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Altitudes AD and BE of acute triangle ABC intersect at point H. If angle AHB=128 and angle BAH=28, then what is angle HCA in degrees?

 Aug 11, 2022
 #1
avatar+2455 
+1

Draw altitude \(CF\). This will intersect at the orthocenter, which is H. This means that \(\triangle AFC \) is a right triangle. 

 

If we know what \(\angle A\) was, we can figure out what \(\angle HCA \) is, because \(\angle FCA = \angle HCA\) and \(\angle F\) is a right angle. 

 

Now, note that \(\triangle ADB \) is a right triangle, and since we know \(\angle BAH = 28\) and \(\angle ADB = 90\), \(\angle ABD = 62\).


Also, we know that \(\angle ABH = 24\), meaning \(\angle HBD = \angle ABD - \angle ABH = 62 - 24 = 38\).

 

Now, \(\triangle BEC\) is a right triangle, so we know that \(\angle BCA = 42\).

 

This means that \(\angle A = 180 - 62 - 42 = 76\), so \(\angle HCA = \angle FCA = 180 - 90 - 76 = \color{brown}\boxed{14}\)

 Aug 12, 2022
 #2
avatar+177 
+2

Sorry, that's wrong. Thanks for trying though.

waterlili  Aug 12, 2022

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