Question

a city park has multiple paths around a green space. what should ∠2 be in order for path c and d to be parallel?

Explanation:

Step1: Find the sum of angles in a quadrilateral

The sum of interior angles in a quadrilateral is \(360^\circ\). Let \(\angle 2 = x\). We know two angles: \(110^\circ\) and \(130^\circ\), and the angle adjacent to \(130^\circ\) (since Path B is a straight line) is \(180^\circ - 130^\circ = 50^\circ\)? Wait, no, actually, for Path C and D to be parallel, we can use the concept of consecutive interior angles or the sum of angles around the figure. Wait, maybe a better approach: the sum of the angles around the intersection (or the quadrilateral formed) should be considered. Wait, actually, the sum of the angles in the quadrilateral (the figure with angles \(110^\circ\), \(130^\circ\), \(\angle 2\), and the angle supplementary to \(\angle 1\) or maybe the angle between Path A and Path C? Wait, no, let's re-express.

Wait, the figure has a quadrilateral with angles \(110^\circ\), \(130^\circ\), \(\angle 2\), and the angle that is supplementary to the angle adjacent to Path C. Wait, actually, when Path C and D are parallel, the sum of the interior angles on the same side should be considered, but maybe using the fact that the sum of angles in a quadrilateral is \(360^\circ\). Wait, the angles given are \(110^\circ\), \(130^\circ\), and we need to find \(\angle 2\) such that Path C and D are parallel. Wait, maybe the angle adjacent to \(130^\circ\) is \(50^\circ\) (since it's a linear pair: \(180 - 130 = 50\)), but no, let's think again.

Wait, the sum of the angles in the quadrilateral (the four angles) should be \(360^\circ\). So we have \(110^\circ + 130^\circ + \angle 2 + \text{the angle opposite to } \angle 2\)? No, maybe the angle that is supplementary to the angle between Path A and Path C. Wait, perhaps a better way: when two lines are parallel, the consecutive interior angles are supplementary. Wait, Path C and D are parallel, and Path B is a transversal? Wait, no, Path B is a straight line, Path A is another line, Path D and C are the two paths we want to be parallel.

Wait, let's consider the sum of angles around the point or the quadrilateral. The sum of interior angles in a quadrilateral is \(360^\circ\). So we have three angles: \(110^\circ\), \(130^\circ\), and the angle that is supplementary to \(\angle 2\)? No, wait, maybe the angle adjacent to \(130^\circ\) is \(50^\circ\) (linear pair), but that's not right. Wait, let's calculate:

Sum of angles in quadrilateral: \(110 + 130 + \angle 2 + (180 - \angle 1)\)? No, this is getting confusing. Wait, maybe the correct approach is: the sum of the angles \(110^\circ\), \(130^\circ\), and \(\angle 2\) and the angle that makes the total \(360^\circ\), but since Path C and D are parallel, the angle \(\angle 2\) should satisfy that the sum of the angles on one side is \(180^\circ\) or something. Wait, no, let's do the math:

\(110 + 130 + \angle 2 + x = 360\), where \(x\) is the angle adjacent to \(\angle 2\). But if Path C and D are parallel, then \(x\) should be equal to the angle opposite? No, maybe the angle \(x\) is \(180 - \angle 2\) (since they are consecutive interior angles). Wait, no, let's try:

\(110 + 130 + \angle 2 + (180 - \angle 2) = 420\), which is not \(360\). So that's wrong.

Wait, maybe the figure is a quadrilateral with angles \(110^\circ\), \(130^\circ\), \(\angle 2\), and the angle that is \(180 - \angle 1\), but \(\angle 1\) is equal to \(180 - 110 = 70^\circ\) (vertical angles or linear pair). Wait, no, \(\angle 1\) and the \(110^\circ\) angle: if Path A and Path B form a triangle? No, the figure is a quadrilateral.

Wait, let's start over. The sum of interior angles in a quadrilateral is \(360^\circ\). So we have three angles: \(110^\circ\), \(130^\circ\), and the angle that is supplementary to \(\angle 2\) (since Path C and D are parallel, consecutive interior angles are supplementary). Wait, no, if Path C and D are parallel, then the angle between Path D and Path B (which is \(130^\circ\)) and the angle between Path C and Path B should be supplementary? No, that's not right.

Wait, maybe the correct calculation is: \(360 - 110 - 130 - (180 - \angle 2) = 0\)? No, this is not working. Wait, let's think of the angles around the quadrilateral. The four angles are \(110^\circ\), \(130^\circ\), \(\angle 2\), and the angle that is \(180 - \angle 2\) (since Path C and D are parallel, so consecutive interior angles are supplementary). Wait, no, consecutive interior angles are supplementary when lines are parallel. So if Path C and D are parallel, and Path B is a transversal, then the angle between Path D and Path B (which is \(130^\circ\)) and the angle between Path C and Path B should be \(180 - 130 = 50^\circ\). But that's not related to \(\angle 2\).

Wait, maybe the figure is a triangle? No, it's a quadrilateral. Wait, the sum of the angles \(110^\circ\), \(130^\circ\), and \(\angle 2\) and the angle opposite to \(\angle 2\) should be \(360^\circ\). But if Path C and D are parallel, then the angle opposite to \(\angle 2\) is equal to \(180 - \angle 2\) (consecutive interior angles). So:

\(110 + 130 + \angle 2 + (180 - \angle 2) = 420\), which is not \(360\). So that's wrong.

Wait, maybe the angle adjacent to \(130^\circ\) is \(50^\circ\) (linear pair: \(180 - 130 = 50\)), and then the sum of \(110 + 50 + \angle 2 = 180\) (triangle)? No, that doesn't make sense.

Wait, I think I made a mistake. Let's use the formula for the sum of angles in a quadrilateral: \(360^\circ\). So we have angles \(110^\circ\), \(130^\circ\), \(\angle 2\), and the angle that is supplementary to \(\angle 2\) (because Path C and D are parallel, so they are consecutive interior angles, hence supplementary). Wait, no, consecutive interior angles are between the two parallel lines and the transversal. So if Path C and D are parallel, and Path A is the transversal, then \(\angle 2\) and the \(110^\circ\) angle would be consecutive interior angles, so they should be supplementary. But \(180 - 110 = 70\), which is not matching.

Wait, maybe the correct approach is: the sum of the angles \(110^\circ\), \(130^\circ\), and \(\angle 2\) and the angle that is \(180 - \angle 2\) (since Path C and D are parallel) should be \(360^\circ\). But that gives \(110 + 130 + \angle 2 + 180 - \angle 2 = 420\), which is not \(360\). So that's wrong.

Wait, maybe the figure is a pentagon? No, it's a quadrilateral. Wait, let's look at the angles again. The angle of \(130^\circ\) is on Path B, so the adjacent angle (linear pair) is \(50^\circ\). Then, in the quadrilateral, we have angles \(110^\circ\), \(50^\circ\), \(\angle 2\), and the angle opposite to \(\angle 2\). But if Path C and D are parallel, then the angle opposite to \(\angle 2\) is equal to \(\angle 2\) (alternate interior angles). Wait, no, alternate interior angles are equal. So if Path C and D are parallel, and Path A is the transversal, then \(\angle 2\) and the angle opposite (let's say \(y\)) are equal. Then the sum of angles in the quadrilateral is \(110 + 50 + \angle 2 + y = 360\). But \(y = \angle 2\) (alternate interior angles), so \(160 + 2\angle 2 = 360\), so \(2\angle 2 = 200\), so \(\angle 2 = 100\). Wait, that makes sense. Let's check:

Linear pair with \(130^\circ\) is \(50^\circ\) (since \(180 - 130 = 50\)). Then the quadrilateral has angles \(110^\circ\), \(50^\circ\), \(\angle 2\), and \(y\) (the angle between Path A and Path C). Since Path C and D are parallel, \(\angle 2\) and \(y\) are alternate interior angles, so \(y = \angle 2\). Sum of angles in quadrilateral: \(110 + 50 + \angle 2 + \angle 2 = 360\). So \(160 + 2\angle 2 = 360\). Subtract 160: \(2\angle 2 = 200\). Divide by 2: \(\angle 2 = 100\). Yes, that works.

Step2: Calculate \(\angle 2\)

First, find the linear pair of \(130^\circ\): \(180 - 130 = 50^\circ\).
Then, use the sum of interior angles in a quadrilateral (\(360^\circ\)):
\(110 + 50 + \angle 2 + \angle 2 = 360\)
Simplify: \(160 + 2\angle 2 = 360\)
Subtract 160: \(2\angle 2 = 200\)
Divide by 2: \(\angle 2 = 100\)

Answer:

\(100\)