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

Let ABC be the right triangle, right angled at B. Let the sides of the triangle be a-d, a and a + d, with a > d >0. Clearly, AC =a + d is the largest side. So, a + d is the side opposite to the right angle.By Pythagoras theorem, we have(a+d)2=a2+(a−d)2⇒4ad=a2⇒a=4dNow, sin⁡A=BCAC=aa+d=4d5d=45and sin⁡C=ABAC=a−da+d=4d−d4d+d=35