If the sides of a right-angled triangle are in A.P., then the sines of the acute angles are

Let ∠C=90∘ being greatest and B=90∘−A.The sides are a - d, a, and a + dWe have (a+d)2=(a−d)2+a2 (using Pythagoras theorem)∴ 4ad−a2=0 or a=4dHence, the sides are 3d, 4d, 5dClearly,        sin⁡A=BCAB=a−da+d=3d5d=35sin⁡B=ACAB=aa+d=4d5d=45