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Altitudes $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at $H$. If $\angle BAC = 46^\circ$ and $\angle ABC = 71^\circ$, then what is $\angle AHB$?

 Apr 1, 2018
 #1
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Angle ADB  = 90°

And angle DBA = 71°

So  angle BAH = 90 - 71  =  19°

 

And angle BEA  = 90°

And angle EAB  = 46°

So  angle ABH = 90 - 46  =  44°

 

So...in triangle AHB,  angle AHB  =  180  - 19 - 44  = 117°

 

Here's a pic  :

 

 

 

cool cool cool

 Apr 1, 2018

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