Moment about O
$M_O = 180$
$xF_y = 180$
Moment about B
$M_B = 90$
$(6 - x)F_y = 90$
$6F_y - xF_y = 90$
Substitute xFy = 180 to the above equation
$6F_y - 180 = 90$
$6F_y = 270$
$F_y = 45 \, \text{ lb}$
$xF_y = 180$
$x(45) = 180$
$x = 4 \, \text{ ft}$
$\tan \theta_x = \dfrac{3}{x}$
$\tan \theta_x = \dfrac{3}{4}$
$\theta_x = 36.87^\circ$
$\sin \theta_x = \dfrac{F_y}{F}$
$F = \dfrac{F_y}{\sin \theta_x} = \dfrac{45}{\sin 36.87^\circ}$
$F = 75 \text{ lb}$
Thus, F = 75 lb downward to the right at θx = 36.87° and x-intercept at (4, 0). answer