+0  
 
0
51
1
avatar

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.

Guest Apr 2, 2018
Sort: 

1+0 Answers

 #1
avatar+85958 
+1

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

CPhill  Apr 2, 2018

22 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details