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?
To solve this problem, we can use the properties of intersecting altitudes in an acute triangle.
Since AD and BE are altitudes of triangle ABC, we know that AH is perpendicular to BC and BH is perpendicular to AC.
First, let's find angle ACH. Angle AHB is given as 128 degrees, and since AH is perpendicular to BC, angle AHB is a right angle. Therefore, angle AHC is 180 - 128 = 52 degrees.
Next, let's find angle BHA. Angle BHA is given as 28 degrees, and since BH is perpendicular to AC, angle BHA is a right angle. Therefore, angle BHC is 180 - 28 = 152 degrees.
Now, we can find angle HCA. Angle HCA is equal to 180 - angle ACH - angle BHC = 180 - 52 - 152 = 180 - 204 = -24 degrees.
However, since angle HCA is an angle in a triangle, it must be a positive value. Therefore, we can find angle HCA by taking the supplement of -24 degrees: 180 - (-24) = 204 degrees.
Therefore, angle HCA is equal to 204 degrees.