1. **State the problem:** We have a right triangle with a right angle at P, a horizontal side PQ = 5.1, a hypotenuse QO = 8.9, and we need to find the angle $x^\circ$ at vertex O. 2. **Formula used:** In a right triangle, the cosine of an angle is the ratio of the adjacent side to the hypotenuse: $$\cos(x) = \frac{\text{adjacent side}}{\text{hypotenuse}}$$ 3. **Identify sides:** The side adjacent to angle $x$ is PQ = 5.1, and the hypotenuse is QO = 8.9. 4. **Calculate cosine:** $$\cos(x) = \frac{5.1}{8.9}$$ 5. **Simplify fraction:** $$\cos(x) = \frac{\cancel{5.1}}{\cancel{8.9}} = 0.5730$$ 6. **Find angle $x$ using inverse cosine:** $$x = \cos^{-1}(0.5730)$$ 7. **Calculate angle:** $$x \approx 54.9^\circ$$ **Final answer:** $$x \approx 54.9^\circ$$