Find the reactions:

1. Draw a free body diagram of the entire truss.
2. Get scalar components of loads $\mathbf{B}$ and $\mathbf{C}$ for later use.

   $\begin{align}\\ B_x &= \var{display(qty(vecB[0],units))} & B_y &= \var{display(qty(vecB[1],units))}\\C_x &= \var{display(qty(vecC[0],units))} & C_y &= \var{display(qty(vecC[1],units))}\end{align}$

3. Take moments at $A$ to find reactions at $D$.  Note 1-2-$\sqrt{3}$ triangle.  

   $\begin{align}\\ \Sigma M_A &= 0\\  2\sqrt{3} \,D &= \var{if(sign(vecB[0])=-1,'– ','')} 1\, B_x +  \var{if(sign(vecB[1])=1,'– ','')} \sqrt{3}\, B_y + \sqrt{3}\,C \\  D &=  \dfrac{ \var{if(sign(vecB[0])=-1,'– ','')} B_x}{2\sqrt{3}}   + \dfrac{\var{if(sign(vecB[1])=1,'– ','')}\, B_y }{2} + \dfrac{C }{2}\\D&= \dfrac{1}{2}\left(\dfrac{ \var{precround(vecB[0],1)} }{\sqrt{3}} + \var{precround( - vecB[1],1) } + \var{scalar(magC)} \right)&= \var{display(magD)}\text{ up.}\end{align}$ 

4. Apply $\Sigma F_x$ and $\Sigma F_y$ equations to get the components of the reaction force at $A$.

   $\begin{align} \\\Sigma F_x &= 0\\ A_x &= B_x \\ &= \var{display(qty(abs(vecA[0]),units))}\quad \var{if(sign(vecA[0])=1,'to the right.','to the left.')} \end{align}$ 

   $\begin{align} \\\Sigma F_y &= 0\\ A_y &= C_y - D_y + B_y  \\ A_y &= \simplify[!collectNumbers]{{siground(-VecC[1],5)} + {siground(-VecD[1],5)} + {siground(-vecB[1],5)}} \\ &= \var{display(qty(abs(vecA[1]),units))} \quad \var{if(sign(vecA[1])=1,'up.','down.')} \end{align}$ 

Analyze the joints:

1. Draw a FBDs of the joints showing your assumed directions for the forces in the members.  Members in tension pull away from the joint, compression push towards the joint.

      

   GeoGebra applet loading...

2. Pick a joint with two unknowns, joint $D$ for example.  Joint $A$ would also work.
3. Apply equations of equilibrium to joint $D$ to find forces in members $BD$ and $CD$.

   $\begin{align}\\D: \Sigma F_y  &= 0\\ BD_y &= D\\ BD &= \dfrac{D}{\sin 30°}\\ &= \dfrac{\var{display(magD)}}{0.5}\\ &= \var{display(BD)} \quad \text{Compression.}\end{align}$

   $\begin{align}\\D: \Sigma F_x &= 0\\ CD &=BD_x\\ &= BD \cos 30° \\&=   \var{display(BD)}  \cos 30° \\&= \var{display(CD)} \quad \text{Tension.}\end{align}$

4. Move to an adjacent joint with two unknowns, joint $C$ in this case.  Joint $C$ is a special case.

   $\begin{align}\\ C: \Sigma F_y &= 0\\BC &= C\\ &= \var{display(BC)} \quad \text{Tension.}\end{align}$ 

   $\begin{align}\\ C: \Sigma F_x &= 0\\AC &= CD\\ &= \var{display(AC)} \quad \text{Tension.}\end{align}$

5. Move to joint $B$ and solve for $AB$.

   $B: \Sigma F_y=0$

   $-AB_y  - BC + BD_y \, \mathbf{\var{if(sign(vecB[1])=1,'+','–')}}\, B_y = 0$

   $AB_y= \var{vecC[1]} + \var{precround(vecD[1],1)} \, \var{if(sign(vecB[1])=-1,'','+')} \, \var{precround(vecB[1],1)}= \var{display(-ABy)}$

   $\begin{align}AB &= \dfrac{AB_y}{\sin 30°}\\ &= \var{display(-2ABy)}\\ &= \var{display(AB)} \var{if(sign(ABy)=-1,' Tension.', ' Compression.')}\end{align}$

Check your work:

At this point you should check your work by verifying that joint $A$ is in equilibrium.  If it is not you have made a mistake somewhere above.

$A: \Sigma F_x \stackrel{?}{=}  0$

$-A_x + AB_x + AC = \var{precround(vecA[0],1)}+ \var{precround(-vecAB[0],1) } + \var{precround(-vecCD[0],1) } =  \var{vecA[0] -vecAB[0] - vecCD[0]}$

Close Enough

$A: \Sigma F_y \stackrel{?}{=}  0$

$A_y + AB_y = \var{precround(vecA[1],1)}+ \var{precround(-vecAB[1],1) } = \var{vecA[1] -vecAB[1]}$

Close Enough