$\sin 30^\circ = \dfrac{a}{3}$
$a = 3 \sin 30^\circ$
$a = 1.5 \, \text{ m}$
$\cos 60^\circ = \dfrac{b}{a}$
$b = a \cos 60^\circ$
$b = 1.5 \cos 60^\circ$
$b = 0.75 \, \text{ m}$
Magnitude of resultant
$R = \Sigma F_v$
$R = 2(890) + 2(1335) + 1780 + 8900$
$R = 15\,130 \, \text{ N}$ downward
Location of resultant
$Rd = \Sigma Fx$
$Rd = 1335(3 - b) + 8900(3) + 1780(4.5) + 1335(6 + b) + 890(9)$
$15\,130d = 1335(3 - 0.75) + 8900(3) + 1780(4.5) + 1335(6 + 0.75) + 890(9)$
$15\,130d = 1335(2.25) + 8900(3) + 1780(4.5) + 1335(6.75) + 890(9)$
$15\,130d = 54\,735$
$d = 3.62 \, \text{ m}$ to the right of A
Thus, R = 15 130 N downward at 3.62 m to the right of left support. answer
