I want an explantion with an answer because I do want to understand plz!

Let M, N, and P be the midpoints of sides \(\overline{ZX}\), \(\overline{XY}\), and \(\overline{YS}\) of triangle YSX, respectively. Let \(\overline{XT} \) be an altitude of the triangle. If \(\angle ZYX = 71^\circ\), \(\angle YZU = 36^\circ\), and \(\angle ZXY = 73^\circ\), then what is \(\angle NTM + \angle NPM \) in degrees?

Thank you in advance

Guest May 6, 2020

#1**+2 **

**Let M, N, and P be the midpoints of sides\( \overline{ZX}\), \(\overline{XY}\), and \(\overline{YZ}\) of triangle YZX, respectively. Let \(\overline{XT}\) be an altitude of the triangle. If \(\angle ZYX = 71^\circ\), \(\angle YZX = 36^\circ\), and \(\angle ZXY = 73^\circ\), then what is \(\angle NTM + \angle NPM\) in degrees?**

\(\text{we have $\overline{XY} \parallel \overline{MP} $ and $\overline{XZ} \parallel \overline{NP}$} \\ \text{so $\angle YZX = \angle NPY$ and $\angle ZYX = \angle MPZ$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{\angle NPM} = \ ? \\ \hline \angle NPM &=& 180^\circ - (36^\circ+71^\circ) \\ \angle NPM &=& 180^\circ - 107^\circ \\ \mathbf{\angle NPM} &=& \mathbf{73^\circ} \\ \hline \end{array}\)

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

\(\text{$\triangle MTZ$ is isosceles, so $\angle TZM = \mathbf{\angle MTZ = 36^\circ}$ }\)

\(\begin{array}{|rcll|} \hline \angle NTM + \angle NPM &=& 73^\circ+73^\circ \\ \mathbf{\angle NTM + \angle NPM} &=& \mathbf{146^\circ} \\ \hline \end{array}\)

heureka May 6, 2020