Assume that member (1) controls. Substitute ${\mathrm{F}}_1=554.4 \mathrm{kN}$ into the equilibrium equations and solve for ${\mathrm{F}}_2$ and $\mathrm{P}$.

$\sum _{}^{}{\mathrm{F}}_{\mathrm{x}}={\mathrm{F}}_{1 }\cos (49.3^{\circ })+{\mathrm{F}}_2 \cos (24.9^{\circ })=0$

$\sum _{}^{}{\mathrm{F}}_{\mathrm{y}}={\mathrm{F}}_{1 }\sin (49.3^{\circ })-{\mathrm{F}}_2 \sin (24.9^{\circ })-\mathrm{P}=0$