1. **Problem statement:** Find angle $A$ in a triangle where side $b=28$, side $a=29$, and angle $C=52^\circ$. We can use the Law of Cosines or Law of Sines. 2. **Choosing the formula:** Since we know two sides and the included angle $C$, the Law of Cosines is appropriate: $$c^2 = a^2 + b^2 - 2ab \cos C$$ But we don't know side $c$, so first find $c$: $$c^2 = 29^2 + 28^2 - 2 \times 29 \times 28 \times \cos 52^\circ$$ 3. **Calculate $c^2$:** $$29^2 = 841$$ $$28^2 = 784$$ $$\cos 52^\circ \approx 0.6157$$ $$c^2 = 841 + 784 - 2 \times 29 \times 28 \times 0.6157$$ $$c^2 = 1625 - 2 \times 29 \times 28 \times 0.6157$$ Calculate the product: $$2 \times 29 \times 28 = 1624$$ $$1624 \times 0.6157 \approx 1000.9$$ So, $$c^2 = 1625 - 1000.9 = 624.1$$ 4. **Find $c$:** $$c = \sqrt{624.1} \approx 24.98$$ 5. **Use Law of Sines to find angle $A$:** $$\frac{a}{\sin A} = \frac{c}{\sin C}$$ Rearranged: $$\sin A = \frac{a \sin C}{c}$$ Calculate: $$\sin 52^\circ \approx 0.7880$$ $$\sin A = \frac{29 \times 0.7880}{24.98} \approx \frac{22.852}{24.98} = 0.9147$$ 6. **Find angle $A$:** $$A = \sin^{-1}(0.9147) \approx 66.4^\circ$$ **Final answer:** Angle $A$ is approximately $66.4^\circ$.