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?
To find the measure of $\angle HZX$, we need to use the given information about the triangle and its altitudes.
First, let's draw the triangle $\triangle XYZ$ and label the given angles and points as described.
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Since $\overline{XD}$ and $\overline{YE}$ are altitudes of triangle $\triangle XYZ$, they are perpendicular to the sides they intersect. Therefore, $\angle YXH$ and $\angle HYX$ are both right angles.
We are given that $\angle YXH = 26^\circ$. Since $\angle YXH$ is a right angle, this means that $\angle YXH = 90^\circ - \angle HYX$. Plugging in the value we know, we have $26^\circ = 90^\circ - \angle HYX$.
Solving for $\angle HYX$, we have $\angle HYX = 90^\circ - 26^\circ = 64^\circ$.
Now, we know that $\angle HYX$ and $\angle YXE$ are supplementary angles, meaning they add up to $180^\circ$. Therefore, $\angle YXE = 180^\circ - \angle HYX = 180^\circ - 64^\circ = 116^\circ$.
Since $\overline{XD}$ and $\overline{YE}$ are altitudes, they intersect at point $H$, forming an angle of $123^\circ$. This means that $\angle YXE + \angle YXH + \angle HXZ = 180^\circ - 123^\circ = 57^\circ$.
We are looking for the measure of $\angle HZX$, which is opposite to $\angle YXE$. Therefore, we have $\angle HZX = 180^\circ - \angle YXE - \angle YXH = 180^\circ - 116^\circ - 26^\circ = 38^\circ$.
Therefore, the measure of $\angle HZX$ is $38^\circ$.