;{\rm{N}}}}} \right)\\ \alpha = {\cos ^{ - 1}}\left( {0.875} \right)\\ \alpha = 28.95^\circ \end{array}\]
 
Step 5

The coordinate direction angle of the resultant force along the y-axis can be calculated as follows:

\[\begin{array}{c} \cos \beta = \frac{{{{\left( {{F_R}} \right)}_y}}}{{{F_R}}}\\ \beta = {\cos ^{ - 1}}\left( {\frac{{166.25\;{\rm{N}}}}{{430.20\;{\rm{N}}}}} \right)\\ \beta = {\cos ^{ - 1}}\left( {0.386} \right)\\ \beta = 67.3^\circ \end{array}\]

Similarly, the coordinate direction angle of the resultant force along the z-axis can be calculated as follows:

\[\begin{array}{c} \cos \gamma = \frac{{{{\left( {{F_R}} \right)}_z}}}{{{F_R}}}\\ \gamma = {\cos ^{ - 1}}\left( {\frac{{ - 125\;{\rm{N}}}}{{430.20\;{\rm{N}}}}} \right)\\ \gamma = {\cos ^{ - 1}}\left( { - 0.290} \right)\\ \gamma = 106.85^\circ \end{array}\]