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 use the property of altitudes in a triangle.
First, let's visualize the triangle $\triangle XYZ$:
```
Z
/ \
/ \
D /_____\
/ \ / \
/ \ / \
X_____\_____\ Y
H E
```
Since $\overline{XD}$ and $\overline{YE}$ are altitudes, $\angle XDH = \angle EHY = 90^\circ$.
We are given that $\angle YXH = 26^\circ$. Since $\overline{XD}$ is an altitude, $\angle YXH = 90^\circ - \angle HXD$.
Therefore, we have $90^\circ - \angle HXD = 26^\circ$, which we can simplify to $\angle HXD = 90^\circ - 26^\circ = 64^\circ$.
Now, let's consider the triangle $\triangle XHZ$:
```
Z
/ \
/ \
D /_____\
/ \ / \
X__________\ Y
H E
```
Here, we know that $\angle XHZ = 90^\circ$ since $\overline{XD}$ is an altitude.
We can find $\angle HZX$ by subtracting the sum of the other two angles in $\triangle XHZ$ from $180^\circ$:
$\angle HZX = 180^\circ - \angle XHZ - \angle HXD$
$\angle HZX = 180^\circ - 90^\circ - 64^\circ$
$\angle HZX = 26^\circ$
Therefore, $\angle HZX$ is equal to $26^\circ$.