$R_x = \Sigma F_x$

$R_x = 390(\frac{12}{13}) + 722(\frac{3}{\sqrt{13}}) - 300 \sin 30^\circ$

$R_x = 810.74 \, \text{ lb to the right}$

$R_y = \Sigma F_y$

$R_y = 390(\frac{5}{13}) - 722(\frac{2}{\sqrt{13}}) - 300 \cos 30^\circ$

$R_y = -510.30 \, \text{ lb}$

$R_y = 510.30 \, \text{ lb downward}$

$R = \sqrt{{R_x}^2 + {R_y}^2}$

$R = \sqrt{810.74^2 + 510.30^2}$

$R = 957.97 \, \text{ lb}$

$\tan \theta_x = \dfrac{R_y}{R_x}$

$\tan \theta_x = \dfrac{510.30}{810.74}$

$\theta_x = 32.19^\circ$

$M_O = \Sigma Fd$

$M_O = -390(\frac{12}{13})(3) + 390(\frac{5}{13})(5) - 722(\frac{2}{\sqrt{13}})(4) + (300 \sin 30^\circ)(3)$

$M_O = -1481.97 \, \text{ lb}\cdot\text{ft}$

$M_O = 1481.97 \, \text{ lb}\cdot\text{ft clockwise}$

$R_xb = M_O$

$810.74b = 1481.97$

$b = 1.83 \, \text{ft above point O}$

$R_ya = M_O$

$510.30a = 1481.97$

$a = 2.90 \, \text{ft to the right of point O}$

Thus, R = 957.97 lb downward to the right at θ_{x} = 32.19°. The x-intercept is at 2.90 ft to the right of O and the y-intercept is 1.83 ft above point O.

Hi, why use sin on the summation of x? And cos on summation of y? Isnt it the other way around?