Question

in this exercise, lines m and n are parallel. find the measure of each numbered angle. the figure is not to scale. m∠1 = ☐°

Explanation:

Step1: Find the measure of ∠7

Since the angle adjacent to 120° on the straight line is supplementary to 120°, we have \( m\angle7 = 180^\circ - 120^\circ = 60^\circ \). But wait, actually, looking at the parallel lines \( m \) and \( n \), and the transversal, we can also use corresponding angles or alternate interior angles. Wait, first, let's correct. The angle marked 120° and ∠7: since they are adjacent and form a linear pair? Wait no, the angle with 120° and ∠7: actually, ∠7 and the 120° angle are supplementary? Wait, no, let's see the diagram. The line \( n \) and the transversal: the angle of 120° and ∠7 are adjacent, so \( m\angle7 = 180 - 120 = 60^\circ \)? Wait, no, maybe I made a mistake. Wait, actually, the angle labeled 80° and the angle related to ∠1: Wait, the problem is to find \( m\angle1 \). Wait, let's re-examine.

Wait, the lines \( m \) and \( n \) are parallel. The angle of 80° and the angle adjacent to ∠1: Wait, maybe vertical angles or corresponding angles. Wait, first, let's find the angle that is supplementary to 120° for the lower line. Wait, the angle with 120° and the angle at the intersection with line \( n \): the angle adjacent to 120° (let's call it ∠8) is \( 180 - 120 = 60^\circ \). Then, since \( m \parallel n \), the angle corresponding to the 80° angle and the angle related to ∠1: Wait, maybe the angle of 80° and the angle that is vertical to ∠1? Wait, no. Wait, the angle of 80° and the angle that is adjacent to ∠1: Wait, let's look at the intersection of the two blue lines and the horizontal line \( m \). The angle of 80° and the angle (let's say ∠5) are adjacent? Wait, maybe I need to use the fact that the sum of angles around a point is 360°, but maybe simpler: the angle of 80° and the angle that is vertical to ∠1? Wait, no. Wait, the angle of 80° and the angle that is supplementary to ∠1? Wait, no, let's think again.

Wait, the problem is to find \( m\angle1 \). Let's see the diagram: the horizontal line \( m \), the other horizontal line \( n \), two blue transversals. The angle of 80° is at the intersection of the left blue transversal and \( m \). The right blue transversal makes a 120° angle with \( n \). Wait, maybe the angle of 80° and the angle that is equal to ∠1 because of vertical angles or corresponding angles. Wait, no, let's use the linear pair. Wait, the angle of 80° and the angle adjacent to it (let's say ∠3) are supplementary? No, 80° and ∠3: if they are adjacent, then \( m\angle3 = 180 - 80 = 100^\circ \)? Wait, no, maybe the angle of 80° and ∠1 are vertical angles? Wait, no, vertical angles are equal. Wait, maybe the angle of 80° and the angle that is equal to ∠1 because of the parallel lines. Wait, I think I made a mistake. Let's start over.

First, find the measure of the angle adjacent to 120° on line \( n \). Since they form a linear pair, that angle is \( 180^\circ - 120^\circ = 60^\circ \). Now, since lines \( m \) and \( n \) are parallel, the angle of 80° and the angle (let's say ∠4) are related? Wait, no, the left blue transversal and the right blue transversal form a triangle? Wait, maybe the angle of 80° and the 60° angle are part of a triangle, so the third angle (∠5) would be \( 180 - 80 - 60 = 40^\circ \)? No, that doesn't seem right. Wait, maybe the angle of 80° and ∠1 are vertical angles? Wait, no, vertical angles are opposite each other. Wait, the angle of 80° and ∠1: if the two blue lines intersect, then the angle of 80° and ∠1 are vertical angles? Wait, no, the angle of 80° is at the intersection of the left blue transversal and \( m \), and ∠1 is at the intersection of the right blue transversal and \( m \). Wait, maybe the angle of 80° and ∠1 are corresponding angles? No, because the transversals are different. Wait, maybe the angle of 80° and the angle that is supplementary to ∠1. Wait, I think I need to look at the linear pair with the 120° angle. Wait, the angle adjacent to 120° is 60°, as \( 180 - 120 = 60 \). Then, since \( m \parallel n \), the angle of 80° and the angle (let's say ∠6) are alternate interior angles? No, ∠6 is on line \( n \). Wait, maybe the angle of 80° and the angle that is equal to ∠1 because of vertical angles. Wait, I'm getting confused. Let's try a different approach.

Wait, the problem is to find \( m\angle1 \). Let's look at the intersection of the two blue lines and the horizontal line \( m \). The angle of 80° and ∠1: if we consider the straight line \( m \), the sum of angles on a straight line is 180°. Wait, no, the two blue lines intersect at a point on \( m \), forming angles. The angle of 80° and ∠1: are they vertical angles? Wait, no, vertical angles are opposite. Wait, the angle of 80° and the angle adjacent to ∠1 (let's say ∠5) are related. Wait, maybe the angle of 80° and the angle that is equal to ∠1 because of the parallel lines. Wait, the angle adjacent to 120° is 60°, as \( 180 - 120 = 60 \). Then, since \( m \parallel n \), the angle of 80° and the angle (let's say ∠4) are alternate interior angles? No, ∠4 is on \( m \). Wait, maybe the angle of 80° and the 60° angle are part of a triangle, so the third angle (∠5) is \( 180 - 80 - 60 = 40 \), but that doesn't help. Wait, maybe the angle of 80° and ∠1 are vertical angles? No, vertical angles are equal. Wait, I think I made a mistake in the diagram. Wait, the angle of 80° and ∠1: if the two blue lines intersect, then the angle of 80° and ∠1 are vertical angles? Wait, no, the angle of 80° is at the intersection of the left blue transversal and \( m \), and ∠1 is at the intersection of the right blue transversal and \( m \). Wait, maybe the angle of 80° and ∠1 are supplementary? No, that would be 100°, but that doesn't match. Wait, maybe the angle adjacent to 120° is 60°, and since \( m \parallel n \), the angle of 80° and the angle (let's say ∠6) are corresponding angles, so ∠6 is 80°, then the angle adjacent to ∠6 and the 120° angle? No, this is getting too confusing. Wait, let's check the linear pair with 120°: the angle is 60°, then since \( m \parallel n \), the angle of 80° and the angle (let's say ∠1) are related by the transversal. Wait, maybe the angle of 80° and ∠1 are vertical angles? No, vertical angles are equal. Wait, maybe the angle of 80° and ∠1 are supplementary? No, 80 + 100 = 180, but where does 100 come from? Wait, the angle of 80° and the angle adjacent to it (∠3) are supplementary, so ∠3 is 100°, then ∠1 is equal to ∠3 because they are vertical angles? Wait, yes! Vertical angles are equal. So if ∠3 is 100°, then ∠1 is also 100°? Wait, no, let's see: the angle of 80° and ∠3 are adjacent and form a linear pair, so \( m\angle3 = 180^\circ - 80^\circ = 100^\circ \). Then, ∠1 and ∠3 are vertical angles, so they are equal. Therefore, \( m\angle1 = 100^\circ \). Wait, but let's confirm with the other angle. The angle adjacent to 120° is 60°, then since \( m \parallel n \), the angle of 80° and the angle (let's say ∠6) are alternate interior angles, so ∠6 is 80°, then the angle adjacent to ∠6 and the 120° angle: no, maybe not. Wait, the key is vertical angles. The angle of 80° and ∠3 are supplementary, so ∠3 is 100°, then ∠1 is vertical to ∠3, so ∠1 is 100°. Yes, that makes sense.

Step1: Find \( m\angle3 \)

The angle of \( 80^\circ \) and \( \angle3 \) form a linear pair, so:
\( m\angle3 = 180^\circ - 80^\circ = 100^\circ \)

Step2: Find \( m\angle1 \)

\( \angle1 \) and \( \angle3 \) are vertical angles, so they are equal:
\( m\angle1 = m\angle3 = 100^\circ \)

Answer:

\( 100 \)