232 Moment of a force about points O and B

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 xF_y = 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 θ_x = 36.87° and x-intercept at (4, 0).           answer