$ \text{Step 1: Problem Statement and Given Data} $

A truss structure is subjected to five vertical loads of $1500\,{\rm{lb}}$ each at points B, C, D, E, and F. The geometric parameters are:
\begin{array}{c}
{EF} = 3\,{\rm{ft}} \\
{GE} = 4\,{\rm{ft}}
\end{array}

$ \text{Step 2: Free Body Diagram and Objective} $

Construct the free body diagram for section EFG to analyze:

\begin{array}{c}
\text{Member HG: } F_{HG} \\
\text{Member HE: } F_{HE} \\
\text{Member DE: } F_{DE}
\end{array}

$ \text{Step 3: Geometric Analysis and Angle Calculation} $

Determine the angle $\theta$ between member HE and horizontal:
\begin{array}{c}
\tan \theta = \frac{GE}{EF} = \frac{4\,{\rm{ft}}}{3\,{\rm{ft}}} \\
\theta = \tan^{-1}\left(\frac{4}{3}\right) \\
\hline
\theta \approx 53.1^\circ
\end{array}

$ \text{Step 4: Force in Member HG (Moment Equilibrium about E)} $

\begin{array}{c}
\sum M_E = 0 \\
F_{HG} \times GE – W_F \times EF = 0 \\
F_{HG} \times 4\,{\rm{ft}} – 1500\,{\rm{lb}} \times 3\,{\rm{ft}} = 0 \\
4F_{HG} = 4500\,{\rm{lb}\cdot{\rm{ft}}} \\
\hline
F_{HG} = 1125\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 5: Force in Member HE (Vertical Equilibrium)} $

\begin{array}{c}
\sum F_y = 0 \\
F_{HE} \sin \theta – W_E – W_F = 0 \\
F_{HE} \sin(53.1^\circ) – 1500\,{\rm{lb}} – 1500\,{\rm{lb}} = 0 \\
F_{HE} = \frac{3000\,{\rm{lb}}}{\sin(53.1^\circ)} \\
\hline
F_{HE} \approx 3751.5\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 6: Force in Member DE (Horizontal Equilibrium)} $

\begin{array}{c}
\sum F_x = 0 \\
-F_{HG} – F_{HE} \cos \theta – F_{DE} = 0 \\
-1125\,{\rm{lb}} – 3751.5\,{\rm{lb}} \times \cos(53.1^\circ) – F_{DE} = 0 \\
F_{DE} = -1125\,{\rm{lb}} – 2252.5\,{\rm{lb}} \\
\hline
F_{DE} = -3377.5\,{\rm{lb}}\,(\rm{C})
\end{array}

$ \text{Step 7: Verification and Summary} $

\begin{array}{c}
\text{Moment Equilibrium: } \sum M_E = 0 \quad \checkmark \\
\text{Force Balance: } \sum F_x = \sum F_y = 0 \quad \checkmark \\
\hline
\text{Final Results:} \\
\text{HG: } 1125\,{\rm{lb}}\,(\rm{T}) \\
\text{HE: } 3751.5\,{\rm{lb}}\,(\rm{T}) \\
\text{DE: } 3377.5\,{\rm{lb}}\,(\rm{C})
\end{array}

$ \text{Step 8: Conclusion} $

\begin{array}{c}
\text{The truss design satisfies equilibrium conditions with:} \\
\text{• HG in tension ($1125\,{\rm{lb}}$)} \\
\text{• HE in tension ($3751.5\,{\rm{lb}}$)} \\
\text{• DE in compression ($3377.5\,{\rm{lb}}$)}
\end{array}