help me help

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! :)