+2669 Altitudes $\overline{AD}$ and $\overline{BE}$ of acute triangle $ABC$ intersect at point $H$. If $\angle AHB = 144^\circ$ and $\angle BAH = 66^\circ$, then what is $\angle HCA$ in degrees?
+1897 Let's focus on triangle AHB.
It's made up of 3 angles. We have \( \angle AHB, \angle BAH, \angle HBA\)
We also have that \(\angle AHB = 144^\circ,\angle BAH = 66^\circ\)
However, note that this would mean triangle AHB has interior angkes summing to
\(144 + 66 = 210°\)
Since a triangle can't sum to above 180, this triangle is therefore impossible and invalid.
Thanks! :)
