Considering the equilibrium condition of beam

$\sum \ F_y=0$

$R_A+R_E-120-50\ sin\ 60^{\circ}=0\\ R_A+R_E=120-50\ sin\ 60^{\circ}=163.30\ N.......(1) $

Taking moment of all forces about 'A'

$\sum M_A=0\\ (120\times 3)+(50\ sin \ 60^{\circ}\times 8)-R_E\times 14=0\\ (120\times 3)+(50\ sin \ 60^{\circ}\times 8)=R_E\times 14\\ R_E=36.17\ N$

Put Value of RE in equation (1) we get, 

$R_A+R_E=163.30\\ R_A=163.30-36.17=127.13\ N$