1. **Problem statement:** In the right triangle ABC with right angle at B, sides AB = 7 units and BC = 6 units, identify the acute angle at vertex C. 2. **Recall:** In a right triangle, the acute angles are the two angles other than the right angle. We can find the measure of angle C using trigonometric ratios. 3. **Use the Pythagorean theorem** to find the hypotenuse AC: $$AC = \sqrt{AB^2 + BC^2} = \sqrt{7^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85}$$ 4. **Find angle C** using cosine, since adjacent side to angle C is BC and hypotenuse is AC: $$\cos C = \frac{BC}{AC} = \frac{6}{\sqrt{85}}$$ 5. **Calculate angle C:** $$C = \cos^{-1}\left(\frac{6}{\sqrt{85}}\right)$$ 6. **Approximate the value:** $$\sqrt{85} \approx 9.22$$ $$\frac{6}{9.22} \approx 0.650$$ $$C \approx \cos^{-1}(0.650) \approx 49.46^\circ$$ **Final answer:** The acute angle at vertex C is approximately $49.46^\circ$.