+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?
+919 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! :)
+919 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! :)
