Determine the horizontal and vertical components of reaction at pin C.
⤹ +\Sigma M_{C}=0;
\begin{aligned} & -\left(\frac{4}{5}\right)\left(F_{A B}\right)(9)+400(6)+500(3)=0 \\ & F_{A B}=541.67 \mathrm{lb} \\ & \stackrel{+}{\rightarrow} F_{x}=0 ;-C_{x}+\frac{3}{5}(541.67)=0 \\ & C_{x}=325 \mathrm{lb} \\ & +\uparrow\Sigma F_{y}=0;C_{x}+{\textstyle{\frac{4}{5}}}\left(541.67\right)-400-500=0 \\& C_{y}=467 \mathrm{lb} \end{aligned}