In \( \triangle M N O, \overline{O M} \cong \overline{N O} \) and \( \mathrm{m} \angle N=19^{\circ} \). Find \( \mathrm{m} \angle O \).

In triangle \( \triangle MNO \), since \( \overline{OM} \cong \overline{NO} \), it is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, we can denote \( \mathrm{m} \angle O \) as \( x \). Both angles \( \mathrm{m} \angle M \) and \( \mathrm{m} \angle O \) will be equal (since \( \overline{OM} \) and \( \overline{NO} \) are congruent). Using the triangle sum property, we know that the sum of the angles in a triangle is \( 180^\circ \). Thus, we can set up the equation: \[ \mathrm{m} \angle O + \mathrm{m} \angle M + \mathrm{m} \angle N = 180^\circ \] \[ x + x + 19^\circ = 180^\circ \] \[ 2x + 19^\circ = 180^\circ \] Now, subtract \( 19^\circ \) from both sides: \[ 2x = 180^\circ - 19^\circ \] \[ 2x = 161^\circ \] Now, divide by 2: \[ x = \frac{161^\circ}{2} = 80.5^\circ \] Thus, \[ \mathrm{m} \angle O = 80.5^\circ \]