The angles of the triangle are 40, 55 and 85. Which side of the triangle is farther from the center of the circumscribed circle?

Let a triangle ABC be given, <A = 40 °, <B = 55 °, <C = 85 °. It is inscribed in a circle with a center O and a radius R, the distance from the center to AB is OT, to AC – OK, BC – OM.
АB⊥OT, АС⊥ОК, ВС⊥ОМ – the distance from a point to a straight line is a perpendicular.
Let’s compare triangles AСM and AСT, they are rectangular and they have a common side AC, AM = OM + R, CT = OT + R.
For triangle AFM:
AC = AM / sin C.
For triangle ACT:
AC = CT / sin A.
AM / sin C = CT / sin A.
AM / CT = sin C / sin A = 0.9962 / 0.6428 = 1.55.
OM + R / OT + R = 1.55.
OM> OT.
Let’s compare triangles ABM and AKB, they are rectangular and they have a common side AB, AM = OM + R, BK = OK + R.
For triangle ABM:
AB = AM / sin B.
For triangle AKB:
AB = BK / sin A.
AM / sin B = BK / sin A.
AM / BK = sin B / sin A = 0.8192 / 0.6428 = 1.274.
OM + R / OK + R = 1.274.
OM> OT.
Answer: the BC side is farther from the center of the circle (it is opposite the smaller angle).