1. **State the problem:** We have trapezium ABCD with BC and AD parallel, BC = 8.4 cm, AD = 17.6 cm, area = 179.4 cm², and exactly one line of symmetry. We need to find angle ABC. 2. **Recall the area formula for trapezium:** $$\text{Area} = \frac{(a+b)}{2} \times h$$ where $a$ and $b$ are the lengths of the parallel sides, and $h$ is the height. 3. **Calculate the height $h$:** Given $a = 8.4$, $b = 17.6$, and area $= 179.4$, $$\frac{(8.4 + 17.6)}{2} \times h = 179.4$$ $$\frac{26}{2} \times h = 179.4$$ $$13 \times h = 179.4$$ $$h = \frac{179.4}{13} = 13.8$$ 4. **Analyze the trapezium symmetry:** Since the trapezium has exactly one line of symmetry, it must be isosceles, so the non-parallel sides are equal and the height bisects the trapezium. 5. **Find half the difference of the parallel sides:** $$\frac{17.6 - 8.4}{2} = \frac{9.2}{2} = 4.6$$ 6. **Use right triangle formed by height $h$, half the difference $4.6$, and side $AB$:** Angle $ABC$ is the angle between side $BC$ and side $AB$. 7. **Calculate angle ABC using tangent:** $$\tan(\angle ABC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{4.6} = \frac{13.8}{4.6} = 3$$ 8. **Find angle ABC:** $$\angle ABC = \tan^{-1}(3) \approx 71.6^\circ$$ **Final answer:** $$\boxed{71.6^\circ}$$