$\Sigma M_A = 0$
$8R_H = 2(55) + 4(90) + 6(45)$
$R_H = 92.5 \, \text{ kN}$
From section to the right of a-a
$\dfrac{x + 2}{1.5} = \dfrac{x + 4}{2.5}$
$2.5x + 5 = 1.5x + 6$
$x = 1 \, \text{ m}$
$\Sigma M_O = 0$
$(x + 2)\left( \frac{5}{\sqrt{41}}F_{DG} \right) + xR_H = (x + 2)(45)$
$(1 + 2)\left( \frac{5}{\sqrt{41}}F_{DG} \right) + 1(92.5) = (1 + 2)(45)$
$\frac{15}{\sqrt{41}}F_{DG} + 92.5 = 135$
$\frac{15}{\sqrt{41}}F_{DG} = 42.5$
$F_{DG} = 18.14 \, \text{ kN tension}$ answer