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?
the answer should be approx 37 degrees.
Whats the exact answer??
To find the measure of $\angle HZX$, we can make use of the fact that the sum of the angles in a triangle is $180^\circ$.
First, let's draw a diagram to better understand the problem.
[Insert diagram here]
Since $\overline{XD}$ and $\overline{YE}$ are altitudes of $\triangle XYZ$, we know that $\overline{XD} \perp \overline{YZ}$ and $\overline{YE} \perp \overline{XZ}$.
The fact that $\angle YXH = 26^\circ$ tells us that $\overline{YX}$ and $\overline{XH}$ create a $26^\circ$ angle.
Since $\overline{XD}$ is an altitude of $\triangle XYZ$, we know that $\overline{XD} \perp \overline{YZ}$. This means that $\angle YXH$ and $\angle DXH$ are complementary angles. Therefore, $\angle DXH = 90^\circ - 26^\circ = 64^\circ$.
Now, we can examine $\triangle HXZ$. We know that $\angle DXH = 64^\circ$, and since $\overline{XD}$ is perpendicular to $\overline{YZ}$, we can conclude that $\angle HZX$ is also $64^\circ$. This is because $\angle HZX$ and $\angle DXH$ are corresponding angles.
Therefore, $\angle HZX = 64^\circ$.
Sally, if you don't stop this cheating, we will have to deactivate your AoPS account. Why don't you ask your peers for some help on our message boards?
~AoPS