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# Triangle problem, last time I didn't get an explanation to the answer.

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

Apr 22, 2020

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Angle NZM + angle NPM = 148 degrees.

Apr 22, 2020
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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 $$\angle TSU = 71^\circ$$, $$\angle STU = 39^\circ$$, and $$\angle TUS = 70^\circ$$, then what is $$\angle NZM + \angle NPM$$ in degrees? $$\text{we have \overline{US} \parallel \overline{MP}  and \overline{UT} \parallel \overline{NP}} \\ \text{so \angle UTS = \angle NPS and \angle UST = \angle MPT }$$

$$\begin{array}{|rcll|} \hline \mathbf{\angle NPM} = \ ? \\ \hline \angle NPM &=& 180^\circ - (39^\circ+71^\circ) \\ \angle NPM &=& 180^\circ - 110^\circ \\ \mathbf{\angle NPM} &=& \mathbf{70^\circ} \\ \hline \end{array}$$ $$\text{\triangle NSZ is isosceles, so \angle NSZ = \mathbf{\angle SZN = 71^\circ} }$$ $$\text{\triangle MZT is isosceles, so \angle MTZ = \mathbf{\angle TZM = 39^\circ} }$$ $$\begin{array}{|rcll|} \hline \angle NZM + \angle NPM &=& 70^\circ+70^\circ \\ \mathbf{\angle NZM + \angle NPM} &=& \mathbf{140^\circ} \\ \hline \end{array}$$ Apr 23, 2020
edited by heureka  Apr 23, 2020