Question

statement reason
1 $m\angle 1 + m\angle a + \text{m}\angle c = 180^{\circ}$ the interior angle measures of a triangle sum to $180^{\circ}$.
2 $m\angle 1 = \boldsymbol{\downarrow}$ substitute and solve for $m\angle 1$.
3 $m\angle 1 + m\angle 2 = 180^{\circ}$ the angle measures of a linear pair sum to $180^{\circ}$.
4 $m\angle 2 = \boldsymbol{\downarrow}$ substitute and solve for $m\angle 2$.
5 $m\angle a + m\angle c = 100^{\circ}$ add given angle measures.
6 $m\angle 2 = m\angle a + m\angle c$ substitution
did we prove the conclusion true for every triangle or only for this specific triangle?

Explanation:

Step1: Find \( m\angle1 \)

We know from step 1 that \( m\angle1 + m\angle A + m\angle C = 180^\circ \) and from step 5 that \( m\angle A + m\angle C = 100^\circ \). Substitute \( m\angle A + m\angle C = 100^\circ \) into the first equation: \( m\angle1 + 100^\circ = 180^\circ \). Solve for \( m\angle1 \): \( m\angle1 = 180^\circ - 100^\circ = 80^\circ \).

Step2: Find \( m\angle2 \)

From step 3, \( m\angle1 + m\angle2 = 180^\circ \). We found \( m\angle1 = 80^\circ \), so substitute: \( 80^\circ + m\angle2 = 180^\circ \). Solve for \( m\angle2 \): \( m\angle2 = 180^\circ - 80^\circ = 100^\circ \). Also, from step 6, \( m\angle2 = m\angle A + m\angle C = 100^\circ \), which confirms this.

Answer:

For step 2: \( m\angle1 = \boldsymbol{80^\circ} \)
For step 4: \( m\angle2 = \boldsymbol{100^\circ} \)