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Altitudes \(\overline{AD}\) and \(\overline{BE}\) of acute triangle $ABC$ intersect at point $H$. If \(\angle AHB = 123^\circ\) and \( \angle BAH = 26^\circ\), then what is \(\angle HCA\) in degrees?

 Apr 19, 2020
 #1
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+1

∠AHB = 123°

∠BAH = 26°

∠ABH = 31°

∠ABD = 64°

∠HBD = 33°

∠BHD = 57°

∠HAE = 33°

∠AHE = 57°

∠ACD = 57°

to be continued...  smiley

 Apr 19, 2020
 #2
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+1

  indecision Why is this???cheeky

Dragan  Apr 19, 2020
edited by Dragan  Apr 19, 2020
edited by Dragan  Apr 19, 2020
edited by Dragan  Apr 19, 2020
edited by Dragan  Apr 19, 2020
 #3
avatar+682 
+2

∠ABH = ∠HCA = 31°  indecision

Dragan  Apr 19, 2020

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