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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?

 Aug 8, 2024
 #1
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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! :)

 Aug 8, 2024
edited by NotThatSmart  Aug 8, 2024

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