$ \text{Step 1: Given Data} $
We are given the force ${F_R} = 500\,{\rm{N}}$ and ${F_1} = 700\,{\rm{N}}$.

We are asked to determine the magnitude of force ${\bf{F}}$ and its direction ${\bf{\theta }}$.

$ \text{Step 2: Diagram of Vector Components} $
The following is the diagram of the vector component.

Vector Diagram

The following is the diagram of the vector component tail to tail.

Diagram of the vector component tail to tail

$ \text{Step 3: Calculate the Magnitude of Force} $
To find the magnitude of force, we will use the relation from the law of cosines.

\begin{array}{c}
{\bf{F}} = \sqrt {{F_1}^2 + {F_R}^2 – \left( 2 {F_1} {F_R} \times \cos 105^\circ \right)}
\end{array}

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

\begin{array}{c}
{\bf{F}} = \sqrt{{(700\,{\rm{N}})}^2 + {(500\,{\rm{N}})}^2 – \left( 2 \times 500\,{\rm{N}} \times 700\,{\rm{N}} \times \cos 105^\circ \right)} \\
{\bf{F}} = \sqrt{921173.33}\, {\rm{N}} \\
{\bf{F}} = 959.7\,{\rm{N}}
\end{array}

To find the direction of the force, we will use the relation from the law of sines.

\begin{array}{c}
\frac{{\sin \left( {90^\circ + {\bf{\theta }}} \right)}}{{{F_1}}} = \frac{{\sin 105^\circ}}{{\bf{F}}}
\end{array}

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

\begin{array}{c}
\frac{{\sin \left( {90^\circ + {\bf{\theta }}} \right)}}{{700\,{\rm{N}}}} = \frac{{\sin 105^\circ}}{{959.7\,{\rm{N}}}} \\
\sin \left( {90^\circ + {\bf{\theta }}} \right) = \frac{{700\,{\rm{N}} \times \sin 105^\circ}}{{959.7\,{\rm{N}}}} \\
\left( {90^\circ + {\bf{\theta }}} \right) = {\sin^{-1}}\left( {0.704} \right) \\
{\bf{\theta }} = 45.26^\circ
\end{array}