1. **Problem statement:** Find angle $A$ in a triangle where side $b=28$, angle $C=52^\circ$, and side $a=29$. 2. **Formula used:** We use the Law of Sines which states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Step 1:** Use the known angle $C$ and side $b$ to find angle $B$ or directly find $A$ using the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin C}$$ 4. **Step 2:** Substitute the known values: $$\frac{29}{\sin A} = \frac{28}{\sin 52^\circ}$$ 5. **Step 3:** Solve for $\sin A$: $$\sin A = \frac{29 \times \sin 52^\circ}{28}$$ Calculate $\sin 52^\circ \approx 0.7880$: $$\sin A = \frac{29 \times 0.7880}{28} = \frac{22.852}{28} \approx 0.8161$$ 6. **Step 4:** Find angle $A$ by taking the inverse sine: $$A = \sin^{-1}(0.8161) \approx 54.7^\circ$$ 7. **Step 5:** Verify the triangle angle sum: $$A + C + B = 180^\circ \Rightarrow B = 180^\circ - 52^\circ - 54.7^\circ = 73.3^\circ$$ **Final answer:** $$\boxed{A \approx 54.7^\circ}$$