Procedure

Start with a joint with two unknowns, in this case $A$.  Draw a free body diagram of the joint and solve for the two unknowns. Move to an adjacent joint with two unknowns and repeat.  Continue in this manner until all the unknowns are found.

When drawing free body diagrams, the force of a member acts along the axis of the member.  A force pushing towards the joint indicates compression, and a force pulling away indicates tension.  An incorrect assumed direction will result in a negative value.

After all the forces in the members are found you can check your answer by calculating the external reactions at $D$ and $E$, and verifying that  joints $D$ and $E$ are in equilibrium.

$\begin{align}A\!: \Sigma F_y &= 0 &  A\!: \Sigma F_x &= 0   \\AB_y &= A&      AC&= AB_x    \\AB \sin 30° &= A  &    &= AB \cos 30°        \\AB &= \dfrac{\var{disp(A)}}{\sin 30°} &   &= \var{disp(AB)} \cos 30°                  \\&= \var{siground(AB,4)}  \text{ (T)} & &= \var{siground(AC,4)}  \text{ (C)}\end{align}$

$\begin{align}B\!: \Sigma F_{y'} &= 0& B\!: \Sigma F_{x'}&= 0\\BC &= B_{y'}&BD&= B_{x'} + AB \\ &= B \cos 30° & &= B \sin 30°+ AB\\ &= {\var{disp(B)}} \cos 30°&&= \var{disp(B)} \sin 30°+ {AB}\\&= \var{siground(BC,4)}  \text{ (C)}& &= \var{siground(BD,4)}  \text{ (T)}\ \\\end{align}$

$\begin{align}C\!: \Sigma {F_y} &= 0 &         C\!: \Sigma {F_x} &= 0\\            -CD_y-BC_{y} -C &=0      &   CE + CD_x &= BC_x + AC\\    CD \sin 60°&= -(BC \sin 60° +C)&    CE &= AC + BC_x - CD_x\\CD&= -\frac{{\var{disp(BC)}} \sin 60° + \var{C}}{\sin 60°} & &= \var{disp(AC)} + \var{disp(BC)} \cos 60° -(\var{disp(-CD)}) \cos 60° \\  &= \var{-disp(CD)}  \text{ (C)}  & &= \var{siground(CE,4)}  \text{ (C)} \\&= \var{siground(CD,4)}  \text{ (T)}  \end{align}$