1. **State the problem:** Given a triangle with $m\angle B = 19^\circ$, side $a = 30$ yd opposite $\angle A$, and side $b = 12$ yd opposite $\angle B$, find the remaining angle $m\angle A$, $m\angle C$, and side $c$. 2. **Recall the Law of Sines:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ This law relates sides and their opposite angles in any triangle. 3. **Find $m\angle A$:** Using the Law of Sines, $$\frac{a}{\sin A} = \frac{b}{\sin B} \implies \frac{30}{\sin A} = \frac{12}{\sin 19^\circ}$$ Cross-multiplied: $$30 \sin 19^\circ = 12 \sin A$$ Divide both sides by 12: $$\frac{30}{\cancel{12}} \sin 19^\circ = \frac{12}{\cancel{12}} \sin A \implies \frac{30}{12} \sin 19^\circ = \sin A$$ Calculate: $$\sin A = 2.5 \times \sin 19^\circ \approx 2.5 \times 0.32557 = 0.8139$$ Find $A$: $$m\angle A = \sin^{-1}(0.8139) \approx 54.7^\circ$$ 4. **Find $m\angle C$:** Sum of angles in a triangle is 180°: $$m\angle C = 180^\circ - m\angle A - m\angle B = 180^\circ - 54.7^\circ - 19^\circ = 106.3^\circ$$ 5. **Find side $c$ using Law of Sines:** $$\frac{c}{\sin C} = \frac{a}{\sin A} \implies c = \frac{a \sin C}{\sin A} = \frac{30 \times \sin 106.3^\circ}{\sin 54.7^\circ}$$ Calculate: $$c = \frac{30 \times 0.9703}{0.8139} \approx \frac{29.11}{0.8139} = 35.77\text{ yd}$$ **Final answers:** $$m\angle A \approx 54.7^\circ$$ $$m\angle C \approx 106.3^\circ$$ $$c \approx 35.8\text{ yd}$$