+0

# I have seen other questions like this but not the same thing

0
84
2

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?

May 6, 2020

#1
+25451
+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?

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

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

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

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

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

May 6, 2020
#2
+1

WOW thx so Much!! I like the way you explaned it step by step! Thank you agian :D

Guest May 6, 2020