Find the acute corners of a right-angled triangle if one of them is 8 times smaller than the other.

Let △ ABC (∠C = 90 °) be given. Let’s denote acute angles ∠A and ∠B as x and y. By hypothesis, ∠A is 8 times less than ∠B, then ∠B = 8 * ∠A (y = 8 * x).

By the theorem on the sum of the angles of a triangle:

∠A + ∠B + ∠C = 180 °;

∠A + ∠B + 90 ° = 180 °;

∠A + ∠B = 180 ° – 90 °;

∠A + ∠B = 90 °.

Therefore, x + y = 90.

Let’s compose a system of linear equations:

x + y = 90;

y = 8 * x.

Substitute y into the first equation:

x + 8 * x = 90;

9 * x = 90;

x = 90/9;

x = 10.

Find the value of y:

y = 8 * x = 8 * 10 = 80.

Thus, ∠A = x = 10 °, ∠B = y = 80 °.

Answer: ∠A = 10 °, ∠B = 80 °.