1. **Problem statement:** We have quadrilateral ABCD with given sides |AD| = 4 cm, |DC| = 6 cm, and angles $\angle ADC = 60^\circ$, $\angle CAB = 30^\circ$, and $\angle ABC = 53^\circ$. We need to find: (i) Length |AC| to two decimal places. (ii) Length |BC| to two decimal places. (b) The area of quadrilateral ABCD to the nearest mm². 2. **Formulas and rules:** - Use the Law of Cosines to find unknown sides in triangles: $$c^2 = a^2 + b^2 - 2ab\cos(\theta)$$ - Use the Law of Sines to find unknown sides or angles: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ - Area of a triangle using two sides and included angle: $$\text{Area} = \frac{1}{2}ab\sin(\theta)$$ 3. **Step (a)(i): Find |AC|** Triangle ADC has sides |AD| = 4 cm, |DC| = 6 cm, and included angle $\angle ADC = 60^\circ$. Using Law of Cosines: $$|AC|^2 = |AD|^2 + |DC|^2 - 2 \times |AD| \times |DC| \times \cos 60^\circ$$ $$= 4^2 + 6^2 - 2 \times 4 \times 6 \times \cos 60^\circ$$ $$= 16 + 36 - 48 \times 0.5$$ $$= 52 - 24 = 28$$ So, $$|AC| = \sqrt{28} = 5.29 \text{ cm (to two decimal places)}$$ 4. **Step (a)(ii): Find |BC|** Consider triangle ABC. We know angles $\angle CAB = 30^\circ$, $\angle ABC = 53^\circ$, and side |AC| = 5.29 cm (from previous step). First, find $\angle ACB$: $$\angle ACB = 180^\circ - 30^\circ - 53^\circ = 97^\circ$$ Using Law of Sines: $$\frac{|BC|}{\sin 30^\circ} = \frac{|AC|}{\sin 53^\circ}$$ Rearranged: $$|BC| = \frac{|AC| \times \sin 30^\circ}{\sin 53^\circ}$$ Calculate: $$|BC| = \frac{5.29 \times 0.5}{0.7986} = \frac{2.645}{0.7986} = 3.31 \text{ cm (to two decimal places)}$$ 5. **Step (b): Find area of quadrilateral ABCD** Quadrilateral ABCD can be split into triangles ADC and ABC. - Area of triangle ADC: $$= \frac{1}{2} \times |AD| \times |DC| \times \sin 60^\circ = \frac{1}{2} \times 4 \times 6 \times 0.8660 = 10.39 \text{ cm}^2$$ - Area of triangle ABC: Use formula with sides |AC| and |BC| and included angle $\angle ACB = 97^\circ$: $$= \frac{1}{2} \times |AC| \times |BC| \times \sin 97^\circ = \frac{1}{2} \times 5.29 \times 3.31 \times 0.9925 = 8.69 \text{ cm}^2$$ 6. **Total area of quadrilateral ABCD:** $$= 10.39 + 8.69 = 19.08 \text{ cm}^2 = 1908 \text{ mm}^2 \text{ (to nearest mm}^2\text{)}$$