AB is an axial member while BC is a three-force member. Thus,

$A_x = R_A \cos 37^\circ$
$A_y = R_A \sin 37^\circ$

$\Sigma M_C = 0$

$A_x(1) + A_y(8) = 4.0(4) + 1.6(2)$

$R_A \cos 37^\circ + (R_A \sin 37^\circ)(8) = 19.2$

$5.6132R_A = 19.2$

$R_A = 3.42 \, \text{ kN}$ *answer*

$\Sigma F_H = 0$

$C_x = A_x$

$C_x = R_A \cos 37^\circ$

$C_x = 3.42 \cos 37^\circ$

$C_x = 2.73 \, \text{ kN}$ *answer*

$\Sigma F_V = 0$

$C_y + A_y = 4.0 + 1.6$

$C_y + R_A \sin 37^\circ = 5.6$

$C_y + 3.42 \sin 37^\circ = 5.6$

$C_y = 3.54 \, \text{ kN}$ *answer*