1. **Problem Statement:** We have a square PQRS and point x is the midpoint of side PQ. We need to find the measure of angle $\angle ZPX$ in degrees. 2. **Understanding the problem:** Since PQRS is a square, all sides are equal and all angles are right angles ($90^\circ$). Point x is the midpoint of PQ, so it divides PQ into two equal segments. 3. **Set up coordinates for clarity:** - Let P be at $(0,0)$. - Since PQRS is a square, Q is at $(a,0)$ for some side length $a$. - R is at $(a,a)$ and S is at $(0,a)$. - Point x, the midpoint of PQ, is at $\left(\frac{a}{2},0\right)$. 4. **Identify points involved in angle $\angle ZPX$:** - $Z$ is not defined in the problem, but assuming $Z$ is point S (common notation for vertices), so $Z = S = (0,a)$. - $P = (0,0)$. - $X = \left(\frac{a}{2},0\right)$. 5. **Calculate vectors:** - Vector $\overrightarrow{PZ} = Z - P = (0,a) - (0,0) = (0,a)$. - Vector $\overrightarrow{PX} = X - P = \left(\frac{a}{2},0\right) - (0,0) = \left(\frac{a}{2},0\right)$. 6. **Find the angle between vectors $\overrightarrow{PZ}$ and $\overrightarrow{PX}$:** - Use the dot product formula: $$\overrightarrow{PZ} \cdot \overrightarrow{PX} = |\overrightarrow{PZ}| |\overrightarrow{PX}| \cos \theta$$ - Calculate dot product: $$0 \times \frac{a}{2} + a \times 0 = 0$$ - Calculate magnitudes: $$|\overrightarrow{PZ}| = \sqrt{0^2 + a^2} = a$$ $$|\overrightarrow{PX}| = \sqrt{\left(\frac{a}{2}\right)^2 + 0^2} = \frac{a}{2}$$ - Substitute into formula: $$0 = a \times \frac{a}{2} \times \cos \theta = \frac{a^2}{2} \cos \theta$$ - Solve for $\cos \theta$: $$\cos \theta = 0$$ 7. **Find angle $\theta$:** - $\cos \theta = 0$ implies $$\theta = 90^\circ$$ **Final answer:** $$\boxed{90^\circ}$$