Altitudes $\overline{AD}$ and $\overline{BE}$ of acute triangle $ABC$ intersect at point $H$. If $\angle AHB = 128^\circ$ and $\angle BAH = 28^\circ$, then what is $\angle HCA$ in degrees?
Let's start by drawing a diagram of triangle ABC with altitudes AD and BE intersecting at H.
[Insert image of triangle ABC with altitudes AD and BE intersecting at H]
Since AD is an altitude, we know that angle AHD = 90 degrees. Similarly, since BE is an altitude, we know that angle BHE = 90 degrees.
We also know that angle AHB = 128 degrees and angle BAH = 28 degrees. Since the sum of the angles in triangle AHB is 180 degrees, we can use these values to find angle ABH:
angle ABH = 180 - angle AHB - angle BAH = 180 - 128 - 28 = 24 degrees
Now, let's consider triangle ABC. Since AD is an altitude, we know that angle CAD = 90 degrees. We want to find angle HCA, so let's focus on triangle HCA. We can use the fact that angles AHC and AHB are supplementary to find angle AHC:
angle AHC = 180 - angle AHB = 180 - 128 = 52 degrees
Next, we can use the fact that angles AHB and ABH are supplementary to find angle BAC:
angle BAC = 180 - angle AHB - angle ABH = 180 - 128 - 24 = 28 degrees
Finally, we can use the fact that the angles in a triangle sum to 180 degrees to find angle HCA:
angle HCA = 180 - angle BAC - angle AHC = 180 - 28 - 52 = 100 degrees
Therefore, angle HCA is 100 degrees.
