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 $\angle TSU = 71^\circ$, $\angle STU = 36^\circ$, and $\angle TUS = 73^\circ$, then what is $\angle NZM + \angle NPM$ in degrees?

<NPM + <NZM = 146°

Step-by-step explanation:

Given:

In triangle STU: M, N and P are the midpoints of the line TU, US and ST respectively.

Line UZ  is the altitude of the triangle STU

<TSU =71°, <TSU = 36°, <TUS = 73°

From the diagram:

N is the midpoint of line SU and M is the mid point of line UT.

∴ Line NM is parallel to line ST

P is the mid point of  line ST and M is the mid point of line UT

∴Line PM is parallel to line SU

N is the mid point of  line SU and P is the mid point of line ST

∴Line NP is parallel to line UT

Δ SPN = Δ STU = 36°

<SPN + <NPM + <MPT (Sum of angles in a triangle = 180°)

<UST = <MPT = 71°

36° + <NPM + 71° = 180°

<NPM + 107° = 180°

<NPM= 180°-107°

<NPM= 73°

In ΔSNZ, line SN = line NZ

∴ side SN = side NZ

< NSZ = <NZS = 71°

<MTZ = <MZT = 36°

<NZS + <NZM <MZT = 180°  (Sum of angles in a triangle = 180°)

71° + <NZM + 36° = 180°

107° + <NZM = 180°

<NZM = 180° - 107°

<NZM = 73°

<NPM + <NZM =73° + 73°

<NPM + <NZM = 146°

1/2*qr*rs

step-by-step explanation:

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