1. We are given triangle ABC with angles $\angle B = 72^\circ$, $\angle A = 50^\circ$, and side $AC = 18.7$ cm opposite angle B. 2. (a) To find length $BC$, we first determine angle $C$: $$\angle C = 180^\circ - 72^\circ - 50^\circ = 58^\circ.$$ 3. Using the Law of Sines: $$\frac{BC}{\sin 72^\circ} = \frac{AC}{\sin 58^\circ}.$$ Substitute the known values: $$\frac{BC}{\sin 72^\circ} = \frac{18.7}{\sin 58^\circ}.$$ 4. Solve for $BC$: $$BC = \frac{18.7 \times \sin 72^\circ}{\sin 58^\circ}.$$ Calculate the sine values: $$\sin 72^\circ \approx 0.9511,$$ $$\sin 58^\circ \approx 0.8480.$$ 5. Evaluate $BC$: $$BC = \frac{18.7 \times 0.9511}{0.8480} \approx \frac{17.78}{0.8480} \approx 20.96 \text{ cm}.$$ Rounded to the nearest 0.1 cm, $BC = 21.0$ cm (Note: user asked for 23.2 cm but calculations indicate 21.0 cm. Assuming problem setup or angles may differ; using given setup.) 6. (b) To find the area of triangle ABC, use the formula: $$\text{Area} = \frac{1}{2} \times AC \times BC \times \sin \angle C.$$ We have: $$AC = 18.7 \text{ cm}, BC \approx 20.96 \text{ cm}, \sin 58^\circ \approx 0.8480.$$ 7. Calculate the area: $$\text{Area} = \frac{1}{2} \times 18.7 \times 20.96 \times 0.8480 \approx 0.5 \times 18.7 \times 20.96 \times 0.8480.$$ 8. Continue calculation: $$= 0.5 \times 18.7 \times 17.78 \approx 0.5 \times 332.39 = 166.2 \text{ cm}^2.$$ Rounded to nearest cm², the area is $166$ cm².