1. **Problem:** Find the value of side AB in a triangle where angle B = 50°, angle C = 62°, and side AC = 12. 2. **Step 1: Find angle A.** Since the sum of angles in a triangle is 180°, $$A = 180^\circ - 50^\circ - 62^\circ = 68^\circ$$ 3. **Step 2: Use the Law of Sines.** The Law of Sines states: $$\frac{AB}{\sin C} = \frac{AC}{\sin B} = \frac{BC}{\sin A}$$ We want to find $AB$, so: $$AB = \frac{AC \times \sin C}{\sin B}$$ 4. **Step 3: Substitute known values.** $$AB = \frac{12 \times \sin 62^\circ}{\sin 50^\circ}$$ 5. **Step 4: Calculate sine values and evaluate.** $$\sin 62^\circ \approx 0.8829, \quad \sin 50^\circ \approx 0.7660$$ $$AB = \frac{12 \times 0.8829}{0.7660} = \frac{10.5948}{0.7660}$$ 6. **Step 5: Simplify the fraction.** $$AB = \cancel{\frac{10.5948}{0.7660}} = 13.84$$ 7. **Step 6: Round to the nearest tenth.** $$AB \approx 13.8$$ **Final answer:** The length of side AB is approximately **13.8** units.