+0  
 
0
3
2
avatar+432 

Altitudes AD and BE of triangle ABC intersect at H . If angle AHB = 120 degrees and angle BAC = 52 degrees, then what is angle AHC?

 Jun 25, 2024
 #1
avatar+1075 
+1

I'm not sure if my logic is correct, but I'll give it a try. 

The point where the 3 altitudes of a triangle meet is the orthocenter. H is the orthocenter in this problem. 

Now, let's extend segment CH until it meets AB. Name this point F. 

 

CF is the altitude now. 

Let's note that \(\angle CFA = 90^\circ\) since CF is now the altitude. 

 

We can write the equation \(\angle ACF = 180 - 90 - 52\) through triangle CAF, so we have \(\angle ACF = 38\)

 

Through triangle EHC, we have \(\angle CHE = 180 - 38 - 90 = 52\). Since angle CHE is equal to angle FHB, angle FHB is also 52 degrees.

 

Now, we can write the equation \(\angle BAC + \angle CFA + \angle BEA + \angle EHF = 360\) since it forms a quadrilateral. 

 

Plugging in what we have, we get that

\(52+90+90+\angle EHF = 360 \\ \angle EHF = 128\)

 

Now, angle EHF and angle AHB share angle AHF. Since FHB is 52, we have that angle AHF is 120 - 52 = 68 degrees. 

EHA is now 128- 68 = 20 degrees. 

 

Thus, angle AHC is equal to 20 + 52 = 72 degrees. 

 

So our answer is 72. 

I'm not sure if this is right....

 

Thanks! :)

 Jun 25, 2024
 #2
avatar+1075 
+1

Sorry, there was a slight error in my math. 

We have angle EHA as \(128-68=60\) degrees. 

 

Therefore, our answer iss

\(60+52=112\) degrees. 

 

Still not sure if this is right...

 

Thanks! :)

NotThatSmart  Jun 25, 2024

2 Online Users

avatar
avatar