+1234 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?
+195 Since ∠TSU=60∘ and ∠STU=45∘, then ∠SUT=75∘. Triangle STU is a 45-45-90 triangle, so SU=UT=2⋅ST. Also, since M and N are midpoints, then SM=MT=NU=UT=2⋅ST and TN=NS=US=2⋅ST.
Triangle UMN is isosceles with UM=MN because they are both half of sides of an isosceles triangle. Similarly, triangle PNM is isosceles with PN=MN. Since ∠SUT=75∘, then ∠SUM=∠SUN=(180∘−75∘)/2=52.5∘. Likewise, ∠PNM=∠PMN=52.5∘.
Looking at triangle ZNM, we see that ∠ZNM+∠ZMN+∠MZN=180∘. Since ∠MZN is a right angle, then ∠ZNM+∠ZMN=90∘. Similarly, for triangle PNM, we have ∠PNM+∠PMN=90∘.
Adding all the angle measures together, we get the answer: $ \angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM = 90^\circ + 90^\circ + 52.5^\circ + 52.5^\circ + \angle NZM + \angle NPM. $
Since ∠NZM and ∠NPM are supplementary angles (they form a straight line), then ∠NZM+∠NPM=180∘. Therefore, the final answer is $ 90^\circ + 90^\circ + 52.5^\circ + 52.5^\circ + 180^\circ = \boxed{465^\circ}$.
