$ \text{Step 1: Given Data} $

The truss is subjected to:
${F_1} = 40\,{\rm{kN}}$ at E, ${F_2} = 50\,{\rm{kN}}$ at D
${F_3} = 30\,{\rm{kN}}$ at C, ${F_4} = 40\,{\rm{kN}}$ at B

Geometric Parameters:
Horizontal spacing ${d} = 2\,{\rm{m}}$
Vertical spacing ${h} = 1.5\,{\rm{m}}$

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

\begin{array}{c}
\tan \theta = \frac{h}{d} = \frac{1.5}{2} = 0.75 \\
\theta = \tan^{-1}(0.75) = 36.87^\circ
\end{array}

$ \text{Step 3: Reaction Force} $

\begin{array}{c}
\sum M_A = 0: \\
F_4(h) + F_3(2h) + F_1(d) – F_y(2d) = 0 \\
(40)(1.5) + (30)(3) + (40)(2) – F_y(4) = 0 \\
\hline
F_y = 57.5\,{\rm{kN}}\,\text{(Upwards)}
\end{array}

$ \text{Step 4: Force in HC} $

\begin{array}{c}
\sum M_D = 0: \\
F_1(d) – F_y(2d) – F_{HC}(h) = 0 \\
(40)(2) – (57.5)(4) – F_{HC}(1.5) = 0 \\
\hline
F_{HC} = -100\,{\rm{kN}}\,\text{(T)}
\end{array}

$ \text{Step 5: Force in DC} $

\begin{array}{c}
\sum F_x = 0: \\
-F_{BC} – F_{CD}\cos\theta = 0 \\
100 – 0.8F_{CD} = 0 \\
\hline
F_{CD} = 125\,{\rm{kN}}\,\text{(C)}
\end{array}

$ \text{Step 6: Force in HI} $

\begin{array}{c}
\sum F_y = 0: \\
F_y + F_{HI} + F_{CD}\sin\theta – F_1 – F_2 = 0 \\
57.5 + F_{HI} + 75 – 40 – 50 = 0 \\
\hline
F_{HI} = -42.5\,{\rm{kN}}\,\text{(T)}
\end{array}

$ \text{Final Results} $

\begin{array}{c}
\text{Member} & \text{Force (kN)} & \text{Type} \\
\hline
DC & 125 & \text{Compression (C)} \\
HC & 100 & \text{Tension (T)} \\
HI & 42.5 & \text{Tension (T)}
\end{array}