解:∵$∠1∶∠2∶∠3 = 28∶5$:$3$
∴设$∠1 = 28x,$则$∠2 = 5x,$$∠3 = 3x$
∵$\triangle ABC$的内角和为$180°$
∴$28x+5x + 3x = 180°,$解得$x = 5°$
∴$∠1 = 28×5°=140°$
∵$\triangle ABE$和$\triangle ADC$是由$\triangle ABC$分别沿着边$AB,$$AC$翻折得到的
∴$∠BAE=∠1 = 140°,$$∠3=∠E=∠G CA$
∴$∠G AC = 360°-∠BAE-∠1 = 360°-140°-140°=80°$
∵$\triangle FGE,$$\triangle AG C$的内角和均为$180°,$$∠FGE=∠AG C,$$∠E=∠G CA$
∴$180°-∠FGE-∠E = 180°-∠AG C-∠G CA,$即$∠α=∠G AC = 80°$