By symmetry

$R_A = R_J = \frac{1}{2}(12 \times 5) = 30 ~ \text{kN}$

By ratio and proportion

$\dfrac{a}{10} = \dfrac{24}{15}$
$a = 16 ~ \text{m}$ |
$\dfrac{b}{15} = \dfrac{a}{20}$
$\dfrac{b}{15} = \dfrac{16}{20}$ |
$\dfrac{c}{5} = \dfrac{24}{15}$
$c = 8 ~ \text{m}$ |

From FBD Through Section M-M

$\Sigma M_H = 0$
$5\left( \frac{8}{\sqrt{89}}F_{GI} \right) + 10(30) = 5(12)$

$F_{GI} = -6\sqrt{89} ~ \text{kN}$

$F_{GI} = 56.60 ~ \text{kN compression}$ *answer*

$\Sigma M_J = 0$

$10F_{GH} = 5(12)$

$F_{GH} = 6 ~ \text{kN tension}$ *answer*

$\Sigma M_G = 0$

$(16 - 8)\left( \frac{5}{\sqrt{41}}F_{EH} \right) + 5(12) = 10(30)$

$F_{EH} = 6\sqrt{41} ~ \text{kN} = 38.42 ~ \text{kN tension}$ *answer*

From FBD Through Section N-N

$\Sigma M_J = 0$

$8F_{HI} + 5(12) = 0$

$F_{HI} = -7.5 ~ \text{kN}$

$F_{HI} = 7.5 ~ \text{kN compression}$ *answer*