1. **Problem statement:** We have points O(0,0), A(8,6), and P(x,10) where P moves along the line $y=10$. We want to show that $$\cos \angle OPA = \frac{x^2 - 8x + 40}{\sqrt{(x^2 - 16x + 80)(x^2 + 100)}}$$ 2. **Step 1: Understand the angle $\angle OPA$.** This angle is at point P formed by points O and A. So vectors involved are $\overrightarrow{PO}$ and $\overrightarrow{PA}$. 3. **Step 2: Write vectors $\overrightarrow{PO}$ and $\overrightarrow{PA}$.** $$\overrightarrow{PO} = O - P = (0 - x, 0 - 10) = (-x, -10)$$ $$\overrightarrow{PA} = A - P = (8 - x, 6 - 10) = (8 - x, -4)$$ 4. **Step 3: Use the cosine formula for angle between two vectors:** $$\cos \theta = \frac{\overrightarrow{PO} \cdot \overrightarrow{PA}}{|\overrightarrow{PO}| |\overrightarrow{PA}|}$$ 5. **Step 4: Calculate the dot product $\overrightarrow{PO} \cdot \overrightarrow{PA}$.** $$(-x)(8 - x) + (-10)(-4) = -8x + x^2 + 40 = x^2 - 8x + 40$$ 6. **Step 5: Calculate magnitudes $|\overrightarrow{PO}|$ and $|\overrightarrow{PA}|$.** $$|\overrightarrow{PO}| = \sqrt{(-x)^2 + (-10)^2} = \sqrt{x^2 + 100}$$ $$|\overrightarrow{PA}| = \sqrt{(8 - x)^2 + (-4)^2} = \sqrt{(8 - x)^2 + 16} = \sqrt{x^2 - 16x + 64 + 16} = \sqrt{x^2 - 16x + 80}$$ 7. **Step 6: Substitute into cosine formula:** $$\cos \angle OPA = \frac{x^2 - 8x + 40}{\sqrt{(x^2 - 16x + 80)(x^2 + 100)}}$$ This matches the expression to be shown. **Final answer:** $$\boxed{\cos \angle OPA = \frac{x^2 - 8x + 40}{\sqrt{(x^2 - 16x + 80)(x^2 + 100)}}}$$