Question

1. find m∠1 and m∠2. justify each answer.

Explanation:

Step1: Identify Parallel Lines and Angles

From the diagram, \( AB \parallel DC \) and \( AD \) and \( BC \) are transversals. For \( \angle 1 \): \( \angle A = 80^\circ \), and \( \angle A \) and \( \angle 1 \) are same - side interior angles? Wait, no, actually, since \( AB \parallel DC \) and \( AD \) is a transversal, but also, let's look at the angle adjacent to \( 70^\circ \). The angle at \( B \) inside the quadrilateral: the linear pair with \( 70^\circ \) is \( 180 - 70=110^\circ \)? Wait, no, first, let's consider the properties of parallel lines.

Wait, actually, \( AB \parallel DC \), so \( \angle A + \angle 1 = 180^\circ \)? No, wait, \( AD \) and \( BC \): Wait, the arrows show \( AB \parallel DC \) and \( AD \) and \( BC \) are sides. Wait, maybe it's a trapezoid with \( AB \parallel DC \). So for \( \angle 2 \): the angle at \( B \) outside is \( 70^\circ \), so the alternate - interior angle with \( \angle 2 \) would be related? Wait, no, the angle adjacent to \( 70^\circ \) on the straight line \( AB \) (extended) is \( 180 - 70 = 110^\circ \)? No, let's start over.

Since \( AB \parallel DC \), the consecutive interior angles between \( AB \) and \( DC \) with transversal \( AD \): \( \angle A \) and \( \angle 1 \): Wait, \( \angle A = 80^\circ \), and if \( AB \parallel DC \), then \( \angle A + \angle 1 = 180^\circ \)? No, that would be if they are same - side interior angles. Wait, \( AD \) is a side, so \( AB \) and \( DC \) are parallel, \( AD \) is a transversal, so \( \angle A \) and \( \angle 1 \) are same - side interior angles, so \( \angle A+\angle 1 = 180^\circ \)? Wait, no, \( \angle A = 80^\circ \), so \( \angle 1=180 - 80 = 100^\circ \)? Wait, no, maybe I got the direction wrong.

Wait, for \( \angle 2 \): the angle at \( B \) is \( 70^\circ \) (the exterior angle). The angle inside the quadrilateral at \( B \) is \( 180 - 70=110^\circ \)? No, wait, the angle marked \( 70^\circ \) is adjacent to \( BC \) on the line \( AB \) (the extension). So the angle between \( AB \) and \( BC \) inside the quadrilateral is \( 180 - 70 = 110^\circ \)? No, that's not right. Wait, actually, the angle of \( 70^\circ \) and \( \angle 2 \) are alternate - interior angles? Wait, \( AB \parallel DC \), and \( BC \) is a transversal. So the alternate - interior angle to \( \angle 2 \) is the angle equal to \( 70^\circ \)? No, that can't be. Wait, no, the angle of \( 70^\circ \) is on the line \( AB \) (the upper parallel line), and \( \angle 2 \) is on the line \( DC \) (the lower parallel line), with transversal \( BC \). So they are alternate - interior angles, so \( \angle 2 = 70^\circ \)? Wait, no, that would be if they are alternate. Wait, no, the angle of \( 70^\circ \) is an exterior angle. Wait, let's use the property of parallel lines: if \( AB \parallel DC \), then \( \angle 2 \) and the \( 70^\circ \) angle are alternate - interior angles, so \( m\angle 2=70^\circ \)? No, that's not correct. Wait, no, the angle adjacent to \( 70^\circ \) on the straight line is \( 180 - 70 = 110^\circ \), and that angle and \( \angle 2 \) are same - side interior angles? No, I think I made a mistake.

Wait, let's look at the quadrilateral \( ABCD \) with \( AB \parallel DC \). So it's a trapezoid. In a trapezoid, consecutive angles between the parallel sides are supplementary. So for \( \angle A \) and \( \angle 1 \): \( \angle A = 80^\circ \), so \( \angle 1=180 - 80 = 100^\circ \) (since \( AB \parallel DC \), same - side interior angles are supplementary). For \( \angle 2 \): the angle at \( B \) outside is \( 70^\circ \), the angle inside at \( B \) is \( 180 - 70 = 110^\circ \)? No, wait, the angle of \( 70^\circ \) and \( \angle 2 \): since \( AB \parallel DC \), \( \angle 2 \) and the \( 70^\circ \) angle are alternate - interior angles? Wait, no, the transversal is \( BC \). So if \( AB \parallel DC \), then \( \angle 2 = 70^\circ \)? No, that's not. Wait, let's calculate the interior angle at \( B \): the linear pair with \( 70^\circ \) is \( 180 - 70=110^\circ \). Then, since \( AB \parallel DC \), the same - side interior angle with \( \angle 2 \) (at \( B \) and \( C \)): \( \angle B_{interior}+ \angle 2 = 180^\circ \)? No, \( \angle B_{interior}=110^\circ \), so \( \angle 2 = 70^\circ \)? Wait, no, that's a contradiction. Wait, I think I messed up the direction.

Wait, let's start again. The diagram has \( AB \parallel DC \) (arrows show parallel). So \( AB \parallel DC \), so for transversal \( BC \), the alternate - interior angles: the angle of \( 70^\circ \) and \( \angle 2 \) are alternate - interior angles, so \( m\angle 2 = 70^\circ \). For transversal \( AD \), \( \angle A = 80^\circ \), and \( \angle 1 \) and \( \angle A \) are same - side interior angles, so \( m\angle 1=180 - 80 = 100^\circ \).

Step2: Verify with Supplementary Angles

- For \( \angle 1 \): Since \( AB \parallel DC \) and \( AD \) is a transversal, \( \angle A \) and \( \angle 1 \) are same - side interior angles. By the same - side interior angles theorem, same - side interior angles are supplementary. So \( m\angle A+m\angle 1 = 180^\circ \). Given \( m\angle A = 80^\circ \), then \( m\angle 1=180 - 80=100^\circ \).
- For \( \angle 2 \): Since \( AB \parallel DC \) and \( BC \) is a transversal, \( \angle 2 \) and the \( 70^\circ \) angle are alternate - interior angles. By the alternate - interior angles theorem, alternate - interior angles are equal. So \( m\angle 2 = 70^\circ \).

Answer:

\( m\angle 1 = 100^\circ \) (supplementary to \( 80^\circ \) as \( AB \parallel DC \)), \( m\angle 2 = 70^\circ \) (alternate - interior angle to \( 70^\circ \) as \( AB \parallel DC \))