Substitute this result into the other equations so that ${\mathrm{C}}_{\mathrm{x}}$ and ${\mathrm{C}}_{\mathrm{y}}$ are also expressed in terms of $\mathrm{w}$:

$\sum _{}{\mathrm{M}}_{\mathrm{C}}=-(3.6 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})(1.8 \mathrm{m})-(5.0 \mathrm{m}){\mathrm{F}}_1\sin (35.0^{\circ })=0$

$\therefore {\mathrm{F}}_1=-(2.2595 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})$

$\sum _{}{\mathrm{F}}_{\mathrm{x}}={\mathrm{C}}_{\mathrm{x}}-{\mathrm{F}}_1\cos (35.0^{\circ })=0$

$\therefore {\mathrm{C}}_{\mathrm{x}}=-(1.8509 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})$

$\sum _{}{\mathrm{F}}_{\mathrm{y}}={\mathrm{C}}_{\mathrm{y}}-(3.6 \mathrm{m})(\mathrm{w} \mathrm{kN}/\mathrm{m})-{\mathrm{F}}_1\sin (35.0^{\circ })=0$

$\therefore {\mathrm{C}}_{\mathrm{y}}=(2.3040 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})$