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In triangle $ABC,$ altitudes $\overline{AD}$ and $\overline{BE}$ intersect at $H$.  If $\angle BAC = 40^\circ$ and $\angle ABC = 40^\circ$, then what is $\angle AHB$?

 Sep 7, 2023
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avatar+128393 
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Angle C =  180  - 40 - 40  = 100°

 

Angles  ADC  and BEC  = 90°   (since AD and BE are altitudes)

 

We have quadrilateral  CEHD where the sum of  its  interior angles  = 360

 

So angle EHD =   360 - 90 - 90  - 100  =  80°

 

And angle AHB is a  vertical angle to angle EHD so it also measures 80°

 

cool cool cool

 Sep 7, 2023

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