If the length of sides of a right triangle are in A.P., then the sines of the acute angle are
- $\dfrac{3}{5}, \dfrac{4}{5}$
- $\sqrt{\dfrac{2}{3}}$ , $\sqrt{\dfrac{1}{3}}$
- $ \sqrt{\dfrac{\sqrt{5}-1}{2}}$ , $\sqrt{\dfrac{\sqrt{5}-1}{2}}$
- $ \sqrt{\dfrac{\sqrt{3}-1}{2}}$ , $\sqrt{\dfrac{\sqrt{3}-1}{2}}$
Let the sides be a-d, a, a+d. By Pythagoras, (a-d)^2 + a^2 = (a+d)^2. a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2. a^2 = 4ad, so a = 4d. Sides are 3d, 4d, 5d. Sines of acute angles are 3/5 and 4/5.
In a right triangle where the side lengths form an arithmetic progression, the sides can be represented as x, x plus d, and x plus 2d, where x plus 2d is the hypotenuse. By applying the Pythagorean theorem, we get x squared plus (x plus d) squared equals (x plus 2d) squared, which expands and simplifies to x equals 3d. This yields a triangle with side lengths of 3d, 4d, and 5d, making it a classic 3-4-5 right triangle. The sines of the two acute angles are the ratios of the opposite sides to the hypotenuse, which calculate to 3 divided by 5 and 4 divided by 5.