Here is an idea, borrowed from linear elastic theory. Solve all the forces in terms of an unknown force (I chose $f_{10}$) and construct a long vector $f$
$$ \boldsymbol f = \begin{bmatrix} f_1 & f_2 & \dots & f_9 \end{bmatrix}^\top $$
where each component is a function of $P$, $\beta$ and $f_{10}$.
Now assemble something resembling the total potential energy by doing this
$$ U = \frac{L}{2\,E\,A} \, \boldsymbol f^\top \boldsymbol f $$
Now by minimizing this with
$$ \frac{{\rm d}U}{{\rm d}f_{10}} = 0 $$
will produce a result for $f_{10}$ and hence all of the values in $\boldsymbol{f}$.
This works because for each element $f_i = \frac{E_i,A_i}{L_i} \delta_i$ where $\delta_i$ is the deformation, and the energy is $U_i = \frac{1}{2} \left(\frac{E_i,A_i}{L_i}\right) \delta_i^2 = \frac{L_i}{2 E_i A_i} f_i^2 $.
So $U = \sum U_i = \frac{L}{2\,E\,A} \, \boldsymbol f^\top \boldsymbol f $ except that not all the elements have the same length. So my method will produce an incorrect result. I just realized this.
To correct this you have to construct the total energy as
$$ U = \sum_{i} \frac{L_i}{2 E_i A_i} f_i^2 $$ and then minimize it with the derivative.
Illustrative Example
The force equilibrium on points A, B and C is
$$\begin{array}{cc}
A_{y}-f_{1}=0\\
-P+f_{1}-f_{2}=0\\
C_{y}+f_{2}=0
\end{array}\biggr\}\begin{array}{cc}
A_{y}=P+f_{2}\\
C_{y}=-f_{2}\\
f_{1}=P+f_{2}
\end{array}$$
which is indeterminate. The total strain energy is
$$U=\frac{L_{1}}{2EA}f_{1}^{2}+\frac{L_{2}}{2EA}f_{2}^{2} \\
=\frac{L_{1}}{2EA}\left(P+f_{2}\right)^{2}+\frac{L_{2}}{2EA}f_{2}^{2}$$
which is minimized by
$$\frac{{\rm d}U}{{\rm d}f_{2}}=\frac{1}{2E\, A}\left[\frac{{\rm d}}{{\rm d}f_{2}}\left(L_{1}\left(P+f_{2}\right)^{2}\right)+\frac{{\rm d}}{{\rm d}f_{2}}\left(L_{2}f_{2}^{2}\right)\right]=0 \\
=\frac{1}{E\, A}\left(L_{1}\left(P+f_{2}\right)+L_{2}f_{2}\right)=0 \\
f_{2}=\mbox{-}\frac{L_{1}}{L_{1}+L_{2}}P$$
back substituting into
$$ A_y = \frac{L_2}{L_1+L_2} P \\ C_y = \frac{L_1}{L_1+L_2} P \\ f_1 = \frac{L_2}{L_1+L_2} P $$
Of course in your case, you need to add the member weights into the node equilibrium equations.
First assume each member force is tensile and assign it a positive quantities $F_1$, $F_2$, .. etc.
Next draw the force balance on each node with all the member forces pointing towards the member. Like a spring being pulled (in tension), the force on ends is pointing towards the center of the spring.
For node 2 see example below:
With equations
$$ F_4 \cos\theta-F_2 \cos\theta = 0
\\ R_2 - F_3 - F_4 \sin\theta - F_2\sin\theta = 0 $$
If in the end the result is a negative number, then the member is in compression. It all follows quite naturally once you have nice free body diagrams for each node.
Best Answer
Like @Doresoom said, the truss member here are "Two Force Members" carrying only tangential forces. Thus the sum of the forces on the triangle tip (Point ?C?) should yield the answer.
$$ T_A \cos \alpha + T_B \cos\beta = 0 \\ T_A \sin \alpha - T_B \sin\beta - F = 0 $$
where $T_A$ and $T_B$ are the tensions on the two members.
Next you construct the strain energy
$$ U = \frac{T_A^2}{2 A E} + \frac{T_B^2}{2 A E} $$ where $A$ is the cross-sectional area and $E$ is the (Young's) Modulus of elasticity.
and solve for the deflection $\delta$ along $F$ as
$$ \delta = \frac{\partial U}{\partial F} = \frac{\cos^2 \alpha + \cos^2 \beta}{\sin^2 (\alpha+\beta)} \frac{F}{A E} $$