$ \text{Step 1: Given Data} $
We are given the value of force $F = 350 \, {\rm{lb}}$.

We are asked to estimate the magnitudes of the two components of $F $.

$ \text{Step 2: Force Conversion} $
The relation to convert the force from pounds into newtons is:

Since $1 \, {\rm{lb}} = {\rm{4.448}} \, {\rm{N}}$, then
\begin{array}{c}
350 \, {\rm{lb}} = \left( 350 \, {\rm{lb}} \times \frac{{4.448 \, {\rm{N}}}}{{1 \, {\rm{lb}}}} \right) \\
= 1556.8 \, {\rm{N}}
\end{array}

The following is the free body diagram to calculate the magnitude of force.

To find the magnitude of force $ F_{AB} $, we will use the law of sines:
\begin{equation}
\frac{{F_{AB}}}{{\sin 60^\circ}} = \frac{F}{{\sin 75^\circ}}
\end{equation}

$ \text{Step 3: Calculate Magnitude of } F_{AB} $
Plugging the known values into the above equation gives:
\begin{array}{c}
\frac{{F_{AB}}}{{\sin 60^\circ}} = \frac{{1556.8 \, {\rm{N}}}}{{\sin 75^\circ}} \\
F_{AB} = 1556.8 \, {\rm{N}} \times \frac{\sin 60^\circ}{\sin 75^\circ} \\
F_{AB} \approx 1395.8 \, {\rm{N}}
\end{array}

$ \text{Step 4: Free Body Diagram for Link AC} $
The following is the free body diagram to calculate the magnitude of force.

To find the magnitude of force $ F_{AC} $, we will use the law of sines:
\begin{equation}
\frac{{F_{AC}}}{{\sin 45^\circ}} = \frac{F}{{\sin 75^\circ}}
\end{equation}

$ \text{Step 5: Calculate Magnitude of } F_{AC} $
Plugging the known values into the above equation gives:
\begin{array}{c}
\frac{{F_{AC}}}{{\sin 45^\circ}} = \frac{{1556.8 \, {\rm{N}}}}{{\sin 75^\circ}} \\
F_{AC} = 1556.8 \, {\rm{N}} \times \frac{\sin 45^\circ}{\sin 75^\circ} \\
F_{AC} \approx 1139.6 \, {\rm{N}}
\end{array}