Altitudes $\overline{XD}$ and $\overline{YE}$ of acute triangle $\triangle XYZ$ intersect at point $H$. If the altitudes intersect at a $123^\circ$ angle, and $\angle YXH = 26^\circ$, then what is $\angle HZX$ in degrees?
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To find $\angle HZX$, we can utilize the property that the sum of the angles in a triangle is $180^\circ$.
First, let's label the given information on the diagram:
[asy]
pair X,Y,Z,H,D,E;
X = (-5,0);
Y = (0,8);
Z = (4,0);
D = (0,0);
E = (0,6);
H = (0,2);
draw(X--Y--Z--cycle);
draw(X--D);
draw(Y--E);
draw(Z--H);
label("$X$",X,SW);
label("$Y$",Y,N);
label("$Z$",Z,SE);
label("$D$",D,S);
label("$E$",E,NW);
label("$H$",H,NE);
[/asy]
We know that $\angle YXH = 26^\circ$. Since $\angle XHY$ and $\angle XYZ$ are right angles (as they are altitudes), we can find $\angle XYZ$ by subtracting the known angles from $180^\circ$:
$\angle XYZ = 180^\circ - \angle YXH - \angle XHY = 180^\circ - 26^\circ - 90^\circ = 64^\circ$.
Now, let's find $\angle HZX$. Since $\angle HZX$ is the remaining angle in $\triangle XHZ$, we can find it by subtracting the known angles from $180^\circ$:
$\angle HZX = 180^\circ - \angle XHZ - \angle HZX$.
However, we don't know $\angle XHZ$ directly. To find it, we can utilize the fact that the sum of the angles in a triangle is $180^\circ$. In $\triangle XYZ$, we have:
$\angle XYZ + \angle XZY + \angle YZX = 180^\circ$.
We already found that $\angle XYZ = 64^\circ$. Also, since $\angle YXH$ and $\angle ZYX$ are right angles (as they are altitudes), $\angle XZY = 90^\circ - \angle YXH = 90^\circ - 26^\circ = 64^\circ$. Substituting these values into the equation above, we have:
$64^\circ + 64^\circ + \angle YZX = 180^\circ$.
Simplifying, we get:
$128^\circ + \angle YZX = 180^\circ$.
Subtracting $128^\circ$ from both sides, we find:
$\angle YZX = 52^\circ$.
Now we can substitute this value into our earlier equation to find $\angle HZX$:
$\angle HZX = 180^\circ - 64^\circ - 52^\circ = \boxed{64^\circ}$.