+1418 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 $\angle TSU = 60^\circ$ and $\angle STU = 45^\circ$, then what is $\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM$ in degrees?
0 \(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM= \) what we need to find.
Lets rearrange what we need to find:
\((\angle ZNM + \angle ZMN + \angle NZM) + (\angle PNM + \angle PMN + \angle NPM)\) we can see they are angles of a triangle so:|
\(\boxed{180+180=360^\circ}\)
0 \(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM= \) what we need to find.
Lets rearrange what we need to find:
\((\angle ZNM + \angle ZMN + \angle NZM) + (\angle PNM + \angle PMN + \angle NPM)\) we can see they are angles of a triangle so:|
\(\boxed{180+180=360^\circ}\)
