1. **State the problem:** Given triangle ABC with angles $A=42^\circ$, $B=58^\circ$, and side $a=15$ cm opposite angle $A$, find sides $b$ and $c$. 2. **Formula and rules:** Use the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Sum of angles in a triangle: $$A + B + C = 180^\circ$$ 3. **Find angle $C$:** $$C = 180^\circ - A - B = 180^\circ - 42^\circ - 58^\circ = 80^\circ$$ 4. **Apply Law of Sines to find $b$:** $$\frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow b = \frac{a \sin B}{\sin A} = \frac{15 \times \sin 58^\circ}{\sin 42^\circ}$$ Calculate sines: $$\sin 58^\circ \approx 0.8480, \quad \sin 42^\circ \approx 0.6691$$ So, $$b = \frac{15 \times 0.8480}{0.6691} = 15 \times \frac{0.8480}{0.6691}$$ Intermediate step with cancellation: $$b = 15 \times \frac{\cancel{0.8480}}{\cancel{0.6691}}$$ (just showing fraction simplification conceptually) Calculate value: $$b \approx 15 \times 1.267 = 19.0 \text{ cm}$$ 5. **Apply Law of Sines to find $c$:** $$\frac{a}{\sin A} = \frac{c}{\sin C} \Rightarrow c = \frac{a \sin C}{\sin A} = \frac{15 \times \sin 80^\circ}{\sin 42^\circ}$$ Calculate sines: $$\sin 80^\circ \approx 0.9848$$ So, $$c = 15 \times \frac{0.9848}{0.6691}$$ Intermediate step with cancellation: $$c = 15 \times \frac{\cancel{0.9848}}{\cancel{0.6691}}$$ Calculate value: $$c \approx 15 \times 1.471 = 22.1 \text{ cm}$$ **Final answers:** $$b \approx 19.0 \text{ cm}, \quad c \approx 22.1 \text{ cm}$$