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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 AHB,BAH,HBA

 

We also have that AHB=144,BAH=66

 

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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