Question

look at this diagram:
if $overleftrightarrow{oq}$ and $overleftrightarrow{rt}$ are parallel lines and $m\angle ops = 110^\circ$, what is $m\angle rsu$?

Explanation:

Step1: Identify the relationship between angles

Since \( OQ \parallel RT \) and \( UN \) is a transversal, \( \angle OPS \) and \( \angle RSU \) are same - side interior angles? Wait, no. Wait, actually, \( \angle OPS \) and \( \angle RSU \) are alternate - exterior angles? Wait, no, let's look again. \( OQ \) and \( RT \) are parallel, and \( UN \) is a transversal. \( \angle OPS \) and \( \angle RSU \): we know that consecutive interior angles are supplementary, but also, \( \angle OPS \) and \( \angle RSU \) are actually alternate - exterior? Wait, no, let's think about linear pairs and parallel lines.

Wait, \( \angle OPS \) and \( \angle SPQ \) are supplementary (linear pair), but \( OQ\parallel RT \), so \( \angle RSU \) and \( \angle SPQ \) are equal (corresponding angles). Wait, \( m\angle OPS = 110^{\circ} \), so \( m\angle SPQ=180^{\circ}- 110^{\circ}=70^{\circ} \) (since they are linear pair). And since \( OQ\parallel RT \), \( \angle RSU=\angle SPQ \) (corresponding angles). So \( m\angle RSU = 70^{\circ} \).

Wait, another way: \( \angle OPS \) and \( \angle RSU \): since \( OQ\parallel RT \), and \( UN \) is a transversal, \( \angle OPS \) and \( \angle RSU \) are same - side interior angles? No, same - side interior angles are supplementary. Wait, \( \angle OPS \) and \( \angle RSP \) would be same - side interior, but maybe I made a mistake. Wait, let's label the points. \( R---S---T \) (vertical line), \( O---P---Q \) (vertical line), and \( U---S---P---N \) (transversal). So \( \angle OPS \) is at point \( P \), between \( OQ \) and \( UN \), and \( \angle RSU \) is at point \( S \), between \( RT \) and \( UN \). Since \( OQ\parallel RT \), \( \angle OPS \) and \( \angle RSU \) are actually alternate - exterior angles? No, alternate - exterior angles are equal. Wait, \( \angle OPS \) is 110 degrees, and \( \angle RSU \): if we consider that \( \angle OPS \) and \( \angle RSU \) are supplementary? No, wait, no. Wait, \( \angle OPS \) and \( \angle RSU \): let's use the fact that when two parallel lines are cut by a transversal, consecutive interior angles are supplementary. Wait, \( \angle OPS \) and \( \angle RSP \) are consecutive interior angles, but \( \angle RSP \) and \( \angle RSU \) are vertical angles? No, \( \angle RSU \) and \( \angle RSP \): no, \( \angle RSU \) is adjacent to \( \angle RSP \)? Wait, maybe a better approach: \( \angle OPS \) and \( \angle RSU \): since \( OQ\parallel RT \), the angle \( \angle OPS \) and \( \angle RSU \) are supplementary? No, that can't be. Wait, I think I messed up the angle relationship. Let's use the linear pair first. \( \angle OPS \) and \( \angle NPQ \) are vertical angles? No, \( \angle OPS \) and \( \angle SPQ \) are linear pair, so \( m\angle SPQ = 180 - 110=70^{\circ} \). Then, since \( OQ\parallel RT \), \( \angle RSU=\angle SPQ \) (corresponding angles), because \( SPQ \) and \( RSU \) are in corresponding positions relative to the parallel lines \( OQ \) and \( RT \) and transversal \( UN \). So \( m\angle RSU = 70^{\circ} \).

Step 1: Find the supplementary angle of \( \angle OPS \)

We know that \( \angle OPS \) and \( \angle SPQ \) form a linear pair, so they are supplementary. The formula for supplementary angles is \( m\angle OPS + m\angle SPQ=180^{\circ} \).

Given \( m\angle OPS = 110^{\circ} \), we can solve for \( m\angle SPQ \):

\( m\angle SPQ=180^{\circ}-m\angle OPS \)

Substitute \( m\angle OPS = 110^{\circ} \) into the formula:

\( m\angle SPQ = 180^{\circ}-110^{\circ}=70^{\circ} \)

Step 2: Use the property of parallel lines (corresponding angles)

Since \( OQ\parallel RT \) and \( UN \) is a transversal, \( \angle RSU \) and \( \angle SPQ \) are corresponding angles. Corresponding angles formed by a transversal cutting two parallel lines are equal.

So \( m\angle RSU=m\angle SPQ \)

Since \( m\angle SPQ = 70^{\circ} \), then \( m\angle RSU = 70^{\circ} \)

Answer:

\( 70^{\circ} \)