$ \text{Step 1: Given Data} $
We are given the value of resultant force ${{\bf{F}}_R} = {{\bf{F}}_1} + {{\bf{F}}_2}$.

We are asked to estimate the magnitudes of the resultant force.

$ \text{Step 2: Diagram from Parallelogram Law} $
The following is the diagram from the parallelogram law of addition.

The diagram of the resultant force is as follows:

$ \text{Step 3: Law of Cosines and Law of Sines} $
The relation from the law of cosines is,
\begin{array}{c}
{{\bf{F}}_R} = \sqrt {{{\bf{F}}_1}^2 + {{\bf{F}}_2}^2 – 2{{\bf{F}}_1}{{\bf{F}}_2}\cos \theta ‘}
\end{array}

Plugging in the known values in the above relation:

\begin{array}{l}
{{\bf{F}}_R} = \sqrt {{{\left( {4\;{\rm{N}}} \right)}^2} + {{\left( {6\;{\rm{N}}} \right)}^2} – 2\left( {4\;{\rm{N}}} \right)\left( {6\;{\rm{N}}} \right)\cos \left( {105^\circ } \right)} \\
{{\bf{F}}_R} = \sqrt {52 + 12.42} \;{\rm{N}}\\
{{\bf{F}}_R} = 8.026\;{\rm{N}}
\end{array}

To find the magnitude of force, we will use the law of sines:
\begin{array}{c}
\frac{{\sin \theta }}{{{{\bf{F}}_2}}} = \frac{{\sin 105^\circ }}{{{{\bf{F}}_R}}}
\end{array}

Plugging the known values in the above relation:

\begin{array}{c}
\frac{{\sin \theta }}{{6\;{\rm{N}}}} = \frac{{\sin 105^\circ }}{{8.026\;{\rm{N}}}}\\
\theta = 46.22^\circ
\end{array}

$ \text{Step 4: Direction of Resultant Force} $
To find the direction of the resultant force, we will use the following relation:
\begin{array}{c}
\phi = \theta – {\theta _R}
\end{array}

Plugging the known values in the above relation:

\begin{array}{l}
\phi = \left( {46.22^\circ } \right) – \left( {45^\circ } \right)\\
\phi = 1.22^\circ
\end{array}