+269 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 $\angle TSU = 62^\circ$ and $\angle STU = 29^\circ$, then what is $\angle TMP + \angle TUZ$ in degrees?
+9664 Note that \(\triangle TMP \sim \triangle TUS\). Then \(\angle TMP = \angle TUS = 180^\circ - \angle TSU - \angle STU = 89^\circ\).
Considering the interior angle sum in triangle TUZ gives
\(90^\circ + \angle STU + \angle TUZ = 180^\circ\\ \angle TUZ = 90^\circ - \angle STU = 61^\circ\)
Therefore,
\(\angle TMP + \angle TUZ = 89^\circ + 61^\circ = \boxed{150^\circ}\)
