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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?

 Feb 20, 2024

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 #1
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\(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM= \) 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:|
\(\boxed{180+180=360^\circ}\)

 Feb 20, 2024
 #1
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+1
Best Answer

\(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM= \) 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:|
\(\boxed{180+180=360^\circ}\)

EnormousBighead Feb 20, 2024

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