Let $M$, $N$, and $P$ be the midpoints of sides $\overline{TU}$, $\overline{US}$, and $\overline{ST}$ of triangle $STU$, respectively. Let $\overline{UZ}$ be an altitude of the triangle. If $\angle TSU = 71^\circ$, $\angle STU = 36^\circ$, and $\angle TUS = 73^\circ$, then what is $\angle NZM + \angle NPM$ in degrees?
To find the values of $\angle NZM$ and $\angle NPM$, we need to determine the measures of angles $\angle NZM$ and $\angle NPM$.
First, let's analyze the given information. We have $\angle TSU = 71^\circ$, $\angle STU = 36^\circ$, and $\angle TUS = 73^\circ$.
To find the measure of angle $\angle NZM$, we need to consider the fact that $M$ is the midpoint of $\overline{TU}$. Since $M$ is the midpoint, we can conclude that $\overline{NZ}$ is parallel to $\overline{TS}$ and $NZ = \frac{1}{2}TS$.
In triangle $STU$, we have the equation $\angle TSU + \angle STU + \angle TUS = 180^\circ$. Substituting the given values, we have $71^\circ + 36^\circ + 73^\circ = 180^\circ$. Simplifying, we get $180^\circ = 180^\circ$, which is true.
Since $NZ$ is parallel to $TS$, we can conclude that $\angle NZM = \angle TSU = 71^\circ$.
Now let's find the measure of angle $\angle NPM$. We know that $N$ is the midpoint of $\overline{US}$. Similarly to before, we have $PM$ parallel to $ST$ and $PM = \frac{1}{2}ST$.
Using the same equation from triangle $STU$, we have $\angle TSU + \angle STU + \angle TUS = 180^\circ$. Since we already know the values of $\angle TSU$ and $\angle STU$, we can substitute them: $71^\circ + 36^\circ + \angle TUS = 180^\circ$. Simplifying, we find $\angle TUS = 73^\circ$.
Since $PM$ is parallel to $ST$, we can conclude that $\angle NPM = \angle TUS = 73^\circ$.
Finally, to find the sum of $\angle NZM + \angle NPM$, we add the values: $\angle NZM + \angle NPM = 71^\circ + 73^\circ = 144^\circ$.
Therefore, $\angle NZM + \angle NPM = 144^\circ$.
∆STU ~ NTM ~ ∆SNP ~ ∆PMU
See where that takes you