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?

LiIIiam0216 Aug 8, 2024

#1**+1 **

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! :)

NotThatSmart Aug 8, 2024