We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
724
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.

 Apr 2, 2018
 #1
avatar+101872 
+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

 Apr 2, 2018

11 Online Users

avatar