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In triangle ABC, altitudes AD, BE, and CF intersect at the orthocenter H. If angle ABC = 49 and angle ACB = 12, then find the measure of BHC, in degrees. 

 Aug 23, 2018
edited by Olpers  Aug 23, 2018

Best Answer 

 #1
avatar+374 
+2

We have that angle \(ABC = 49\), and that \(ACB = 12\), so we can see that \(BAC = 119\)\(FHE = BHE\) and we can find \(FHE\) by calculating the angle degrees in \(FHEA\), by doing \(360 - 90(AFH) - 90(AEH) - 119(FAE) = 61 = FHE = BHC\), so your answer would be \(61\).

 

- Daisy

 Aug 23, 2018
edited by dierdurst  Aug 23, 2018
 #1
avatar+374 
+2
Best Answer

We have that angle \(ABC = 49\), and that \(ACB = 12\), so we can see that \(BAC = 119\)\(FHE = BHE\) and we can find \(FHE\) by calculating the angle degrees in \(FHEA\), by doing \(360 - 90(AFH) - 90(AEH) - 119(FAE) = 61 = FHE = BHC\), so your answer would be \(61\).

 

- Daisy

dierdurst Aug 23, 2018
edited by dierdurst  Aug 23, 2018
 #2
avatar+96 
+1

Thank you!!! smiley

Olpers  Aug 23, 2018
edited by Olpers  Aug 23, 2018
 #3
avatar+374 
+2

Haha, no problem!

 

- Daisy

dierdurst  Aug 23, 2018
 #4
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0

Can any of u guys help the problem that is P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) . It is the only problem posted today not answered. Here is the link: https://web2.0calc.com/questions/heeeeelp-asap-this-is-due-tomorrow

Guest Aug 23, 2018
 #5
avatar+98090 
+1

Thanks, Daisy...!!!

 

Here's a pic of the situation :

 

 

cool cool cool

 Aug 23, 2018
 #6
avatar+96 
0

Thank you! The picture is very clear!

Olpers  Aug 24, 2018

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