1. **Problem statement:** We have a right triangle PON with a right angle at O. Side PO is 7 units, side PN is 9.3 units, and we want to find the angle $x^\circ$ at vertex N. 2. **Identify the sides relative to angle $x$:** - Hypotenuse: PN = 9.3 - Adjacent side to angle $x$: PO = 7 - Opposite side to angle $x$: ON (unknown) 3. **Formula used:** To find angle $x$, we use the cosine function because we know the adjacent side and hypotenuse: $$\cos(x) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PO}{PN}$$ 4. **Calculate cosine of $x$:** $$\cos(x) = \frac{7}{9.3}$$ 5. **Evaluate the fraction:** $$\cos(x) = 0.7527$$ 6. **Find angle $x$ by taking the inverse cosine:** $$x = \cos^{-1}(0.7527)$$ 7. **Calculate $x$ using a calculator:** $$x \approx 41.0^\circ$$ **Final answer:** $$x \approx 41.0^\circ$$