Altitudes AD and BE of acute triangle ABC intersect at point H. If AHB = 128^ degrees and BAH = 28 degrees, then what is HCA in degrees?
Since AD is the altitude of triangle ABC and BAH is also an altitude, we have HAB = 90° - BAH = 90° - 28° = 62°.
Similarly, we have HB = HD = HC, so triangle HBC is isosceles.
Therefore, HCB = HBC = (180° - 128°)/2 = 52°.
Finally, we have HCA = 180° - (62° + 52°) = 180° - 114° = <<180-114=66>>66°. Answer: \boxed{66}.