Altitudes AD and BE of acute triangle ABC intersect at point H. If angle AHB=128 and angle BAH=28, then what is angle HCA in degrees?

Draw altitude $CF$. This will intersect at the orthocenter, which is H. This means that $\triangle AFC$ is a right triangle. 

If we know what $\angle A$ was, we can figure out what $\angle HCA$ is, because $\angle FCA = \angle HCA$ and $\angle F$ is a right angle. 

Now, note that $\triangle ADB$ is a right triangle, and since we know $\angle BAH = 28$ and $\angle ADB = 90$, $\angle ABD = 62$.

Also, we know that $\angle ABH = 24$, meaning $\angle HBD = \angle ABD - \angle ABH = 62 - 24 = 38$.

Now, $\triangle BEC$ is a right triangle, so we know that $\angle BCA = 42$.

This means that $\angle A = 180 - 62 - 42 = 76$, so $\angle HCA = \angle FCA = 180 - 90 - 76 = \color{brown}\boxed{14}$