$ \text{Step 1: Given Data} $
We are given the force ${F_R} = {F_1} + {F_2}$, where ${F_1} = 250\;{\rm{lb}}$ and ${F_2} = 375\;{\rm{lb}}$.

We are asked to determine the magnitude and direction of the resultant force.

$ \text{Step 2: Diagram of Forces} $
The following is the diagram to calculate the projected component of the force.

To find the magnitude of the resultant force, we will use the formula:
\begin{array}{c}
{F_R} = \sqrt{{F_1}^2 + {F_2}^2 – 2{F_1}{F_2}\cos \theta}
\end{array}

On plugging the values into the above relation, we get:

\begin{array}{l}
{F_R} = \sqrt{{(250\;{\rm{lb}})}^2 + {(375\;{\rm{lb}})}^2 – 2\left( {250\;{\rm{lb}}} \right)\left( {375\;{\rm{lb}}} \right)\cos 75^\circ} \\
{F_R} = \sqrt{\left( 62500\;{\rm{lb}^2} \right) + \left( 140625\;{\rm{lb}^2} \right) – \left( 48528.57\;{\rm{lb}^2} \right)} \\
{F_R} = \sqrt{203125 – 48528.57} = 393.2\;{\rm{lb}}
\end{array}

$ \text{Step 3: Finding Direction of Resultant Force} $
To find the direction of the resultant force, we will use the law of sines:
\begin{array}{c}
\frac{{F_R}}{{\sin 75^\circ}} = \frac{{F_1}}{{\sin \phi}}
\end{array}

On plugging the values into the above relation, we get:

\begin{array}{c}
\frac{{393.2\;{\rm{lb}}}}{{\sin 75^\circ}} = \frac{{250\;{\rm{lb}}}}{{\sin \phi}} \\
\sin \phi = \frac{{250\;{\rm{lb}}}}{{393.2\;{\rm{lb}} \cdot \sin 75^\circ}} \\
\phi = \sin^{-1}\left( \frac{{250\;{\rm{lb}}}}{{407.07\;{\rm{lb}}}} \right) \\
\phi = 37.9^\circ
\end{array}

The direction of the resultant force is calculated as:

\begin{array}{l}
{\theta_R} = 360^\circ – 45^\circ + \phi \\
{\theta_R} = 360^\circ – 45^\circ + 37.9^\circ \\
{\theta_R} = 352.9^\circ \; \text{(CCW)}
\end{array}