For vertical resultant, R
x = 0 and R
y = R
$M_R = \Sigma M_E$
$R(2) = T_x(4) - T_y(2)$
$R(2) = 361(\frac{3}{\sqrt{13}})(4) - 361(\frac{2}{\sqrt{13}})(2)$
$R = 400.49 \, \text{ lb downward at point A}$
$\Sigma M_B = M_R$
$F_x(2) + F_y(1) = R(1)$
$F(\frac{1}{\sqrt{10}})(2) + F(\frac{3}{\sqrt{10}})(1) = 400.49(1)$
$F = 253.29 \, \text{ lb}$ answer
$\Sigma F_V = R$
$P_y + T_y - F_y = R$
$P(\frac{2}{\sqrt{5}}) + T(\frac{2}{\sqrt{13}}) - F(\frac{3}{\sqrt{10}}) = R$
$P(\frac{2}{\sqrt{5}}) + 361(\frac{2}{\sqrt{13}}) - 253.29(\frac{3}{\sqrt{10}}) = 400.49$
$P = 492.53 \, \text{ lb}$ answer