Explanation:
Okay, so I need to figure out the measure of angle PQR in this diagram. Let me start by trying to visualize the diagram based on the Asymptote code provided.

First, there's a horizontal line from (0,0) to (10,0), which is labeled with a MidArrow. Then, another segment from (10,0) to (20,0), also with a MidArrow. So that seems like two horizontal segments connected at (10,0), maybe forming a straight line? Wait, but the first segment is from (0,0) to (10,0), then another from (10,0) to (20,0). So that's just a longer horizontal line from (0,0) to (20,0), but with arrows on both parts? Not sure if that matters yet.

Then there are other lines: from (0,0) to (-7,10), then from (-7,10) to (7,10), then from (7,10) to (10,0). So that makes a quadrilateral? Let me think. Starting at Q (0,0), going up to P (-7,10), then to T (7,10), then down to R (10,0). Then there's a line from (-5,10) to (7,10) with a MidArrow. Hmm. The labels are: x degrees at (-6,10) which is near point P, 2x degrees at (7,10) which is point T, 128 degrees at (10,0) which is point R, and labels for points P, T, R, Q.

So, angle PQR is the angle at point Q, between points P, Q, and R. So, that's the angle at Q between QP and QR. So, I need to find angle PQR.

Given that PT is parallel to QR. Let me note that. So, PT is parallel to QR. PT is the segment from P (-7,10) to T (7,10). QR is the segment from Q (0,0) to R (10,0). Wait, but QR is along the x-axis from (0,0) to (10,0). PT is from (-7,10) to (7,10), so that's a horizontal line segment as well. So, PT is parallel to QR because both are horizontal. That makes sense.

So, PT and QR are both horizontal lines, hence parallel.

Now, the angles labeled: x degrees at (-6,10), which is near point P. Since P is at (-7,10), the label x° is near (-6,10), which is slightly to the right of P. Similarly, 2x° is at (7,10), which is point T. Then 128° is at (10,0), which is point R.

So, looking at the diagram, maybe there are some angles formed by the transversals? Since PT is parallel to QR, maybe there's a transversal cutting them, creating some corresponding angles or alternate interior angles?

Let me try to sketch mentally. The segment from P (-7,10) to T (7,10) is horizontal. Then segment from T (7,10) to R (10,0). Also, segment from Q (0,0) to P (-7,10). So, triangle PQT? Wait, but there's also a segment from Q to R, which is QR.

Perhaps there are some triangles here. Let me try to figure out the angles.

First, the angle at point P is labeled x°. Wait, but which angle? The label is at (-6,10), which is near the segment PT. Since PT is horizontal, maybe that's an angle between some lines. Let me think.

Looking at the Asymptote code: the segment from (-7,10) to (7,10) is drawn, and then there is a MidArrow on (-5,10) to (7,10). So, that's the segment PT. Then, the label x° is at (-6,10), SE, which would be southeast of that point, so maybe near the segment from P to Q? Wait, the segment from P (-7,10) to Q (0,0). So, maybe the angle at P between PT and PQ?

Similarly, the label 2x° is at (7,10), SW, so southwest of that point, which would be near the segment from T to R? Since T is connected to R (from (7,10) to (10,0)), so maybe the angle at T between PT and TR?

And then the label 128° is at (10,0), NE, which would be northeast of that point, so near the segment from R to Q? Since R is connected to Q (from (10,0) to (0,0)).

So, perhaps angle at P between PT and PQ is x°, angle at T between PT and TR is 2x°, and angle at R between QR and something is 128°? Wait, the 128° is labeled at R, which is connected to Q and T? Wait, R is connected to Q (QR) and to T (TR). So, maybe angle at R between QR and TR? But that angle is labeled as 128°, so that's angle TRQ?

But how do these angles relate? Since PT is parallel to QR, and maybe there are some transversals. For example, the segment PQ and TR might be transversals? Or TR and PQ?

Alternatively, maybe triangle PQT has some angles. Let me think again.

First, let me note all given angles. Let me denote them:

At point P: angle between PT and PQ is x°. Since PT is horizontal, going from P (-7,10) to T (7,10), and PQ goes from P (-7,10) to Q (0,0). So, angle between PT (horizontal right) and PQ (going down to Q). Similarly, at point T, angle between PT (horizontal left to right) and TR (going down to R (10,0)) is 2x°. At point R, angle between QR (horizontal left to right) and TR is 128°.

Since PT is parallel to QR, and if we can find some transversal, maybe PQ or TR? Let me try to consider the transversal TR. Since TR connects T to R, which is from (7,10) to (10,0). Let me find the slope of TR. The slope is (0 - 10)/(10 -7) = -10/3. Similarly, the line PQ connects P (-7,10) to Q (0,0), so slope is (0 - 10)/(0 - (-7)) = -10/7.

But maybe that's not helpful yet.

Alternatively, since PT is parallel to QR, and if we can find some corresponding angles. For example, if we have a transversal that cuts PT and QR, then the corresponding angles would be equal. Let me think about the transversal that connects P to R? Wait, PR? Is PR a transversal? Let me see. If we consider PR, which connects P (-7,10) to R (10,0). Then PT is parallel to QR, so maybe PR is a transversal cutting PT and QR.

Alternatively, maybe considering triangle PQT? Let me think.

But perhaps it's better to use the given angles. Let me note that at point T, angle between PT and TR is 2x°, and since PT is parallel to QR, then maybe angle between TR and QR is equal? But angle at R is given as 128°, which is angle between QR and TR? If that is the case, then maybe angle at R is corresponding to angle at T?

Wait, if PT is parallel to QR, and if TR is a transversal, then the angle at T (which is 2x°) and the angle at R (which is 128°) might be same-side interior angles? Because if TR is a transversal intersecting PT and QR, then the angles on the same side would add up to 180°?

Wait, let me think. If PT is parallel to QR, and TR is a transversal, then the angle at T (between PT and TR) is 2x°, and the angle at R (between QR and TR) is 128°. If these are same-side interior angles, then their sum would be 180°.

So, 2x + 128 = 180? Then, 2x = 52°, so x = 26°?

If that is the case, then x is 26°, and then angle at P is x = 26°, and angle at T is 2x = 52°, and angle at R is 128°.

But then, how does that help us find angle PQR?

Angle PQR is the angle at Q between PQ and QR. So, maybe if we can find other angles in triangle PQR or something.

Wait, let's consider triangle PQR. But wait, in the diagram, point T is connected to P and R, and Q is connected to P and R. Maybe triangle PQT? Wait, perhaps not. Let me think again.

Alternatively, maybe considering triangle PQT. Let me see. If we consider triangle PQT, we have angle at P is x°, angle at T is 2x°, so maybe we can find some angles? Wait, but is T connected to Q? Wait, in the Asymptote code, the segments are from Q (0,0) to P (-7,10), then P to T (7,10), then T to R (10,0). So, Q is connected to P and R. T is connected to P and R. There's no direct connection between Q and T.

So, perhaps triangle PQT is not a triangle? Wait, P, Q, T? If we connect Q to T? Wait, in the Asymptote code, there is a segment from (-5,10) to (7,10) with a MidArrow, which is the segment PT. But there is no segment from Q to T.

Alternatively, maybe triangle QTR? Q to T? Not sure.

Alternatively, maybe focusing on angles around point Q. The angle PQR is at Q, between PQ and QR. Let me think about the lines at Q. PQ goes up to P, QR goes to R. So, if we can find some angles adjacent to angle PQR.

But perhaps triangle PQR. If we can find the other angles in triangle PQR, then we can find angle PQR.

But to do that, we need more information.

Wait, maybe if we consider that PT is parallel to QR, so if we can find some relation. For example, if we can find the angles at P and T, maybe relate them to angles at Q?

Alternatively, note that PT is parallel to QR. Let me see. If we have a transversal PQ intersecting PT and QR. Wait, PQ starts at P, which is on PT, and goes to Q, which is on QR. So, PQ is a transversal that starts on PT and ends on QR.

So, at point P, the angle between PT and PQ is x°, and at point Q, the angle between PQ and QR... Wait, since PT is parallel to QR, then maybe the angle between PQ and PT at P is equal to the angle between PQ and QR at Q? But since PT is parallel to QR, then PQ would be a transversal, so the corresponding angles would be equal?

Wait, if two parallel lines are cut by a transversal, then corresponding angles are equal. So, if PT || QR and PQ is a transversal, then angle between PT and PQ at P is equal to the angle between QR and PQ at Q.

But angle between PT and PQ at P is x°, so that would mean the angle between QR and PQ at Q is also x°? But angle between QR and PQ at Q is angle PQR. Wait, is that right?

Wait, let me draw it mentally. At point P, the line PT is parallel to QR. Then PQ is a transversal. The angle between PT and PQ at P is x°, which is like an angle inside the transversal. Then, the corresponding angle at Q would be the angle between QR and QP. Because QR is parallel to PT, so the corresponding angle would be the angle between QR and the transversal PQ.

Therefore, angle PQR is equal to x°? Wait, but if that's the case, then angle PQR would be x°, which we found earlier as 26°, but then if we can check with other angles?

Wait, but if angle PQR is x°, which is 26°, then in triangle PQR, we would also need to know another angle. Maybe angle at R?

Wait, angle at R is given as 128°, which is the angle between QR and TR. But angle at R in triangle PQR would be angle QRP? Wait, maybe not. Let me clarify.

Point R is connected to Q and to T. So, angle at R between QR and TR is 128°, but angle at R in triangle PQR is angle QRP. Unless TR is colinear with something? Wait, not sure.

Alternatively, maybe triangle PQR has angles at P, Q, R. But we have angles at P (angle QPR?), and angle at R?

Wait, if angle at P is x°, angle at Q is x°, then angle at R would be 180 - x - x = 180 - 2x. But we also have another angle at R, which is 128°, which is angle TRQ. Maybe these are supplementary?

Wait, angle at R in triangle PQR is angle QRP, which is different from angle TRQ. Let me think. If we have point T connected to R, then at R, there's angle TRQ (which is 128°) and angle QRP (which is part of triangle PQR). Are these angles adjacent?

If points Q, R, and T are connected, then maybe there is a triangle or something. Let me think. If we consider line TR, which goes from T (7,10) to R (10,0), then angle between TR and QR at R is 128°, which is angle TRQ. Then, if we look at triangle TRQ, we have angles at T, R, and Q. Wait, but we don't have point Q connected to T.

Alternatively, if we can find angle QRP (angle at R in triangle PQR) in terms of x, and relate it with angle TRQ (which is 128°)?

Alternatively, since we have some information about angles at T and R. Earlier, I considered that angle at T (2x°) and angle at R (128°) are same-side interior angles. Hence, adding to 180.

If that is the case, then 2x + 128 = 180 => 2x = 52 => x=26.

So then, angle PQR would be x°, which is 26°? But wait, is that correct?

Wait, let me verify. If PT is parallel to QR, and TR is a transversal, then the same-side interior angles would be angle at T (between PT and TR) and angle at R (between QR and TR). If they are same-side interior angles, then they add to 180°, which gives 2x + 128° = 180°, so x=26°.

But then angle PQR is supposed to be x°, which would be 26°, but is that accurate?

Wait, earlier I thought that angle PQR is equal to x° because of corresponding angles. But let me confirm that.

If PQ is a transversal cutting PT (parallel to QR), then angle between PT and PQ at P (x°) corresponds to angle between QR and PQ at Q.

Yes, since PT || QR, and PQ is a transversal, then corresponding angles are equal. So, angle between PT and PQ at P is equal to the angle between QR and PQ at Q.

But angle between QR and PQ at Q is angle PQR. Therefore, angle PQR = x° = 26°.

But let me check if that's consistent with the rest of the diagram.

If angle PQR is 26°, and angle at R in triangle PQR (angle QRP) is then 180° - angle at P - angle at Q. Wait, but angle at P in triangle PQR is not necessarily x°. Wait, angle at P in triangle PQR is angle QPR. Earlier, we had angle between PT and PQ at P is x°, but angle QPR is that angle?

Yes, because PT is the line from P to T, which is horizontal, and PQ is going from P to Q. So, angle between PQ and PT is x°, which is angle QPT. But in triangle PQR, angle at P is angle QPR. However, if we consider that PT is a line from P to T, which is parallel to QR. So, is angle QPT (x°) the same as angle QPR?

Wait, unless point T lies on PR or something? But in the Asymptote code, PT is from (-7,10) to (7,10), and R is at (10,0). So, point T is not on PR.

Therefore, angle at P in triangle PQR is angle QPR, which is different from angle QPT (x°). Because angle QPT is between PQ and PT, while angle QPR is between PQ and PR. So, unless PR is aligned with PT, which it isn't.

So my earlier assumption that angle PQR is x° might be incorrect.

Therefore, I need to re-examine.

Let me think again. If PT is parallel to QR, and PQ is a transversal. Then, the corresponding angles would be angle between PT and PQ at P (which is x°) and angle between QR and PQ at Q. But what is that angle?

If QR is the line from Q (0,0) to R (10,0), and PQ is the line from Q (0,0) to P (-7,10). Then, the angle between QR and PQ at Q is indeed angle PQR. Therefore, angle PQR is equal to x°, because they are corresponding angles.

Therefore, if that is correct, then angle PQR is x°, which is 26°, as found earlier.

But then, in triangle PQR, we have angle at Q = 26°, angle at R? Let's see. If we can find angle at R.

But angle at R is given as 128°, but that is angle TRQ (angle between TR and QR). So, angle TRQ is 128°, which is different from angle QRP (angle between QR and PR in triangle PQR).

However, maybe we can find angle QRP?

Let me consider triangle TRQ. In triangle TRQ, we have angle at R is 128°, angle at T is 2x° = 52°, and angle at Q is angle TQR.

Wait, if we consider triangle TRQ, angles sum to 180°. So, angle at T (2x) + angle at R (128°) + angle at Q (TQR) = 180°.

So, angle TQR = 180° - 2x - 128° = 180° - 128° - 2x = 52° - 2x.

So angle TQR is 52° - 2x.

But angle TQR is also part of angle PQR? Wait, angle PQR is at point Q between PQ and QR. If there is a line QT? Wait, but QT is not drawn in the diagram. Wait, in the Asymptote code, there is a segment from Q (0,0) to P (-7,10), then P to T (7,10), then T to R. There's no segment from Q to T.

So angle TQR is the angle at Q between TQ and QR, but since TQ is not present, unless we assume that QT is a line? Wait, but the Asymptote code draws a segment from (-5,10) to (7,10) with a MidArrow, which is PT. The other segments are QP, PT, TR, QR. So, unless there is a line from Q to T?

Wait, actually, in the Asymptote code, the points are Q (0,0), P (-7,10), T (7,10), R (10,0). The segments are QP, PT, TR, QR. Also, there is a segment from (-7,10) to (7,10) which is PT.

Therefore, there's no segment connecting Q to T. So, angle TQR is not directly present.

But earlier, in triangle TRQ, angle at Q is angle TQR, which would require that line QT exists. However, since QT is not drawn, maybe there's an implied line? Or maybe I'm overcomplicating.

Alternatively, since we have angle TQR = 52° - 2x, and angle PQR is the angle at Q between PQ and QR. If there is a line from Q to T, then angle PQR would be split into angle PQT and angle TQR? But unless we know something about that.

Alternatively, maybe angle PQT is equal to something?

Wait, but without knowing more, perhaps there's another approach.

Earlier, I thought angle PQR is x°, which is 26°, but if we consider triangle PQR, we need to ensure that the angles add up.

But angle PQR is x° = 26°, angle at R is something, angle at P is something.

Wait, maybe considering triangle PQT?

In triangle PQT, we have angle at P is x°, angle at T is 2x°, so the third angle at Q is 180° - x° - 2x° = 180° - 3x°.

So, angle at Q in triangle PQT is 180° - 3x°.

But angle at Q in triangle PQT is angle PQT.

But in the diagram, angle at Q is also angle PQR and angle PQT? Unless points R and T are on the same side.

Wait, if we consider that line QT is drawn, then angle PQR would be part of angle PQT?

But since QT is not drawn, maybe we need to think differently.

Alternatively, if we consider that line TR is a transversal, and we found angle at Q in triangle TRQ is 52° - 2x.

But how does that relate to angle PQR?

Alternatively, maybe considering that angle PQR is equal to angle PQT + angle TQR? If there is a line QT.

Assuming that, then angle PQR = angle PQT + angle TQR.

From triangle PQT, angle PQT is 180° - 3x°, and angle TQR is 52° - 2x°.

So, angle PQR = (180 - 3x) + (52 - 2x) = 232 - 5x.

But earlier, we thought angle PQR was equal to x°. Therefore, if angle PQR is both x° and 232 - 5x, then:

x = 232 - 5x
6x = 232
x = 232 / 6 ≈ 38.666...

But earlier, we had from the same-side interior angles that x = 26°.

This inconsistency suggests that there is a mistake in my reasoning.

Hmm, this is getting complicated. Maybe my assumption that angle PQR is x° is wrong. Let me re-examine that.

If PT is parallel to QR, and PQ is a transversal, then the corresponding angles should be equal. The angle between PT and PQ at P is x°, so the corresponding angle would be the angle between QR and PQ at Q.

But which angle is that?

Imagine PQ going from P to Q, and QR going from Q to R. Since PT is parallel to QR, then the line PQ intersects PT at P and QR at Q. Therefore, the corresponding angles would be the angle between PT and PQ at P (x°) and the angle between QR and PQ at Q.

But angle between QR and PQ at Q is the angle between QR and QP, which is exactly angle PQR.

Therefore, angle PQR = x°. So that should be valid.

But then why the other approach gives inconsistency?

Let me check my calculation again.

From triangle TRQ, angles sum to 180:

angle T (2x) + angle R (128) + angle Q (TQR) = 180
So angle TQR = 180 - 2x - 128 = 52 - 2x.

Assuming angle PQR = angle PQT + angle TQR.

But angle PQT is part of triangle PQT, which has angles x°, 2x°, and angle PQT = 180 - 3x.

But if angle PQT is part of angle PQR, then angle PQR = angle PQT + angle TQR
=> angle PQR = (180 - 3x) + (52 - 2x) = 232 - 5x
But angle PQR is also x, so:

x = 232 - 5x
6x = 232
x = 232 / 6 ≈ 38.666...

But earlier, from the same-side interior angles, we found x = 26°. There's a contradiction here. Therefore, one of the assumptions is wrong.

Where is the mistake?

Possibly in the assumption that angle PQR = angle PQT + angle TQR.

If there is no segment QT, then angle PQR is not necessarily composed of angle PQT and angle TQR. Therefore, that assumption was incorrect.

Therefore, need another approach.

Let me try to consider that we have two different expressions for angle PQR. From the corresponding angles, angle PQR = x°. From triangle TRQ, angle TQR = 52 - 2x. If there's a relation between angle TQR and angle PQR?

Maybe angle PQR and angle TQR are supplementary or something?

Wait, if we consider line TR and line PQ, but without knowing more, it's hard to tell.

Alternatively, maybe using the fact that PT is parallel to QR, and using other transversals.

Let me try another approach.

Given that PT || QR, and looking at the angles formed by another transversal, say PR.

Wait, PR connects P (-7,10) to R (10,0). Let me calculate its slope. Slope is (0 - 10)/(10 - (-7)) = -10/17.

Not sure. Maybe not helpful.

Alternatively, consider triangle PQT and triangle QTR? Not sure.

Wait, another thought. Since PT is parallel to QR, then the slope of PT is 0 (since it's horizontal), and slope of QR is 0 as well. Therefore, they are parallel.

Now, consider the angles at P and T. At point P, angle between PT and PQ is x°, and at point T, angle between PT and TR is 2x°.

If I can find the measure of angle between PQ and TR or something.

Alternatively, using the fact that the sum of angles around point T?

Wait, at point T, we have angle between PT and TR is 2x°, and if we can find other angles at T.

But point T is connected to P and R. Also, there is a segment from (-5,10) to (7,10) with a MidArrow, which is PT.

Wait, maybe there are more angles given? The label of x° is near (-6,10), which is on PT. Maybe that is angle between PT and another line.

Wait, the Asymptote code draws:

draw((-7,10)--(-5,10)--(7,10)--(10,0),black+linewidth(1));

Wait, no, original code:

draw((-7,10)--(7,10)--(10,0),black+linewidth(1));

Wait, no, the original Asymptote code is:

draw((0,0)--(10,0),black+linewidth(1));
draw((0,0)--(10,0),MidArrow);
draw((10,0)--(20,0),black+linewidth(1));
draw((0,0)--(-7,10)--(7,10)--(10,0),black+linewidth(1));
draw((-5,10)--(7,10),MidArrow);
label("$x^{\circ}$",(-6,10),SE);
label("$2x^{\circ}$",(7,10),SW);
label("$128^{\circ}$",(10,0),NE);
label("$P$",(-7,10),N);
label("$T$",(7,10),N);
label("$R$",(10,0),S);
label("$Q$",(0,0),S);

So, the diagram has:

- The base line from Q(0,0) to R(10,0), then to S(20,0).

- Then from Q(0,0) to P(-7,10), then P to T(7,10), then T to R(10,0).

- The segment PT is drawn from (-7,10) to (7,10), with a MidArrow from (-5,10) to (7,10), meaning that perhaps the arrow is on the segment PT from (-5,10) to (7,10)?

- The label x° is at (-6,10), SE, so south-east of (-6,10). That would be near point P(-7,10), but slightly to the right. So, maybe the angle between PQ and PT at point P?

Similarly, label 2x° at (7,10), SW, which is at point T. So, the angle at T between PT and TR?

Label 128° at R(10,0), NE, which is northeast of R, so near the segment from R to T or R to Q?

Therefore, the diagram has:

At point P(-7,10): angle between PQ and PT is x°, which is given.

At point T(7,10): angle between PT and TR is 2x°, given.

At point R(10,0): angle between QR and... well, since TR connects to R, and QR is the base, the angle between QR and TR is 128°, which is given.

So, with those angles, and PT || QR, how can we find angle PQR?

Let me try to use the fact that PT || QR, and transversal TR creates angles.

As before, angle at T (2x°) and angle at R (128°) are on the same side of the transversal TR, so they are same-side interior angles and add to 180°.

Hence, 2x + 128 = 180 => 2x = 52 => x = 26°.

So now, we know x = 26°.

Now, to find angle PQR, which is at Q, between PQ and QR.

Let me think about triangle PQR. To find angle at Q, I need more information about the triangle. Maybe if I can find other angles or sides?

Alternatively, consider triangle PQT. In triangle PQT, we know two angles: at P is x°, at T is 2x°, so angle at Q (which is angle PQT) is 180 - x - 2x = 180 - 3x.

So angle PQT = 180 - 3x.

If we can relate angle PQT to angle PQR.

Assuming that line QT is drawn (even though it's not in the original diagram), angle PQR would be angle between PQ and QR, which might be related to angle PQT and angle TQR.

If we assume that line QT is drawn, forming angle PQT and TQR, then angle PQR = angle PQT + angle TQR.

But from before, in triangle TRQ, angle at Q is angle TQR = 180 - 2x - 128 = 52 - 2x.

So angle PQR = (180 - 3x) + (52 - 2x) = 232 - 5x.

But we found x = 26°, so substituting:

angle PQR = 232 - 5*26 = 232 - 130 = 102°.

Wait, but earlier I thought angle PQR = x = 26°, but this new result gives 102°. Which one is correct?

But where was the mistake in the previous assumption?

The assumption that angle PQR is x° was based on corresponding angles, but that might not hold if there's more to the diagram. Let me verify that again.

If PT || QR and PQ is a transversal, then corresponding angles should be equal. The angle between PT and PQ at P is x°, which should correspond to the angle between QR and PQ at Q.

But what is the angle between QR and PQ at Q? It is indeed angle PQR. So angle PQR should be x°.

However, according to the other calculation, angle PQR = 102°. This contradiction suggests that there's a flaw in one of the reasonings.

Let me try to find out.

If x = 26°, then angle PQT is 180 - 3x = 180 - 78 = 102°, and angle TQR is 52 - 2x = 52 - 52 = 0°? Wait, that can't be.

Wait, if x = 26°, then angle TQR = 52 - 2*26 = 52 - 52 = 0°, which is impossible.

So this shows that my previous assumption that angle TQR = 52 - 2x is wrong.

How did I get that?

I considered triangle TRQ, where angle at T is 2x°, angle at R is 128°, and then angle at Q (TQR) = 180 - 2x - 128.

But if angle at T is 2x°, but is that the case?

Wait, earlier, I thought that angle at T is between PT and TR, which is 2x°. But if PT is parallel to QR, and TR is a transversal, then the angle at T (between PT and TR) is equal to the angle at R (between QR and TR)? No, they are same-side interior angles, which add to 180°.

Wait, if angle at T is between PT and TR, which is 2x°, and angle at R is between QR and TR, which is 128°, then those two angles are on the same side of the transversal TR, so they should add to 180°. Hence, 2x + 128 = 180 => x=26°.

But then, in triangle TRQ, angle at T is not 2x°, but angle between TR and QT? Wait, no, if we consider triangle TRQ, then it depends on where point Q is.

Wait, triangle TRQ has vertices at T, R, and Q.

So, to find the angles in triangle TRQ, we need to know the angles at each vertex.

Angle at T is between TR and TQ. But we only know the angle between PT and TR at T, which is 2x°. However, PT is not the same as TQ.

Therefore, my earlier assumption that angle at T in triangle TRQ is 2x° is wrong.

That was the mistake.

Instead, the angle at T between PT and TR is 2x°, but in triangle TRQ, the angle at T is between TR and TQ, not TR and PT.

Therefore, the angle at T in triangle TRQ is different from 2x°.

Therefore, the previous calculation of angle TQR = 52 - 2x is incorrect.

So, I need to discard that approach.

So, going back.

We know x = 26° from the same-side interior angles.

Now, how to find angle PQR?

Let me re-examine the corresponding angles idea.

PT is parallel to QR, PQ is a transversal. The angle between PT and PQ at P is x°, and the corresponding angle between QR and PQ at Q should be equal.

So angle PQR = x° = 26°.

But then, if I consider triangle PQR, with angle at Q = 26°, angle at R = ?

Wait, at point R, we have angle between QR and TR is 128°, which is not angle at R in triangle PQR.

Angle at R in triangle PQR is angle PRQ.

To find angle PRQ, maybe we can consider triangle PRT or something.

Alternatively, since we know x = 26°, we can calculate other angles.

From triangle PQT, which has angles x° (at P), angle at T (which is 2x°?), wait no.

Wait, earlier, I considered angle at T as 2x°, which is the angle between PT and TR. But in triangle PQT, angle at T is between PT and QT. Unless QT is aligned with TR?

If QT is aligned with TR, then angle at T in triangle PQT would be same as angle between PT and TR, which is 2x°. But unless T, Q, R are colinear?

But in the diagram, point Q is at (0,0), T is at (7,10), and R is at (10,0). So, are Q, T, R colinear?

Let me check. The slope from Q(0,0) to T(7,10) is (10-0)/(7-0) = 10/7. The slope from T(7,10) to R(10,0) is (0-10)/(10-7) = -10/3. These slopes are different, so Q, T, R are not colinear.

Therefore, angle at T in triangle PQT is different from angle between PT and TR.

So, I can't assume that.

This is getting really tangled. Let me try a different approach using coordinates.

Since coordinates are given, perhaps I can calculate the angles using coordinates.

Points:

Q(0,0), R(10,0), P(-7,10), T(7,10).

First, confirm that PT is parallel to QR.

Slope of PT: since PT goes from (-7,10) to (7,10), it's horizontal, slope 0.

Slope of QR: from (0,0) to (10,0), also horizontal, slope 0. So they are parallel, as given.

Now, need to find angle PQR, which is the angle at Q between QP and QR.

Vector QP goes from Q(0,0) to P(-7,10), so vector is (-7,10).

Vector QR goes from Q(0,0) to R(10,0), vector is (10,0).

The angle between vectors QP and QR is angle PQR.

We can calculate this angle using the dot product formula.

The formula is:

cos(theta) = (QPR ⋅ QR) / (|QP| |QR|)

First, compute the dot product:

QP ⋅ QR = (-7)(10) + (10)(0) = -70 + 0 = -70

|QP| = sqrt((-7)^2 + 10^2) = sqrt(49 + 100) = sqrt(149)

|QR| = sqrt(10^2 + 0^2) = 10

Thus,

cos(theta) = -70 / (sqrt(149)*10) = -7 / sqrt(149)

Therefore, theta = arccos(-7 / sqrt(149))

Let me calculate that.

First, sqrt(149) ≈ 12.20655

So, -7 / 12.20655 ≈ -0.5735

Then arccos(-0.5735) ≈ 125 degrees? Let me check with calculator.

But wait, the problem gives specific angles x° and 2x°, and angle at R as 128°, so maybe this coordinate method is not accounting for the given angles. Wait, but according to the diagram, the given angles are x, 2x, and 128°, which should be consistent with the coordinates.

But according to the coordinates, angle PQR is arccos(-7 / sqrt(149)) ≈ 125 degrees, but I need to check if that matches with our previous calculations.

But earlier, from the same-side interior angles, we found x = 26°, which would imply angle PQR = x = 26°, but that contradicts the coordinate calculation.

This suggests that there is inconsistency.

But why? Because the diagram might not be drawn to scale, or maybe I misunderstood the given angles.

Wait, the problem says "in the diagram, PT is parallel to QR." But in the Asymptote code, the coordinates are given, so maybe the diagram is accurate, and we need to use the given angles to find x, then find angle PQR.

But according to the Asymptote code, the diagram has specific angles labeled: x° at (-6,10) SE, 2x° at (7,10) SW, and 128° at (10,0) NE.

So, these angles are not just arbitrary; they are part of the diagram. So, we need to use those to find x, then use that to find angle PQR.

So, going back, we had:

From the same-side interior angles: 2x + 128 = 180 => x = 26°

But if we calculate angle PQR using coordinates, we get ≈ 125°, which doesn't match x = 26°. So, there must be something wrong here.

Wait, maybe the angle labeled x° is not the angle between PQ and PT, but another angle? Let me check the diagram again.

The Asymptote code labels "$x^{\circ}$" at (-6,10), SE. Point (-6,10) is between P(-7,10) and T(7,10), closer to P. The label is at (-6,10) SE, which is to the southeast of that point, so it's near the segment from P(-7,10) to Q(0,0). So, it's likely the angle between PT and PQ at point P.

Similarly, "$2x^{\circ}$" at (7,10), SW, which is at point T, southwest of T, so near the segment from T(7,10) to R(10,0). So, that's the angle between PT and TR at point T.

And angle at R is 128°, between QR and TR.

So, with that, our previous calculation of x = 26° is correct.

But according to coordinate calculation, angle PQR is approximately 125°, which would mean that our assumption of angle PQR = x° is wrong.

But why?

Let me recalculate the angle PQR using coordinates, but with x = 26°.

If x = 26°, then the angle between PQ and PT at P is 26°. Let me verify this with coordinates.

Vector PT is from P(-7,10) to T(7,10), which is (14, 0).

Vector PQ is from P(-7,10) to Q(0,0), which is (7, -10).

The angle between vectors PT and PQ at P should be x = 26°.

Let's calculate this angle using dot product.

Vector PT is (14, 0). Vector PQ is (7, -10).

Dot product: 14*7 + 0*(-10) = 98 + 0 = 98.

|PT| = sqrt(14² + 0²) = 14.

|PQ| = sqrt(7² + (-10)²) = sqrt(49 + 100) = sqrt(149) ≈ 12.2066.

cos(theta) = 98 / (14 * 12.2066) = 98 / (170.892) ≈ 0.5735

So theta ≈ arccos(0.5735) ≈ 55°, not 26°.

Wait, this contradicts our earlier calculation where we thought x = 26°.

This suggests that our assumption that same-side interior angles give 2x + 128 = 180 is incorrect.

So, where is the error?

Let me re-examine the same-side interior angles.

We have PT || QR, and transversal TR. The angles at T and R.

At T, the angle between PT and TR is given as 2x°, and at R, the angle between QR and TR is given as 128°.

If these are same-side interior angles, then their sum should be 180°, but according to coordinate calculation, angle at T (between PT and TR) is not 2x°, but something else.

Wait, let me calculate the angle at T between PT and TR using coordinates.

Point T is at (7,10). PT is from T(7,10) to P(-7,10), vector (-14, 0).

TR is from T(7,10) to R(10,0), vector (3, -10).

The angle between PT and TR at T is the angle between vectors (-14, 0) and (3, -10).

Let me calculate this angle using dot product.

Dot product: (-14)(3) + 0*(-10) = -42 + 0 = -42

|PT| = 14, |TR| = sqrt(3² + (-10)^2) = sqrt(9 + 100) = sqrt(109) ≈ 10.4403

cos(theta) = -42 / (14 * 10.4403) ≈ -42 / 146.1642 ≈ -0.2874

theta ≈ arccos(-0.2874) ≈ 107°.

But according to the diagram, this angle is labeled as 2x°, which with x = 26° would be 52°, but actual angle is ≈107°.

This indicates a major inconsistency.

Therefore, my initial assumption about which angles are same-side interior angles is wrong.

Let me reconsider the diagram.

At point T, the angle between PT and TR is labeled as 2x°, but according to calculation, it's ≈107°. If this is supposed to be 2x°, then x ≈ 53.5°, but then angle at R is 128°.

If PT || QR, and TR is a transversal, then the angle at R between QR and TR is 128°. Let me calculate that angle.

Vector QR is from R(10,0) to Q(0,0), vector (-10, 0).

Vector TR is from R(10,0) to T(7,10), vector (-3, 10).

The angle between QR and TR at R is between vectors (-10, 0) and (-3, 10).

Dot product: (-10)(-3) + 0*10 = 30 + 0 = 30

|QR| = 10, |TR| = sqrt(9 + 100) = sqrt(109) ≈ 10.4403

cos(theta) = 30 / (10 * 10.4403) ≈ 30 / 104.403 ≈ 0.2874

theta ≈ arccos(0.2874) ≈ 73°.

But the diagram says this angle is 128°, which is not matching.

So, this suggests that the diagram as per coordinates does not match the given angles, which means the diagram is not to scale, and we need to rely on the given angles to solve the problem, not the coordinates.

So, going back, the given angles are: at T, 2x°, at R, 128°, and at P, x°.

With PT || QR, and need to find angle PQR.

Let me try again.

Assume that the diagram is not to scale, and the coordinates are just for drawing.

With PT || QR, and angles at P, T, R given.

At point P: angle between PT and PQ is x°.

At point T: angle between PT and TR is 2x°.

At point R: angle between QR and TR is 128°.

Since PT || QR, and TR is a transversal, the angle at T (between PT and TR) and the angle at R (between QR and TR) are same-side interior angles, so they should add up to 180°.

Thus, 2x + 128 = 180 → 2x = 52 → x = 26°.

Now, with x = 26°, we need to find angle PQR.

Now, let's consider triangle PQR. To find angle at Q, we need more information. Let's consider triangle PQT.

In triangle PQT, we know angle at P is x° = 26°, angle at T is...?

Wait, angle at T in triangle PQT is the angle between PT and QT. But we know the angle between PT and TR is 2x = 52°, but unless QT is related to TR.

Alternatively, if we consider that line QT is such that it creates some relations.

Alternatively, use the fact that PT || QR, and PQ is a transversal, so angle between PT and PQ (x°) is equal to the corresponding angle between QR and PQ, which is angle PQR.

Therefore, angle PQR = x = 26°.

But earlier, the coordinate calculation gave ≈125°, but that was under the assumption of specific coordinates, which might not match the given angles.

Therefore, trusting the diagram and the given angles, the answer is 26°.

But let me verify with another method.

Let me consider extending some lines or using alternate interior angles.

Since PT || QR, and if we can find a transversal that creates alternate interior angles.

For example, if we consider the line PQ, which intersects PT at P and QR at Q. As before, corresponding angles would make angle PQR = x = 26°.

Alternatively, consider triangle PQR. If I can find two other angles, I can find the third.

At point Q, angle PQR = ?

At point P, angle QPR. Let's find that.

From the diagram, at point P, the angle between PT and PQ is x° = 26°, and since PT is parallel to QR, the angle between PQ and QR at Q is equal to x°, as corresponding angles.

But what about the angle at P in triangle PQR? It's angle QPR.

If we can find angle QPR, then we can use triangle angle sum.

Let me think.

At point P, angle between PQ and PT is x° = 26°. Since PT is parallel to QR, and PQ is a transversal, the angle between PQ and QR at Q is x° = 26°, which is angle PQR.

Now, if we consider the line PR, which connects P to R. Let me try to find the angles at P and R in triangle PQR.

But without more information, this is difficult.

However, if we consider that in triangle PQR, the sum of angles is 180°, and we know angle PQR = x = 26°, and if we can find another angle, say angle PRQ or angle QPR.

Wait, at point R, we have angle between QR and TR is 128°. If we can relate that to angle PRQ.

If we consider triangle PRQ, angle at R is angle PRQ, and angle between QR and TR is 128°. If lines PR and TR are related, but without knowing more, it's hard to say.

Alternatively, if we consider that PT || QR, and TR is a transversal, then the alternate interior angles would be...

At point T, the angle between PT and TR is 2x° = 52°, and the alternate interior angle would be the angle between QR and... some line. If there is a line parallel to TR, but not sure.

This is getting too convoluted.

Given that the corresponding angle logic gives angle PQR = x = 26°, and the same-side interior angles give x = 26°, and despite the coordinate calculation suggesting otherwise (which might be due to the diagram not being to scale), I think the correct answer is 26°.

However, I need to check for any other possible relationships.

Wait, let's consider the sum of angles around point T. At point T, there is angle between PT and TR (2x°), and if we can find other angles.

But without knowing more, it's difficult.

Alternatively, if we consider triangle PQT, and with x = 26°, we can find other angles.

Angle at P is 26°, angle at T is some angle, and angle at Q is 180 - 26 - angle at T.

But without knowing angle at T, it's not helpful.

Wait, but if we consider that at point T, the angle between PT and TR is 2x = 52°, and if QT is such that it forms a straight line with something?

Alternatively, since PT is parallel to QR, and if we draw a line from Q parallel to TR, but this is too vague.

Given the time I've spent and the consistency in the same-side interior angle calculation, I think the measure of angle PQR is 26°, so the answer is 26°.

But let me check with another approach.

If PT || QR, and PQ is a transversal, then angle between PT and PQ at P is equal to the angle between QR and PQ at Q.

This is the definition of corresponding angles, so angle PQR = x.

Since we found x = 26°, then angle PQR = 26°.

Therefore, the answer is 26°, so \boxed{26}.

**Final Answer**
\boxed{26}

To determine the measure of $\angle PQR$ in the given diagram, we are provided with certain angle relationships and the fact that segments $PT$ and $QR$ are parallel. Here's a step-by-step breakdown:

---

### **Understanding the Diagram and Given Angles**

From the Asymptote diagram:

- $PT$ and $QR$ are horizontal line segments and thus **parallel**.
- Point $P$ is at the left end of segment $PT$, point $T$ is its right end, and point $R$ is on the line $QR$ (at the right end of the base horizontal segment).
- Segments $QP$ and $QR$ form the angle $\angle PQR$, which is the angle of interest.
- The diagram labeled:

  - $x^\circ$ at point $P$, likely the angle between segment $PT$ (horizontal) and segment $QP$ (going down to $Q$).
  - $2x^\circ$ at point $T$, likely the angle between segment $PT$ and segment $TR$.
  - $128^\circ$ at point $R$, between segments $QR$ and $TR$.

From this, we can infer that the angle at point $T$ is between parallel lines (since $PT \parallel QR$) and a transversal $TR$. This gives us a chance to apply the **same-side interior angles** theorem.

---

### **Using the Same-Side Interior Angles Theorem**

Since $PT \parallel QR$ and the segment $TR$ is a transversal intersecting these lines:

- The angle at $T$ between $PT$ and $TR$ is $2x^\circ$.
- The angle at $R$ between $QR$ and $TR$ is $128^\circ$.
- These are *same-side interior angles*, and they must **add up to $180^\circ$** if the lines are parallel.

Therefore,

$$
2x + 128 = 180
$$

$$
2x = 52 \Rightarrow x = 26^\circ
$$

---

### **Corresponding Angles to Find $\angle PQR$**

We are now asked to find the measure of $\angle PQR$, which is the angle at point $Q$ between segments $QP$ and $QR$.

- The transversal $PQ$ intersects the parallel lines $PT$ and $QR$.
- The angle at $P$ between $PT$ and $PQ$ is $x = 26^\circ$.
- By the **corresponding angle theorem**, this angle must be equal to the angle at $Q$ between $QR$ and $PQ$, which is $\angle PQR$.

Thus,

$$
\angle PQR = x = 26^\circ
$$

---

### **Final Answer**

$$
\boxed{26}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the measure of angle PQR in this diagram. Let me start by trying to visualize the diagram based on the Asymptote code provided.

First, there's a horizontal line from (0,0) to (10,0), which is labeled with a MidArrow. Then, another segment from (10,0) to (20,0), also with a MidArrow. So that seems like two horizontal segments connected at (10,0), maybe forming a straight line? Wait, but the first segment is from (0,0) to (10,0), then another from (10,0) to (20,0). So that's just a longer horizontal line from (0,0) to (20,0), but with arrows on both parts? Not sure if that matters yet.

Then there are other lines: from (0,0) to (-7,10), then from (-7,10) to (7,10), then from (7,10) to (10,0). So that makes a quadrilateral? Let me think. Starting at Q (0,0), going up to P (-7,10), then to T (7,10), then down to R (10,0). Then there's a line from (-5,10) to (7,10) with a MidArrow. Hmm. The labels are: x degrees at (-6,10) which is near point P, 2x degrees at (7,10) which is point T, 128 degrees at (10,0) which is point R, and labels for points P, T, R, Q.

So, angle PQR is the angle at point Q, between points P, Q, and R. So, that's the angle at Q between QP and QR. So, I need to find angle PQR.

Given that PT is parallel to QR. Let me note that. So, PT is parallel to QR. PT is the segment from P (-7,10) to T (7,10). QR is the segment from Q (0,0) to R (10,0). Wait, but QR is along the x-axis from (0,0) to (10,0). PT is from (-7,10) to (7,10), so that's a horizontal line segment as well. So, PT is parallel to QR because both are horizontal. That makes sense.

So, PT and QR are both horizontal lines, hence parallel.

Now, the angles labeled: x degrees at (-6,10), which is near point P. Since P is at (-7,10), the label x° is near (-6,10), which is slightly to the right of P. Similarly, 2x° is at (7,10), which is point T. Then 128° is at (10,0), which is point R.

So, looking at the diagram, maybe there are some angles formed by the transversals? Since PT is parallel to QR, maybe there's a transversal cutting them, creating some corresponding angles or alternate interior angles?

Let me try to sketch mentally. The segment from P (-7,10) to T (7,10) is horizontal. Then segment from T (7,10) to R (10,0). Also, segment from Q (0,0) to P (-7,10). So, triangle PQT? Wait, but there's also a segment from Q to R, which is QR.

Perhaps there are some triangles here. Let me try to figure out the angles.

First, the angle at point P is labeled x°. Wait, but which angle? The label is at (-6,10), which is near the segment PT. Since PT is horizontal, maybe that's an angle between some lines. Let me think.

Looking at the Asymptote code: the segment from (-7,10) to (7,10) is drawn, and then there is a MidArrow on (-5,10) to (7,10). So, that's the segment PT. Then, the label x° is at (-6,10), SE, which would be southeast of that point, so maybe near the segment from P to Q? Wait, the segment from P (-7,10) to Q (0,0). So, maybe the angle at P between PT and PQ?

Similarly, the label 2x° is at (7,10), SW, so southwest of that point, which would be near the segment from T to R? Since T is connected to R (from (7,10) to (10,0)), so maybe the angle at T between PT and TR?

And then the label 128° is at (10,0), NE, which would be northeast of that point, so near the segment from R to Q? Since R is connected to Q (from (10,0) to (0,0)).

So, perhaps angle at P between PT and PQ is x°, angle at T between PT and TR is 2x°, and angle at R between QR and something is 128°? Wait, the 128° is labeled at R, which is connected to Q and T? Wait, R is connected to Q (QR) and to T (TR). So, maybe angle at R between QR and TR? But that angle is labeled as 128°, so that's angle TRQ?

But how do these angles relate? Since PT is parallel to QR, and maybe there are some transversals. For example, the segment PQ and TR might be transversals? Or TR and PQ?

Alternatively, maybe triangle PQT has some angles. Let me think again.

First, let me note all given angles. Let me denote them:

At point P: angle between PT and PQ is x°. Since PT is horizontal, going from P (-7,10) to T (7,10), and PQ goes from P (-7,10) to Q (0,0). So, angle between PT (horizontal right) and PQ (going down to Q). Similarly, at point T, angle between PT (horizontal left to right) and TR (going down to R (10,0)) is 2x°. At point R, angle between QR (horizontal left to right) and TR is 128°.

Since PT is parallel to QR, and if we can find some transversal, maybe PQ or TR? Let me try to consider the transversal TR. Since TR connects T to R, which is from (7,10) to (10,0). Let me find the slope of TR. The slope is (0 - 10)/(10 -7) = -10/3. Similarly, the line PQ connects P (-7,10) to Q (0,0), so slope is (0 - 10)/(0 - (-7)) = -10/7.

But maybe that's not helpful yet.

Alternatively, since PT is parallel to QR, and if we can find some corresponding angles. For example, if we have a transversal that cuts PT and QR, then the corresponding angles would be equal. Let me think about the transversal that connects P to R? Wait, PR? Is PR a transversal? Let me see. If we consider PR, which connects P (-7,10) to R (10,0). Then PT is parallel to QR, so maybe PR is a transversal cutting PT and QR.

Alternatively, maybe considering triangle PQT? Let me think.

But perhaps it's better to use the given angles. Let me note that at point T, angle between PT and TR is 2x°, and since PT is parallel to QR, then maybe angle between TR and QR is equal? But angle at R is given as 128°, which is angle between QR and TR? If that is the case, then maybe angle at R is corresponding to angle at T?

Wait, if PT is parallel to QR, and if TR is a transversal, then the angle at T (which is 2x°) and the angle at R (which is 128°) might be same-side interior angles? Because if TR is a transversal intersecting PT and QR, then the angles on the same side would add up to 180°?

Wait, let me think. If PT is parallel to QR, and TR is a transversal, then the angle at T (between PT and TR) is 2x°, and the angle at R (between QR and TR) is 128°. If these are same-side interior angles, then their sum would be 180°.

So, 2x + 128 = 180? Then, 2x = 52°, so x = 26°?

If that is the case, then x is 26°, and then angle at P is x = 26°, and angle at T is 2x = 52°, and angle at R is 128°.

But then, how does that help us find angle PQR?

Angle PQR is the angle at Q between PQ and QR. So, maybe if we can find other angles in triangle PQR or something.

Wait, let's consider triangle PQR. But wait, in the diagram, point T is connected to P and R, and Q is connected to P and R. Maybe triangle PQT? Wait, perhaps not. Let me think again.

Alternatively, maybe considering triangle PQT. Let me see. If we consider triangle PQT, we have angle at P is x°, angle at T is 2x°, so maybe we can find some angles? Wait, but is T connected to Q? Wait, in the Asymptote code, the segments are from Q (0,0) to P (-7,10), then P to T (7,10), then T to R (10,0). So, Q is connected to P and R. T is connected to P and R. There's no direct connection between Q and T.

So, perhaps triangle PQT is not a triangle? Wait, P, Q, T? If we connect Q to T? Wait, in the Asymptote code, there is a segment from (-5,10) to (7,10) with a MidArrow, which is the segment PT. But there is no segment from Q to T.

Alternatively, maybe triangle QTR? Q to T? Not sure.

Alternatively, maybe focusing on angles around point Q. The angle PQR is at Q, between PQ and QR. Let me think about the lines at Q. PQ goes up to P, QR goes to R. So, if we can find some angles adjacent to angle PQR.

But perhaps triangle PQR. If we can find the other angles in triangle PQR, then we can find angle PQR.

But to do that, we need more information.

Wait, maybe if we consider that PT is parallel to QR, so if we can find some relation. For example, if we can find the angles at P and T, maybe relate them to angles at Q?

Alternatively, note that PT is parallel to QR. Let me see. If we have a transversal PQ intersecting PT and QR. Wait, PQ starts at P, which is on PT, and goes to Q, which is on QR. So, PQ is a transversal that starts on PT and ends on QR.

So, at point P, the angle between PT and PQ is x°, and at point Q, the angle between PQ and QR... Wait, since PT is parallel to QR, then maybe the angle between PQ and PT at P is equal to the angle between PQ and QR at Q? But since PT is parallel to QR, then PQ would be a transversal, so the corresponding angles would be equal?

Wait, if two parallel lines are cut by a transversal, then corresponding angles are equal. So, if PT || QR and PQ is a transversal, then angle between PT and PQ at P is equal to the angle between QR and PQ at Q.

But angle between PT and PQ at P is x°, so that would mean the angle between QR and PQ at Q is also x°? But angle between QR and PQ at Q is angle PQR. Wait, is that right?

Wait, let me draw it mentally. At point P, the line PT is parallel to QR. Then PQ is a transversal. The angle between PT and PQ at P is x°, which is like an angle inside the transversal. Then, the corresponding angle at Q would be the angle between QR and QP. Because QR is parallel to PT, so the corresponding angle would be the angle between QR and the transversal PQ.

Therefore, angle PQR is equal to x°? Wait, but if that's the case, then angle PQR would be x°, which we found earlier as 26°, but then if we can check with other angles?

Wait, but if angle PQR is x°, which is 26°, then in triangle PQR, we would also need to know another angle. Maybe angle at R?

Wait, angle at R is given as 128°, which is the angle between QR and TR. But angle at R in triangle PQR would be angle QRP? Wait, maybe not. Let me clarify.

Point R is connected to Q and to T. So, angle at R between QR and TR is 128°, but angle at R in triangle PQR is angle QRP. Unless TR is colinear with something? Wait, not sure.

Alternatively, maybe triangle PQR has angles at P, Q, R. But we have angles at P (angle QPR?), and angle at R?

Wait, if angle at P is x°, angle at Q is x°, then angle at R would be 180 - x - x = 180 - 2x. But we also have another angle at R, which is 128°, which is angle TRQ. Maybe these are supplementary?

Wait, angle at R in triangle PQR is angle QRP, which is different from angle TRQ. Let me think. If we have point T connected to R, then at R, there's angle TRQ (which is 128°) and angle QRP (which is part of triangle PQR). Are these angles adjacent?

If points Q, R, and T are connected, then maybe there is a triangle or something. Let me think. If we consider line TR, which goes from T (7,10) to R (10,0), then angle between TR and QR at R is 128°, which is angle TRQ. Then, if we look at triangle TRQ, we have angles at T, R, and Q. Wait, but we don't have point Q connected to T.

Alternatively, if we can find angle QRP (angle at R in triangle PQR) in terms of x, and relate it with angle TRQ (which is 128°)?

Alternatively, since we have some information about angles at T and R. Earlier, I considered that angle at T (2x°) and angle at R (128°) are same-side interior angles. Hence, adding to 180.

If that is the case, then 2x + 128 = 180 => 2x = 52 => x=26.

So then, angle PQR would be x°, which is 26°? But wait, is that correct?

Wait, let me verify. If PT is parallel to QR, and TR is a transversal, then the same-side interior angles would be angle at T (between PT and TR) and angle at R (between QR and TR). If they are same-side interior angles, then they add to 180°, which gives 2x + 128° = 180°, so x=26°.

But then angle PQR is supposed to be x°, which would be 26°, but is that accurate?

Wait, earlier I thought that angle PQR is equal to x° because of corresponding angles. But let me confirm that.

If PQ is a transversal cutting PT (parallel to QR), then angle between PT and PQ at P (x°) corresponds to angle between QR and PQ at Q.

Yes, since PT || QR, and PQ is a transversal, then corresponding angles are equal. So, angle between PT and PQ at P is equal to the angle between QR and PQ at Q.

But angle between QR and PQ at Q is angle PQR. Therefore, angle PQR = x° = 26°.

But let me check if that's consistent with the rest of the diagram.

If angle PQR is 26°, and angle at R in triangle PQR (angle QRP) is then 180° - angle at P - angle at Q. Wait, but angle at P in triangle PQR is not necessarily x°. Wait, angle at P in triangle PQR is angle QPR. Earlier, we had angle between PT and PQ at P is x°, but angle QPR is that angle?

Yes, because PT is the line from P to T, which is horizontal, and PQ is going from P to Q. So, angle between PQ and PT is x°, which is angle QPT. But in triangle PQR, angle at P is angle QPR. However, if we consider that PT is a line from P to T, which is parallel to QR. So, is angle QPT (x°) the same as angle QPR?

Wait, unless point T lies on PR or something? But in the Asymptote code, PT is from (-7,10) to (7,10), and R is at (10,0). So, point T is not on PR.

Therefore, angle at P in triangle PQR is angle QPR, which is different from angle QPT (x°). Because angle QPT is between PQ and PT, while angle QPR is between PQ and PR. So, unless PR is aligned with PT, which it isn't.

So my earlier assumption that angle PQR is x° might be incorrect.

Therefore, I need to re-examine.

Let me think again. If PT is parallel to QR, and PQ is a transversal. Then, the corresponding angles would be angle between PT and PQ at P (which is x°) and angle between QR and PQ at Q. But what is that angle?

If QR is the line from Q (0,0) to R (10,0), and PQ is the line from Q (0,0) to P (-7,10). Then, the angle between QR and PQ at Q is indeed angle PQR. Therefore, angle PQR is equal to x°, because they are corresponding angles.

Therefore, if that is correct, then angle PQR is x°, which is 26°, as found earlier.

But then, in triangle PQR, we have angle at Q = 26°, angle at R? Let's see. If we can find angle at R.

But angle at R is given as 128°, but that is angle TRQ (angle between TR and QR). So, angle TRQ is 128°, which is different from angle QRP (angle between QR and PR in triangle PQR).

However, maybe we can find angle QRP?

Let me consider triangle TRQ. In triangle TRQ, we have angle at R is 128°, angle at T is 2x° = 52°, and angle at Q is angle TQR.

Wait, if we consider triangle TRQ, angles sum to 180°. So, angle at T (2x) + angle at R (128°) + angle at Q (TQR) = 180°.

So, angle TQR = 180° - 2x - 128° = 180° - 128° - 2x = 52° - 2x.

So angle TQR is 52° - 2x.

But angle TQR is also part of angle PQR? Wait, angle PQR is at point Q between PQ and QR. If there is a line QT? Wait, but QT is not drawn in the diagram. Wait, in the Asymptote code, there is a segment from Q (0,0) to P (-7,10), then P to T (7,10), then T to R. There's no segment from Q to T.

So angle TQR is the angle at Q between TQ and QR, but since TQ is not present, unless we assume that QT is a line? Wait, but the Asymptote code draws a segment from (-5,10) to (7,10) with a MidArrow, which is PT. The other segments are QP, PT, TR, QR. So, unless there is a line from Q to T?

Wait, actually, in the Asymptote code, the points are Q (0,0), P (-7,10), T (7,10), R (10,0). The segments are QP, PT, TR, QR. Also, there is a segment from (-7,10) to (7,10) which is PT.

Therefore, there's no segment connecting Q to T. So, angle TQR is not directly present.

But earlier, in triangle TRQ, angle at Q is angle TQR, which would require that line QT exists. However, since QT is not drawn, maybe there's an implied line? Or maybe I'm overcomplicating.

Alternatively, since we have angle TQR = 52° - 2x, and angle PQR is the angle at Q between PQ and QR. If there is a line from Q to T, then angle PQR would be split into angle PQT and angle TQR? But unless we know something about that.

Alternatively, maybe angle PQT is equal to something?

Wait, but without knowing more, perhaps there's another approach.

Earlier, I thought angle PQR is x°, which is 26°, but if we consider triangle PQR, we need to ensure that the angles add up.

But angle PQR is x° = 26°, angle at R is something, angle at P is something.

Wait, maybe considering triangle PQT?

In triangle PQT, we have angle at P is x°, angle at T is 2x°, so the third angle at Q is 180° - x° - 2x° = 180° - 3x°.

So, angle at Q in triangle PQT is 180° - 3x°.

But angle at Q in triangle PQT is angle PQT.

But in the diagram, angle at Q is also angle PQR and angle PQT? Unless points R and T are on the same side.

Wait, if we consider that line QT is drawn, then angle PQR would be part of angle PQT?

But since QT is not drawn, maybe we need to think differently.

Alternatively, if we consider that line TR is a transversal, and we found angle at Q in triangle TRQ is 52° - 2x.

But how does that relate to angle PQR?

Alternatively, maybe considering that angle PQR is equal to angle PQT + angle TQR? If there is a line QT.

Assuming that, then angle PQR = angle PQT + angle TQR.

From triangle PQT, angle PQT is 180° - 3x°, and angle TQR is 52° - 2x°.

So, angle PQR = (180 - 3x) + (52 - 2x) = 232 - 5x.

But earlier, we thought angle PQR was equal to x°. Therefore, if angle PQR is both x° and 232 - 5x, then:

x = 232 - 5x
6x = 232
x = 232 / 6 ≈ 38.666...

But earlier, we had from the same-side interior angles that x = 26°.

This inconsistency suggests that there is a mistake in my reasoning.

Hmm, this is getting complicated. Maybe my assumption that angle PQR is x° is wrong. Let me re-examine that.

If PT is parallel to QR, and PQ is a transversal, then the corresponding angles should be equal. The angle between PT and PQ at P is x°, so the corresponding angle would be the angle between QR and PQ at Q.

But which angle is that?

Imagine PQ going from P to Q, and QR going from Q to R. Since PT is parallel to QR, then the line PQ intersects PT at P and QR at Q. Therefore, the corresponding angles would be the angle between PT and PQ at P (x°) and the angle between QR and PQ at Q.

But angle between QR and PQ at Q is the angle between QR and QP, which is exactly angle PQR.

Therefore, angle PQR = x°. So that should be valid.

But then why the other approach gives inconsistency?

Let me check my calculation again.

From triangle TRQ, angles sum to 180:

angle T (2x) + angle R (128) + angle Q (TQR) = 180
So angle TQR = 180 - 2x - 128 = 52 - 2x.

Assuming angle PQR = angle PQT + angle TQR.

But angle PQT is part of triangle PQT, which has angles x°, 2x°, and angle PQT = 180 - 3x.

But if angle PQT is part of angle PQR, then angle PQR = angle PQT + angle TQR
=> angle PQR = (180 - 3x) + (52 - 2x) = 232 - 5x
But angle PQR is also x, so:

x = 232 - 5x
6x = 232
x = 232 / 6 ≈ 38.666...

But earlier, from the same-side interior angles, we found x = 26°. There's a contradiction here. Therefore, one of the assumptions is wrong.

Where is the mistake?

Possibly in the assumption that angle PQR = angle PQT + angle TQR.

If there is no segment QT, then angle PQR is not necessarily composed of angle PQT and angle TQR. Therefore, that assumption was incorrect.

Therefore, need another approach.

Let me try to consider that we have two different expressions for angle PQR. From the corresponding angles, angle PQR = x°. From triangle TRQ, angle TQR = 52 - 2x. If there's a relation between angle TQR and angle PQR?

Maybe angle PQR and angle TQR are supplementary or something?

Wait, if we consider line TR and line PQ, but without knowing more, it's hard to tell.

Alternatively, maybe using the fact that PT is parallel to QR, and using other transversals.

Let me try another approach.

Given that PT || QR, and looking at the angles formed by another transversal, say PR.

Wait, PR connects P (-7,10) to R (10,0). Let me calculate its slope. Slope is (0 - 10)/(10 - (-7)) = -10/17.

Not sure. Maybe not helpful.

Alternatively, consider triangle PQT and triangle QTR? Not sure.

Wait, another thought. Since PT is parallel to QR, then the slope of PT is 0 (since it's horizontal), and slope of QR is 0 as well. Therefore, they are parallel.

Now, consider the angles at P and T. At point P, angle between PT and PQ is x°, and at point T, angle between PT and TR is 2x°.

If I can find the measure of angle between PQ and TR or something.

Alternatively, using the fact that the sum of angles around point T?

Wait, at point T, we have angle between PT and TR is 2x°, and if we can find other angles at T.

But point T is connected to P and R. Also, there is a segment from (-5,10) to (7,10) with a MidArrow, which is PT.

Wait, maybe there are more angles given? The label of x° is near (-6,10), which is on PT. Maybe that is angle between PT and another line.

Wait, the Asymptote code draws:

draw((-7,10)--(-5,10)--(7,10)--(10,0),black+linewidth(1));

Wait, no, original code:

draw((-7,10)--(7,10)--(10,0),black+linewidth(1));

Wait, no, the original Asymptote code is:

draw((0,0)--(10,0),black+linewidth(1));
draw((0,0)--(10,0),MidArrow);
draw((10,0)--(20,0),black+linewidth(1));
draw((0,0)--(-7,10)--(7,10)--(10,0),black+linewidth(1));
draw((-5,10)--(7,10),MidArrow);
label("$x^{\circ}$",(-6,10),SE);
label("$2x^{\circ}$",(7,10),SW);
label("$128^{\circ}$",(10,0),NE);
label("$P$",(-7,10),N);
label("$T$",(7,10),N);
label("$R$",(10,0),S);
label("$Q$",(0,0),S);

So, the diagram has:

- The base line from Q(0,0) to R(10,0), then to S(20,0).

- Then from Q(0,0) to P(-7,10), then P to T(7,10), then T to R(10,0).

- The segment PT is drawn from (-7,10) to (7,10), with a MidArrow from (-5,10) to (7,10), meaning that perhaps the arrow is on the segment PT from (-5,10) to (7,10)?

- The label x° is at (-6,10), SE, so south-east of (-6,10). That would be near point P(-7,10), but slightly to the right. So, maybe the angle between PQ and PT at point P?

Similarly, label 2x° at (7,10), SW, which is at point T. So, the angle at T between PT and TR?

Label 128° at R(10,0), NE, which is northeast of R, so near the segment from R to T or R to Q?

Therefore, the diagram has:

At point P(-7,10): angle between PQ and PT is x°, which is given.

At point T(7,10): angle between PT and TR is 2x°, given.

At point R(10,0): angle between QR and... well, since TR connects to R, and QR is the base, the angle between QR and TR is 128°, which is given.

So, with those angles, and PT || QR, how can we find angle PQR?

Let me try to use the fact that PT || QR, and transversal TR creates angles.

As before, angle at T (2x°) and angle at R (128°) are on the same side of the transversal TR, so they are same-side interior angles and add to 180°.

Hence, 2x + 128 = 180 => 2x = 52 => x = 26°.

So now, we know x = 26°.

Now, to find angle PQR, which is at Q, between PQ and QR.

Let me think about triangle PQR. To find angle at Q, I need more information about the triangle. Maybe if I can find other angles or sides?

Alternatively, consider triangle PQT. In triangle PQT, we know two angles: at P is x°, at T is 2x°, so angle at Q (which is angle PQT) is 180 - x - 2x = 180 - 3x.

So angle PQT = 180 - 3x.

If we can relate angle PQT to angle PQR.

Assuming that line QT is drawn (even though it's not in the original diagram), angle PQR would be angle between PQ and QR, which might be related to angle PQT and angle TQR.

If we assume that line QT is drawn, forming angle PQT and TQR, then angle PQR = angle PQT + angle TQR.

But from before, in triangle TRQ, angle at Q is angle TQR = 180 - 2x - 128 = 52 - 2x.

So angle PQR = (180 - 3x) + (52 - 2x) = 232 - 5x.

But we found x = 26°, so substituting:

angle PQR = 232 - 5*26 = 232 - 130 = 102°.

Wait, but earlier I thought angle PQR = x = 26°, but this new result gives 102°. Which one is correct?

But where was the mistake in the previous assumption?

The assumption that angle PQR is x° was based on corresponding angles, but that might not hold if there's more to the diagram. Let me verify that again.

If PT || QR and PQ is a transversal, then corresponding angles should be equal. The angle between PT and PQ at P is x°, which should correspond to the angle between QR and PQ at Q.

But what is the angle between QR and PQ at Q? It is indeed angle PQR. So angle PQR should be x°.

However, according to the other calculation, angle PQR = 102°. This contradiction suggests that there's a flaw in one of the reasonings.

Let me try to find out.

If x = 26°, then angle PQT is 180 - 3x = 180 - 78 = 102°, and angle TQR is 52 - 2x = 52 - 52 = 0°? Wait, that can't be.

Wait, if x = 26°, then angle TQR = 52 - 2*26 = 52 - 52 = 0°, which is impossible.

So this shows that my previous assumption that angle TQR = 52 - 2x is wrong.

How did I get that?

I considered triangle TRQ, where angle at T is 2x°, angle at R is 128°, and then angle at Q (TQR) = 180 - 2x - 128.

But if angle at T is 2x°, but is that the case?

Wait, earlier, I thought that angle at T is between PT and TR, which is 2x°. But if PT is parallel to QR, and TR is a transversal, then the angle at T (between PT and TR) is equal to the angle at R (between QR and TR)? No, they are same-side interior angles, which add to 180°.

Wait, if angle at T is between PT and TR, which is 2x°, and angle at R is between QR and TR, which is 128°, then those two angles are on the same side of the transversal TR, so they should add to 180°. Hence, 2x + 128 = 180 => x=26°.

But then, in triangle TRQ, angle at T is not 2x°, but angle between TR and QT? Wait, no, if we consider triangle TRQ, then it depends on where point Q is.

Wait, triangle TRQ has vertices at T, R, and Q.

So, to find the angles in triangle TRQ, we need to know the angles at each vertex.

Angle at T is between TR and TQ. But we only know the angle between PT and TR at T, which is 2x°. However, PT is not the same as TQ.

Therefore, my earlier assumption that angle at T in triangle TRQ is 2x° is wrong.

That was the mistake.

Instead, the angle at T between PT and TR is 2x°, but in triangle TRQ, the angle at T is between TR and TQ, not TR and PT.

Therefore, the angle at T in triangle TRQ is different from 2x°.

Therefore, the previous calculation of angle TQR = 52 - 2x is incorrect.

So, I need to discard that approach.

So, going back.

We know x = 26° from the same-side interior angles.

Now, how to find angle PQR?

Let me re-examine the corresponding angles idea.

PT is parallel to QR, PQ is a transversal. The angle between PT and PQ at P is x°, and the corresponding angle between QR and PQ at Q should be equal.

So angle PQR = x° = 26°.

But then, if I consider triangle PQR, with angle at Q = 26°, angle at R = ?

Wait, at point R, we have angle between QR and TR is 128°, which is not angle at R in triangle PQR.

Angle at R in triangle PQR is angle PRQ.

To find angle PRQ, maybe we can consider triangle PRT or something.

Alternatively, since we know x = 26°, we can calculate other angles.

From triangle PQT, which has angles x° (at P), angle at T (which is 2x°?), wait no.

Wait, earlier, I considered angle at T as 2x°, which is the angle between PT and TR. But in triangle PQT, angle at T is between PT and QT. Unless QT is aligned with TR?

If QT is aligned with TR, then angle at T in triangle PQT would be same as angle between PT and TR, which is 2x°. But unless T, Q, R are colinear?

But in the diagram, point Q is at (0,0), T is at (7,10), and R is at (10,0). So, are Q, T, R colinear?

Let me check. The slope from Q(0,0) to T(7,10) is (10-0)/(7-0) = 10/7. The slope from T(7,10) to R(10,0) is (0-10)/(10-7) = -10/3. These slopes are different, so Q, T, R are not colinear.

Therefore, angle at T in triangle PQT is different from angle between PT and TR.

So, I can't assume that.

This is getting really tangled. Let me try a different approach using coordinates.

Since coordinates are given, perhaps I can calculate the angles using coordinates.

Points:

Q(0,0), R(10,0), P(-7,10), T(7,10).

First, confirm that PT is parallel to QR.

Slope of PT: since PT goes from (-7,10) to (7,10), it's horizontal, slope 0.

Slope of QR: from (0,0) to (10,0), also horizontal, slope 0. So they are parallel, as given.

Now, need to find angle PQR, which is the angle at Q between QP and QR.

Vector QP goes from Q(0,0) to P(-7,10), so vector is (-7,10).

Vector QR goes from Q(0,0) to R(10,0), vector is (10,0).

The angle between vectors QP and QR is angle PQR.

We can calculate this angle using the dot product formula.

The formula is:

cos(theta) = (QPR ⋅ QR) / (|QP| |QR|)

First, compute the dot product:

QP ⋅ QR = (-7)(10) + (10)(0) = -70 + 0 = -70

|QP| = sqrt((-7)^2 + 10^2) = sqrt(49 + 100) = sqrt(149)

|QR| = sqrt(10^2 + 0^2) = 10

Thus,

cos(theta) = -70 / (sqrt(149)*10) = -7 / sqrt(149)

Therefore, theta = arccos(-7 / sqrt(149))

Let me calculate that.

First, sqrt(149) ≈ 12.20655

So, -7 / 12.20655 ≈ -0.5735

Then arccos(-0.5735) ≈ 125 degrees? Let me check with calculator.

But wait, the problem gives specific angles x° and 2x°, and angle at R as 128°, so maybe this coordinate method is not accounting for the given angles. Wait, but according to the diagram, the given angles are x, 2x, and 128°, which should be consistent with the coordinates.

But according to the coordinates, angle PQR is arccos(-7 / sqrt(149)) ≈ 125 degrees, but I need to check if that matches with our previous calculations.

But earlier, from the same-side interior angles, we found x = 26°, which would imply angle PQR = x = 26°, but that contradicts the coordinate calculation.

This suggests that there is inconsistency.

But why? Because the diagram might not be drawn to scale, or maybe I misunderstood the given angles.

Wait, the problem says "in the diagram, PT is parallel to QR." But in the Asymptote code, the coordinates are given, so maybe the diagram is accurate, and we need to use the given angles to find x, then find angle PQR.

But according to the Asymptote code, the diagram has specific angles labeled: x° at (-6,10) SE, 2x° at (7,10) SW, and 128° at (10,0) NE.

So, these angles are not just arbitrary; they are part of the diagram. So, we need to use those to find x, then use that to find angle PQR.

So, going back, we had:

From the same-side interior angles: 2x + 128 = 180 => x = 26°

But if we calculate angle PQR using coordinates, we get ≈ 125°, which doesn't match x = 26°. So, there must be something wrong here.

Wait, maybe the angle labeled x° is not the angle between PQ and PT, but another angle? Let me check the diagram again.

The Asymptote code labels "$x^{\circ}$" at (-6,10), SE. Point (-6,10) is between P(-7,10) and T(7,10), closer to P. The label is at (-6,10) SE, which is to the southeast of that point, so it's near the segment from P(-7,10) to Q(0,0). So, it's likely the angle between PT and PQ at point P.

Similarly, "$2x^{\circ}$" at (7,10), SW, which is at point T, southwest of T, so near the segment from T(7,10) to R(10,0). So, that's the angle between PT and TR at point T.

And angle at R is 128°, between QR and TR.

So, with that, our previous calculation of x = 26° is correct.

But according to coordinate calculation, angle PQR is approximately 125°, which would mean that our assumption of angle PQR = x° is wrong.

But why?

Let me recalculate the angle PQR using coordinates, but with x = 26°.

If x = 26°, then the angle between PQ and PT at P is 26°. Let me verify this with coordinates.

Vector PT is from P(-7,10) to T(7,10), which is (14, 0).

Vector PQ is from P(-7,10) to Q(0,0), which is (7, -10).

The angle between vectors PT and PQ at P should be x = 26°.

Let's calculate this angle using dot product.

Vector PT is (14, 0). Vector PQ is (7, -10).

Dot product: 14*7 + 0*(-10) = 98 + 0 = 98.

|PT| = sqrt(14² + 0²) = 14.

|PQ| = sqrt(7² + (-10)²) = sqrt(49 + 100) = sqrt(149) ≈ 12.2066.

cos(theta) = 98 / (14 * 12.2066) = 98 / (170.892) ≈ 0.5735

So theta ≈ arccos(0.5735) ≈ 55°, not 26°.

Wait, this contradicts our earlier calculation where we thought x = 26°.

This suggests that our assumption that same-side interior angles give 2x + 128 = 180 is incorrect.

So, where is the error?

Let me re-examine the same-side interior angles.

We have PT || QR, and transversal TR. The angles at T and R.

At T, the angle between PT and TR is given as 2x°, and at R, the angle between QR and TR is given as 128°.

If these are same-side interior angles, then their sum should be 180°, but according to coordinate calculation, angle at T (between PT and TR) is not 2x°, but something else.

Wait, let me calculate the angle at T between PT and TR using coordinates.

Point T is at (7,10). PT is from T(7,10) to P(-7,10), vector (-14, 0).

TR is from T(7,10) to R(10,0), vector (3, -10).

The angle between PT and TR at T is the angle between vectors (-14, 0) and (3, -10).

Let me calculate this angle using dot product.

Dot product: (-14)(3) + 0*(-10) = -42 + 0 = -42

|PT| = 14, |TR| = sqrt(3² + (-10)^2) = sqrt(9 + 100) = sqrt(109) ≈ 10.4403

cos(theta) = -42 / (14 * 10.4403) ≈ -42 / 146.1642 ≈ -0.2874

theta ≈ arccos(-0.2874) ≈ 107°.

But according to the diagram, this angle is labeled as 2x°, which with x = 26° would be 52°, but actual angle is ≈107°.

This indicates a major inconsistency.

Therefore, my initial assumption about which angles are same-side interior angles is wrong.

Let me reconsider the diagram.

At point T, the angle between PT and TR is labeled as 2x°, but according to calculation, it's ≈107°. If this is supposed to be 2x°, then x ≈ 53.5°, but then angle at R is 128°.

If PT || QR, and TR is a transversal, then the angle at R between QR and TR is 128°. Let me calculate that angle.

Vector QR is from R(10,0) to Q(0,0), vector (-10, 0).

Vector TR is from R(10,0) to T(7,10), vector (-3, 10).

The angle between QR and TR at R is between vectors (-10, 0) and (-3, 10).

Dot product: (-10)(-3) + 0*10 = 30 + 0 = 30

|QR| = 10, |TR| = sqrt(9 + 100) = sqrt(109) ≈ 10.4403

cos(theta) = 30 / (10 * 10.4403) ≈ 30 / 104.403 ≈ 0.2874

theta ≈ arccos(0.2874) ≈ 73°.

But the diagram says this angle is 128°, which is not matching.

So, this suggests that the diagram as per coordinates does not match the given angles, which means the diagram is not to scale, and we need to rely on the given angles to solve the problem, not the coordinates.

So, going back, the given angles are: at T, 2x°, at R, 128°, and at P, x°.

With PT || QR, and need to find angle PQR.

Let me try again.

Assume that the diagram is not to scale, and the coordinates are just for drawing.

With PT || QR, and angles at P, T, R given.

At point P: angle between PT and PQ is x°.

At point T: angle between PT and TR is 2x°.

At point R: angle between QR and TR is 128°.

Since PT || QR, and TR is a transversal, the angle at T (between PT and TR) and the angle at R (between QR and TR) are same-side interior angles, so they should add up to 180°.

Thus, 2x + 128 = 180 → 2x = 52 → x = 26°.

Now, with x = 26°, we need to find angle PQR.

Now, let's consider triangle PQR. To find angle at Q, we need more information. Let's consider triangle PQT.

In triangle PQT, we know angle at P is x° = 26°, angle at T is...?

Wait, angle at T in triangle PQT is the angle between PT and QT. But we know the angle between PT and TR is 2x = 52°, but unless QT is related to TR.

Alternatively, if we consider that line QT is such that it creates some relations.

Alternatively, use the fact that PT || QR, and PQ is a transversal, so angle between PT and PQ (x°) is equal to the corresponding angle between QR and PQ, which is angle PQR.

Therefore, angle PQR = x = 26°.

But earlier, the coordinate calculation gave ≈125°, but that was under the assumption of specific coordinates, which might not match the given angles.

Therefore, trusting the diagram and the given angles, the answer is 26°.

But let me verify with another method.

Let me consider extending some lines or using alternate interior angles.

Since PT || QR, and if we can find a transversal that creates alternate interior angles.

For example, if we consider the line PQ, which intersects PT at P and QR at Q. As before, corresponding angles would make angle PQR = x = 26°.

Alternatively, consider triangle PQR. If I can find two other angles, I can find the third.

At point Q, angle PQR = ?

At point P, angle QPR. Let's find that.

From the diagram, at point P, the angle between PT and PQ is x° = 26°, and since PT is parallel to QR, the angle between PQ and QR at Q is equal to x°, as corresponding angles.

But what about the angle at P in triangle PQR? It's angle QPR.

If we can find angle QPR, then we can use triangle angle sum.

Let me think.

At point P, angle between PQ and PT is x° = 26°. Since PT is parallel to QR, and PQ is a transversal, the angle between PQ and QR at Q is x° = 26°, which is angle PQR.

Now, if we consider the line PR, which connects P to R. Let me try to find the angles at P and R in triangle PQR.

But without more information, this is difficult.

However, if we consider that in triangle PQR, the sum of angles is 180°, and we know angle PQR = x = 26°, and if we can find another angle, say angle PRQ or angle QPR.

Wait, at point R, we have angle between QR and TR is 128°. If we can relate that to angle PRQ.

If we consider triangle PRQ, angle at R is angle PRQ, and angle between QR and TR is 128°. If lines PR and TR are related, but without knowing more, it's hard to say.

Alternatively, if we consider that PT || QR, and TR is a transversal, then the alternate interior angles would be...

At point T, the angle between PT and TR is 2x° = 52°, and the alternate interior angle would be the angle between QR and... some line. If there is a line parallel to TR, but not sure.

This is getting too convoluted.

Given that the corresponding angle logic gives angle PQR = x = 26°, and the same-side interior angles give x = 26°, and despite the coordinate calculation suggesting otherwise (which might be due to the diagram not being to scale), I think the correct answer is 26°.

However, I need to check for any other possible relationships.

Wait, let's consider the sum of angles around point T. At point T, there is angle between PT and TR (2x°), and if we can find other angles.

But without knowing more, it's difficult.

Alternatively, if we consider triangle PQT, and with x = 26°, we can find other angles.

Angle at P is 26°, angle at T is some angle, and angle at Q is 180 - 26 - angle at T.

But without knowing angle at T, it's not helpful.

Wait, but if we consider that at point T, the angle between PT and TR is 2x = 52°, and if QT is such that it forms a straight line with something?

Alternatively, since PT is parallel to QR, and if we draw a line from Q parallel to TR, but this is too vague.

Given the time I've spent and the consistency in the same-side interior angle calculation, I think the measure of angle PQR is 26°, so the answer is 26°.

But let me check with another approach.

If PT || QR, and PQ is a transversal, then angle between PT and PQ at P is equal to the angle between QR and PQ at Q.

This is the definition of corresponding angles, so angle PQR = x.

Since we found x = 26°, then angle PQR = 26°.

Therefore, the answer is 26°, so \boxed{26}.

**Final Answer**
\boxed{26}

To determine the measure of $\angle PQR$ in the given diagram, we are provided with certain angle relationships and the fact that segments $PT$ and $QR$ are parallel. Here's a step-by-step breakdown:

---

### **Understanding the Diagram and Given Angles**

From the Asymptote diagram:

- $PT$ and $QR$ are horizontal line segments and thus **parallel**.
- Point $P$ is at the left end of segment $PT$, point $T$ is its right end, and point $R$ is on the line $QR$ (at the right end of the base horizontal segment).
- Segments $QP$ and $QR$ form the angle $\angle PQR$, which is the angle of interest.
- The diagram labeled:

  - $x^\circ$ at point $P$, likely the angle between segment $PT$ (horizontal) and segment $QP$ (going down to $Q$).
  - $2x^\circ$ at point $T$, likely the angle between segment $PT$ and segment $TR$.
  - $128^\circ$ at point $R$, between segments $QR$ and $TR$.

From this, we can infer that the angle at point $T$ is between parallel lines (since $PT \parallel QR$) and a transversal $TR$. This gives us a chance to apply the **same-side interior angles** theorem.

---

### **Using the Same-Side Interior Angles Theorem**

Since $PT \parallel QR$ and the segment $TR$ is a transversal intersecting these lines:

- The angle at $T$ between $PT$ and $TR$ is $2x^\circ$.
- The angle at $R$ between $QR$ and $TR$ is $128^\circ$.
- These are *same-side interior angles*, and they must **add up to $180^\circ$** if the lines are parallel.

Therefore,

$$
2x + 128 = 180
$$

$$
2x = 52 \Rightarrow x = 26^\circ
$$

---

### **Corresponding Angles to Find $\angle PQR$**

We are now asked to find the measure of $\angle PQR$, which is the angle at point $Q$ between segments $QP$ and $QR$.

- The transversal $PQ$ intersects the parallel lines $PT$ and $QR$.
- The angle at $P$ between $PT$ and $PQ$ is $x = 26^\circ$.
- By the **corresponding angle theorem**, this angle must be equal to the angle at $Q$ between $QR$ and $PQ$, which is $\angle PQR$.

Thus,

$$
\angle PQR = x = 26^\circ
$$

---

### **Final Answer**

$$
\boxed{26}
$$

Solution:
Since $PT$ and $RQ$ are parallel, then $2x^\circ=128^\circ,$ so $x=64,$ so $\angle TPQ=64^\circ.$

[asy]
draw((0,0)--(10,0),black+linewidth(1));
draw((0,0)--(10,0),MidArrow);
draw((10,0)--(20,0),black+linewidth(1));
draw((0,0)--(-7,10)--(7,10)--(10,0),black+linewidth(1));
draw((-5,10)--(7,10),MidArrow);
label("$x^{\circ}$",(-6,10),SE);
label("$2x^{\circ}$",(7,10),SW);
label("$128^{\circ}$",(10,0),NE);
label("$P$",(-7,10),N);
label("$T$",(7,10),N);
label("$R$",(10,0),S);
label("$Q$",(0,0),S);
[/asy]

Since $PT$ and $QR$ are parallel, then $\angle TPQ$ and $\angle PQR$ are supplementary. Thus, $\angle PQR + 64^\circ = 180^\circ,$ so $\angle PQR = \boxed{116} \text{ degrees}.$