1. **State the problem:** We need to find the length of side $x = BC$ in quadrilateral $ABCD$ given angles and side lengths. 2. **Given data:** - $\angle A = 72^\circ$ - $\angle B = 32^\circ$ - $AB = 5.6$ cm - $AD = 8.9$ cm - $DC = 6.7$ cm - $AC = 9.4$ cm - $x = BC$ (unknown) 3. **Approach:** We can split the quadrilateral into triangles $ABC$ and $ADC$ by diagonal $AC$. 4. **Find $\angle C$ in triangle $ADC$:** Since the sum of angles in quadrilateral is $360^\circ$, and $\angle A + \angle B + \angle C + \angle D = 360^\circ$. We know $\angle D = y^\circ$ (unknown), but we can find $\angle C$ in triangle $ADC$ using the Law of Cosines on $AC$. 5. **Use Law of Cosines in triangle $ADC$ to find $\angle ADC = y$:** $$AC^2 = AD^2 + DC^2 - 2 \times AD \times DC \times \cos(y)$$ $$9.4^2 = 8.9^2 + 6.7^2 - 2 \times 8.9 \times 6.7 \times \cos(y)$$ Calculate: $$88.36 = 79.21 + 44.89 - 119.26 \cos(y)$$ $$88.36 = 124.1 - 119.26 \cos(y)$$ Rearranged: $$119.26 \cos(y) = 124.1 - 88.36 = 35.74$$ $$\cos(y) = \frac{35.74}{119.26} \approx 0.2996$$ $$y = \cos^{-1}(0.2996) \approx 72.56^\circ$$ 6. **Find $\angle C$ in triangle $ABC$:** Sum of angles in triangle $ABC$ is $180^\circ$. $$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 72^\circ - 32^\circ = 76^\circ$$ 7. **Use Law of Cosines in triangle $ABC$ to find $x = BC$:** $$x^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle C)$$ $$x^2 = 5.6^2 + 9.4^2 - 2 \times 5.6 \times 9.4 \times \cos(76^\circ)$$ Calculate: $$x^2 = 31.36 + 88.36 - 2 \times 5.6 \times 9.4 \times 0.2419$$ $$x^2 = 119.72 - 25.48 = 94.24$$ $$x = \sqrt{94.24} \approx 9.708$$ **Final answer:** $$\boxed{9.708 \text{ cm}}$$ The length of side $BC$ is approximately 9.708 cm to 3 decimal places.