1. **State the problem:** We need to determine which of the two acute angles in a right triangle has a cosine value of $\frac{4}{5}$. The triangle has sides $AB=6$, $BC=8$, and hypotenuse $AC=10$. 2. **Recall the cosine definition:** For an angle $\theta$ in a right triangle, $\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}$. 3. **Identify the angles:** The right angle is at $B$, so the acute angles are at $A$ and $C$. 4. **Calculate $\cos A$:** Adjacent to angle $A$ is side $AB=6$, hypotenuse is $AC=10$, so $$\cos A = \frac{6}{10} = \frac{3}{5}.$$ This is not $\frac{4}{5}$. 5. **Calculate $\cos C$:** Adjacent to angle $C$ is side $BC=8$, hypotenuse is $AC=10$, so $$\cos C = \frac{8}{10} = \frac{4}{5}.$$ This matches the given cosine value. 6. **Conclusion:** The angle with cosine $\frac{4}{5}$ is angle $C$. **Final answer:** Angle $C$ has $\cos C = \frac{4}{5}$.