1. **State the problem:** We have a right triangle PON with sides PO = 7, PN = 9.3, and a right angle at O. We need to find the angle $x$ at point N. 2. **Formula used:** In a right triangle, the cosine of an angle is the ratio of the adjacent side to the hypotenuse. Here, angle $x$ is at N, so the adjacent side to $x$ is PO, and the hypotenuse is PN. 3. **Apply the cosine formula:** $$\cos(x) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PO}{PN} = \frac{7}{9.3}$$ 4. **Calculate the ratio:** $$\cos(x) = \frac{7}{9.3} \approx 0.7527$$ 5. **Find the angle $x$ by taking the inverse cosine:** $$x = \cos^{-1}(0.7527)$$ 6. **Evaluate the inverse cosine:** $$x \approx 41.0^\circ$$ **Final answer:** $$x \approx 41.0^\circ$$