1. **Problem Statement:** We are given a triangle ABC with point D on segment BC such that AD bisects angle A, meaning $\angle DAC = \angle BAD$. We know the lengths $AC = 5.9$, $CD = 2$, and $DB = 2.8$. We need to find the length of $AB$ rounded to one decimal place. 2. **Formula Used:** The Angle Bisector Theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. Mathematically, $$\frac{AB}{AC} = \frac{DB}{DC}$$ 3. **Apply the theorem:** Substitute the known values: $$\frac{AB}{5.9} = \frac{2.8}{2}$$ 4. **Solve for $AB$:** $$AB = 5.9 \times \frac{2.8}{2} = 5.9 \times 1.4 = 8.26$$ 5. **Round the answer:** $AB \approx 8.3$ units. **Final answer:** The length of $\overline{AB}$ is approximately **8.3** units.