In acute triangle ABC, AB = c, BC = a, CA = b, and ac = 2b. Find the numerical value of cos(A)/a + cos(C)/c.

Let the side lengths of the triangle ABC be: a=1, c=1, and b= ac/2 = 0.5

cos(A) /a + cos(C)/c=(cos75.522°/1)+(cos75.522°/1) = 0.25 + 0.25 = 0.5

What makes you think it is an isosceles triangle? How about sides a = 3, side b= 6 and side c = 4

Yoy still have: a x c =2b =3 x 4 = 2 x 6 ??

Do all 3 sides have to be different in length?