By symmetry

$R_A = R_H = \frac{1}{2}(3 \times 8)$

$R_A = R_H = 12 \, \text{kN}$

**At Joint A**

$\Sigma F_V = 0$

$F_{AB} \sin \alpha = 12$

$F_{AB} (\frac{3}{5}) = 12$

$F_{AB} = 20 \, \text{kN}$ compression

$\Sigma F_H = 0$

$F_{AC} = F_{AB} \cos \alpha$

$F_{AC} = 20 (\frac{4}{5})$

$F_{AC} = 16 \, \text{kN}$ tension

**At Joint B**

By inspection

$F_{BD} = 20 \, \text{kN}$ compression

$F_{BC} = 8 \, \text{kN}$ compression

**At Joint C**

$\Sigma F_V = 0$

$F_{CD} \sin \beta = 8$

$F_{CD} (\frac{3}{\sqrt{13}}) = 8$

$F_{CD} = 9.6148 \, \text{kN}$ tension

$\Sigma F_H = 0$

$F_{CE} + F_{CD} \cos \beta = 16$

$F_{CE} + 9.6148(\frac{2}{\sqrt{13}}) = 16$

$F_{CE} = 10.6667 \, \text{kN}$ tension

**At Joint E**

By inspection

$F_{DE} = 0$

**Summary**

AB = FH = 20 kN compression

AC = GH = 16 kN tension

BC = FG = 8 kN compression

BD = DF = 20 kN compression

CD = DG = 9.6148 kN tension

CE = EG = 10.6667 kN tension

DE = 0

## how were you able to get the

how were you able to get the triangle? the one on point A