Find the angles of a right-angled triangle, legs and hypotenuse of which are equal to 1 and √2.

Dan △ ABC: ∠C = 90 °, AC = 1 and BC – legs, AB = √2 – hypotenuse.

The sine of an acute angle of a right triangle is the ratio of the length of the leg opposite to the given angle to the length of the hypotenuse.

Find the sine ∠B (∠B lies opposite the leg AC):

sin∠B = AC / AB = 1 / √2.

Let’s get rid of the irrationality in the denominator:

sin∠B = 1 / √2 * √2 / √2 = (1 * √2) / (√2²) = √2 / 2.

∠B = 45 °.

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

∠A + ∠B + ∠C = 180 °;

∠A + 45 ° + 90 ° = 180 °;

∠A = 180 ° – 135 °;

∠A = 45 °.

Answer: ∠A = 45 °, ∠B = 45 °, ∠C = 90 °.