### Instruction: In right triangle $ABC$ with $\angle A = 90^\circ$, we have $AB = 6$ and $BC = 10$.  Find $\cos C$.### Response: I need to find the cosine of the angle at $C$, which is the ratio of the adjacent side to the hypotenuse.
The adjacent side is $AC$, which I can find using the Pythagorean theorem.
$AC^2 + AB^2 = BC^2$, so $AC^2 = BC^2 - AB^2 = 10^2 - 6^2 = 64$.
Taking the square root of both sides, I get $AC = \sqrt{64} = 8$.