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In triangle $ABC$, altitudes $AD$, $BE$, and $CF$ intersect at the orthocenter $H$. If $\angle ABC = 49^\circ$ and $\angle ACB = 12^\circ$, then find the measure of $\angle BHC$, in degrees.

 Apr 2, 2018
 #1
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See the pic, here :

 

 

Since ABC = 49  and ACB  = 12

Then BAC  =  180  - 49 - 12 =  119

 

Then  we have quadrilateral HFAE  

Angle HFA  = 90

Angle FAE  =  Angle BAC = 119

Angle AEH  = 90

 

So  angle FHE  = Angle BHC  =   360 - 90  - 90 - 119   =  61°

 

 

cool cool cool

 Apr 2, 2018

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