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 \(\angle TSU = 71^\circ\), \(\angle STU = 39^\circ\), and \(\angle TUS = 70^\circ\), then what is \(\angle NZM + \angle NPM\) in degrees?

Guest Apr 22, 2020

#2**+2 **

**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 \(\angle TSU = 71^\circ\), \(\angle STU = 39^\circ\), and \(\angle TUS = 70^\circ\), then what is \(\angle NZM + \angle NPM\) in degrees?**

\(\text{we have $\overline{US} \parallel \overline{MP} $ and $\overline{UT} \parallel \overline{NP}$} \\ \text{so $\angle UTS = \angle NPS$ and $\angle UST = \angle MPT$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{\angle NPM} = \ ? \\ \hline \angle NPM &=& 180^\circ - (39^\circ+71^\circ) \\ \angle NPM &=& 180^\circ - 110^\circ \\ \mathbf{\angle NPM} &=& \mathbf{70^\circ} \\ \hline \end{array}\)

\(\text{$\triangle NSZ$ is isosceles, so $\angle NSZ = \mathbf{\angle SZN = 71^\circ}$ }\)

\(\text{$\triangle MZT$ is isosceles, so $\angle MTZ = \mathbf{\angle TZM = 39^\circ}$ }\)

\(\begin{array}{|rcll|} \hline \angle NZM + \angle NPM &=& 70^\circ+70^\circ \\ \mathbf{\angle NZM + \angle NPM} &=& \mathbf{140^\circ} \\ \hline \end{array}\)

heureka Apr 23, 2020