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Altitudes AD and BE of triangle ABC intersect at H . If angle BAC = 54 degrees and angle ABC = 52 degrees, then what is angle AHB?

 May 27, 2020
 #1
avatar+21953 
0

Since  angle(BAC) = 54o  and  angle(ABC) = 52o,  angle(BCA)  =  180o - 54o - 52o  =  74o.

 

Look at quadrilateral(DCEH):

  --  since AD is an altitude, angle(CDA) = 90o

  --  since BE is an altitude, angle(CEB) = 90o 

  --  since there are 360o in a quadrilateral:  angle(CDA) + angle(CEB) + angle(BCA) + angle(DHE)  =  360o

       --  substituting, we get angle(DHE) = 106o

 

angle(AHB) and angle(DHE) are vertical angles   --->   angle(DHE)  =  106o

 May 28, 2020
 #2
avatar+1326 
+1

∠AHB = ∠EHD

 

∠DCE = 180° - 54° - 52° = 74°

 

To find the angle EHD, we can use the interior angles of a quadrilateral EHDC ( 360° )

 

∠EHD = 360° - ( 90° + 74° + 90° )   smiley

 May 28, 2020

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