1. **State the problem:** We have a right pyramid with a square base of side length 10 cm and sloping edges of length 17 cm. The perpendicular height drops from the apex to the center of the base. We need to find the angle between a sloping edge and the base. 2. **Identify the elements:** The base is a square with side 10 cm, so half the diagonal of the base is the horizontal distance from the center to a corner where the sloping edge meets the base. 3. **Calculate the half diagonal of the base:** The diagonal $d$ of the square base is given by $$d = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}.$$ Half the diagonal is $$\frac{d}{2} = 5\sqrt{2}.$$ This is the horizontal distance from the center to the base corner. 4. **Set up the right triangle:** The sloping edge (length 17 cm) is the hypotenuse of a right triangle formed by the perpendicular height (vertical leg) and the half diagonal (horizontal leg). Let the perpendicular height be $h$. 5. **Find the perpendicular height $h$ using Pythagoras:** $$h = \sqrt{17^2 - (5\sqrt{2})^2} = \sqrt{289 - 50} = \sqrt{239}.$$ 6. **Find the angle $\theta$ between the sloping edge and the base:** This angle is between the sloping edge (hypotenuse) and the base (horizontal leg). Using cosine, $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5\sqrt{2}}{17}.$$ 7. **Calculate $\theta$:** $$\theta = \cos^{-1}\left(\frac{5\sqrt{2}}{17}\right) = \cos^{-1}(0.416).$$ 8. **Evaluate $\theta$:** Using a calculator, $$\theta \approx 65^\circ.$$ **Final answer:** The angle between a sloping edge and the base is approximately **65 degrees**.