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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^he was probably searching up the answers too
To find the measure of $\angle HZX$, we need to consider the relationship between the angles of triangle $\triangle XYZ$.
First, let's draw a diagram to help visualize the problem. We have triangle $\triangle XYZ$ with altitudes $\overline{XD}$ and $\overline{YE}$ intersecting at point $H$. Angle $\angle YXH$ is given as $26^\circ$, and the angle formed by the intersection of the altitudes is $123^\circ$.
Since $\overline{XD}$ and $\overline{YE}$ are altitudes, they are perpendicular to the sides $\overline{YZ}$ and $\overline{XZ}$, respectively. This means $\angle XYZ = \angle YXH = 26^\circ$.
The sum of all angles in a triangle is $180^\circ$, so we can find $\angle HZX$ by subtracting the angles we know from $180^\circ$:
$\angle HZX = 180^\circ - \angle XYZ - \angle YXH$
Substituting the known values, we have:
$\angle HZX = 180^\circ - 26^\circ - 26^\circ$
Calculating the expression, we get:
$\angle HZX = 128^\circ$
Therefore, $\angle HZX$ measures $128^\circ$.