Geometry worksheet focusing on angle relationships and parallel lines, with problems involving corresponding angles, angle measures, and solving for x.

Let’s solve each part step by step.

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Problem 3:

We are given a figure with two parallel lines AB and CD, cut by two transversals (PS and RQ) intersecting at point O above line AB. Points X and Y are on AB; M and N are on CD. Angles are labeled accordingly.

(a) ∠POR = ∠ ?

∠POR is the angle at point O between points P, O, R — that’s the topmost angle formed by the two crossing lines PS and RQ. The vertically opposite angle to ∠POR is ∠XOY (or ∠MON if we consider the lower intersection). But looking at standard notation in such figures, since PS and RQ cross at O, then ∠POR and ∠SOQ are vertically opposite angles → they are equal.

But wait — let’s check the diagram logic:
Lines PS and RQ intersect at O. So vertical angles:
→ ∠POR = ∠SOQ
Also, ∠POQ = ∠ROS? No — actually, ∠POR and ∠SOQ are one pair of vertical angles.

But perhaps the expected answer is based on corresponding or alternate positions? Let’s think differently.

Actually, in many textbooks, when two lines intersect, vertical angles are equal. Since PS and RQ intersect at O, then:

∠POR and ∠SOQ are vertical angles → so ∠POR = ∠SOQ

But maybe the question expects an angle from the same “position” relative to the parallels? Hmm.

Wait — look again: The figure shows two transversals cutting two parallel lines. Point O is above AB, where PS and RQ meet. Then X and Y are intersections of PS and RQ with AB. M and N are intersections with CD.

So for (a): ∠POR — this is the angle at O between rays OP and OR. Its vertical opposite is ∠SOQ (between OS and OQ). So likely:

(a) ∠POR = ∠SOQ

But let’s hold off and see other parts.

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(b) ∠ROQ = ∠ ?

∠ROQ is the angle at O between R, O, Q — that’s the right-side angle at the top intersection. Its vertical opposite would be ∠POS (left side). But also, note that ∠ROQ and ∠XOY might be related? Actually, no — because X and Y are below.

Wait — perhaps it's asking for an angle that is equal due to parallel lines?

Actually, ∠ROQ is at point O. If we consider triangle XOY or something... Maybe not.

Alternatively, since AB || CD, and PS and RQ are transversals, then some angles correspond.

But ∠ROQ is at the top vertex. Perhaps its vertical angle is ∠POS? Yes.

Vertical angles: when two lines intersect, opposite angles are equal.

Lines PS and RQ intersect at O → so:

- ∠POR = ∠SOQ
- ∠ROQ = ∠POS

So probably:

(b) ∠ROQ = ∠POS

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(c) Write the angle which forms a corresponding angle pair with:

Corresponding angles are in matching corners when a transversal cuts two parallel lines.

(i) ∠BYQ — this is at point Y on line AB, on the right side, below AB? Wait — point Y is on AB, and ray YQ goes down-right toward Q. So ∠BYQ is the angle between BY (which is along AB to the right) and YQ (going down-right).

Since AB || CD, and YQ is part of transversal RQ, then the corresponding angle on line CD would be at point N, on the same side (right), and same relative position.

At point N, the angle between DN (along CD to the right) and NQ (same direction as YQ) would be ∠DNQ.

Because:

- Transversal RQ cuts AB at Y and CD at N.
- ∠BYQ is at Y, between AB (direction B) and transversal (down to Q).
- Corresponding angle at N should be between CD (direction D) and transversal (down to Q) → that’s ∠DNQ.

So (i) ∠BYQ corresponds to ∠DNQ

(ii) ∠SMC — point M is on CD, S is down-left, C is left along CD. So ∠SMC is angle at M between SM (transversal going down-left) and MC (along CD to the left).

Transversal PS cuts AB at X and CD at M.

∠SMC is at M, on the left side of transversal, below CD? Wait — let’s define:

Point M is on CD. Ray MS goes down-left to S. Ray MC goes left along CD.

So ∠SMC is the angle inside, between the transversal and the parallel line, on the lower-left side.

The corresponding angle on AB would be at point X, on the same side (left), and same relative position.

At X, the angle between AX (left along AB) and XS (down-left along transversal) is ∠AXS.

So (ii) ∠SMC corresponds to ∠AXS

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(d) ∠AXR = ? = ? = ?

∠AXR is at point X, between A (left on AB) and XR (up-right along transversal RQ? Wait — XR is part of RQ? Actually, from X to R — but R is up-right, so yes, XR is along RQ upward.

Wait — actually, looking at the figure: From point X, we have rays to A (left), to B (right), to S (down-left via PS), and to R (up-right via RQ)? That doesn’t make sense — only two transversals: PS and RQ.

Actually, point X is intersection of PS and AB. Point Y is intersection of RQ and AB.

So at point X, the angles are formed by AB and PS.

Similarly, at Y, by AB and RQ.

So ∠AXR — R is not connected to X directly? Unless XR means the ray from X through Y to R? Because RQ passes through Y and goes to R.

Ah! Probably, XR means the ray starting at X going through Y to R — i.e., along the transversal RQ extended.

But typically, we denote angles using three points where the middle is the vertex.

So ∠AXR: vertex at X, armsXA and XR.

If XR is along RQ, then yes — so ∠AXR is the angle at X between leftward AB and the transversal RQ going up-right.

This angle will have several equals due to parallel lines and vertical/corresponding/alternate angles.

First, since AB || CD, and RQ is a transversal, then:

∠AXR and ∠CNX are corresponding angles? Let’s see:

- At X: between AB (A-direction) and RQ (up to R)
- At N: between CD (C-direction) and RQ (up to R) — but N is on CD, and RQ goes up to R, so from N, going up to R is the same direction.

Actually, ∠AXR and ∠CNX: both are on the "upper" side of the transversal, and on the "left" side of the respective parallels? Not exactly.

Better to use alternate interior or corresponding.

Note that ∠AXR and ∠BXR are adjacent on straight line AB → sum to 180°.

But we need equal angles.

Since AB || CD, and RQ is transversal, then:

∠AXR and ∠CNX are corresponding angles? Let’s map:

- Line AB and CD parallel.
- Transversal RQ intersects AB at Y? Wait — earlier I thought X is on PS, Y on RQ.

I think I made a mistake here.

Let me reassign based on standard labeling:

Typically in such diagrams:

- Two horizontal parallel lines: AB (top) and CD (bottom)
- Two transversals: one from top-left to bottom-right (say PS), intersecting AB at X, CD at M
- Other transversal from top-right to bottom-left (RQ), intersecting AB at Y, CD at N
- They intersect each other at O above AB.

So:

- PS: P-O-X-M-S
- RQ: R-O-Y-N-Q

Therefore, at point X: only PS and AB intersect → so angles at X involve PS and AB.

Similarly at Y: RQ and AB.

So ∠AXR — if R is on the other transversal, then XR is not a single ray unless we mean from X to R passing through Y? But geometrically, X and R are not collinear with any named point except possibly O, but O is above.

Actually, ray XR would go from X through Y to R, since Y is on RQ and X is on AB, and RQ passes through Y.

So yes, ray XR is the same as ray XY extended to R.

So ∠AXR is the angle at X between ray XA (left along AB) and ray XR (which is along RQ, going up-right through Y to R).

Now, since AB || CD, and RQ is a transversal, then the corresponding angle to ∠AXR on line CD would be at point N, between ray NC (left along CD) and ray NR (up along RQ) — which is ∠CNR.

But ∠CNR is the same as ∠CNQ? No — Q is down, R is up.

Ray NR is toward R, up.

So ∠CNR is angle at N between C (left) and R (up).

Yes, so ∠AXR and ∠CNR are corresponding angles → equal.

Additionally, vertically opposite or other relations.

Also, at point Y, ∠AYR or something.

Another way: ∠AXR and ∠XYR are related? Not directly.

Note that ∠AXR and ∠BXR are supplementary.

But we need equal angles.

Since AB || CD, and PS is another transversal, but ∠AXR involves RQ, not PS.

Perhaps ∠AXR = ∠DNY or something.

Let’s list all angles equal to ∠AXR.

First, corresponding angle on CD: as said, ∠CNR (at N, between CN and NR)

Second, vertically opposite? At point Y, what angle corresponds?

At Y, the angle between AY (left along AB) and YR (up along RQ) is ∠AYR.

Is ∠AXR equal to ∠AYR? Only if X and Y are symmetric, which they’re not necessarily.

Actually, no — because X and Y are different points.

Unless the transversals are symmetric, but generally not.

Another approach: since PS and RQ intersect at O, then at O, we have vertical angles.

But ∠AXR is at X.

Perhaps use the fact that ∠AXR and ∠OXY are the same angle? No.

Let’s think of alternate exterior or something.

I recall that in such configurations, there are multiple equal angles due to parallel lines.

Specifically, for transversal RQ cutting parallel lines AB and CD:

- Corresponding angles:
- ∠AYR and ∠DNR (both on the right side, upper)
- ∠BYR and ∠CNR (on the left side? Let's define properly)

Standard correspondence:

When transversal RQ cuts AB at Y and CD at N:

- Angle at Y above AB on the right: ∠RYB
- Corresponding angle at N above CD on the right: ∠RND

But our angle is ∠AXR, which is at X, not Y.

X is on PS, not on RQ.

So ∠AXR is formed by AB and the ray to R, which is along RQ, but from X, not from Y.

This is confusing.

Perhaps in the diagram, "XR" means the ray from X to R, but since R is on the other transversal, and X is on PS, then XR is not along a straight line with any transversal unless specified.

I think there might be a misinterpretation.

Looking back at the problem: in part (d), it says ∠AXR = ___ = ___ = ___ , implying three angles equal to it.

In standard geometry problems like this, often ∠AXR refers to the angle at X between A, X, and R, but if R is not on the same line, it might be a typo or convention.

Perhaps "R" here is meant to be "S" or something else.

Another possibility: in some notations, the letter after X indicates the direction. For example, in ∠AXR, R might indicate the ray towards R, but in the context, from X, the ray towards R is not defined unless we consider the line.

Perhaps it's ∠AXS or ∠AXP.

Let's look at part (e): ∠BYQ = ___ = ___ = ___

∠BYQ is at Y, between B, Y, Q. B is right on AB, Q is down-right on RQ. So this is the angle at Y between AB (right) and RQ (down to Q).

This is a standard angle.

For ∠BYQ, since AB || CD, and RQ is transversal, then corresponding angle on CD is at N, between D (right) and NQ (down to Q) — so ∠DNQ.

Also, vertically opposite or other.

At point Y, ∠BYQ and ∠AYP or something.

Also, since PS and RQ intersect at O, then at O, there might be equal angles.

For example, ∠BYQ and ∠ROY or something.

Let's calculate for (e) first.

(e) ∠BYQ = ? = ? = ?

∠BYQ is at Y, between BY (right on AB) and YQ (down along RQ).

This angle has:

- Corresponding angle on CD: at N, between DN (right on CD) and NQ (down along RQ) — so ∠DNQ

- Also, vertically opposite angle? At Y, the vertical angle to ∠BYQ would be the angle between AY and YP or something, but YP is not defined.

At point Y, the two lines are AB and RQ, so the vertical angle to ∠BYQ is ∠AYR (between AY and YR, where YR is up along RQ).

Because when two lines intersect, vertical angles are equal.

Lines AB and RQ intersect at Y, so:

- ∠BYQ and ∠AYR are vertical angles → equal

- Similarly, ∠AYQ and ∠BYR are vertical angles.

So ∠BYQ = ∠AYR

Additionally, since AB || CD, and RQ is transversal, then ∠BYQ and ∠DNQ are corresponding angles → equal

Also, ∠DNQ and its vertical angle or other.

At N, ∠DNQ and ∠CNM or something.

Moreover, since PS is another transversal, but for now, for ∠BYQ, we have:

∠BYQ = ∠AYR (vertical) = ∠DNQ (corresponding)

And perhaps another one.

At point O, the angle ∠ROY or something.

For example, ∠BYQ and ∠ROY: are they equal? Only if corresponding or alternate.

Actually, ∠BYQ and ∠ROY are not directly related.

Note that ∠BYQ and ∠OYR are adjacent.

Perhaps ∠BYQ = ∠ONM or something.

Let's list:

From above:

1. Vertical angle at Y: ∠AYR

2. Corresponding angle on CD: ∠DNQ

3. Now, at N, ∠DNQ has a vertical angle? When RQ and CD intersect at N, the vertical angle to ∠DNQ is ∠CNQ? No.

Lines RQ and CD intersect at N, so vertical angles are:

- ∠DNQ and ∠CNQ? No, those are adjacent.

Actually, the four angles at N:

- Between DN and NQ: ∠DNQ

- Between NQ and NC: ∠QNC

- Between NC and NM: etc.

Vertical to ∠DNQ is the angle between CN and NQ? No.

When two lines intersect, vertical angles are opposite.

So at N, lines CD and RQ intersect, so:

- Angle between DN and NQ: let's call it α

- Vertical to it is the angle between CN and the extension of NQ beyond N, but since NQ is a ray, the opposite ray is NR.

So the vertical angle to ∠DNQ is ∠CNR (between CN and NR)

Because DN and CN are opposite rays on line CD, and NQ and NR are opposite rays on line RQ.

So yes, ∠DNQ and ∠CNR are vertical angles → equal.

So ∠DNQ = ∠CNR

Therefore, for ∠BYQ, we have:

∠BYQ = ∠AYR (vertical at Y) = ∠DNQ (corresponding) = ∠CNR (vertical at N)

So three angles: ∠AYR, ∠DNQ, ∠CNR

But the problem asks for three blanks, so likely these.

Similarly for (d) ∠AXR.

Assume ∠AXR is at X, between A (left on AB) and XR. If XR is along RQ, but from X, it's not direct.

Perhaps in the diagram, "R" is a typo, and it's meant to be "S", since PS passes through X.

Let me check common problems.

Often, ∠AXS or ∠AXP is used.

Suppose ∠AXR is meant to be ∠AXS, where S is on PS down-left.

Then ∠AXS is at X, between A (left) and S (down-left along PS).

Then, since AB || CD, and PS is transversal, corresponding angle on CD is at M, between C (left) and S (down-left) — so ∠CMS.

Also, vertical angle at X: when AB and PS intersect at X, vertical to ∠AXS is ∠BXP (between B and P, up-right).

Also, at M, ∠CMS and its vertical angle.

So ∠AXS = ∠BXP (vertical) = ∠CMS (corresponding) = ∠DMN or something.

At M, vertical to ∠CMS is ∠DMN? Let's see.

Lines AB and PS intersect at X, CD and PS intersect at M.

At X: vertical angles: ∠AXS and ∠BXP

At M: vertical angles: ∠CMS and ∠DMN (since CS and DS are opposite, and MS and NS are opposite, but NS is not defined; usually, the vertical to ∠CMS is the angle between DM and the opposite ray of MS, which is MP or something.

Ray MS is down-left, so opposite ray is up-right, say to P.

So at M, angle between DM (right on CD) and MP (up-right along PS) is ∠DMP, and this is vertical to ∠CMS.

Yes.

So if ∠AXR is actually ∠AXS, then:

∠AXS = ∠BXP = ∠CMS = ∠DMP

But the problem writes ∠AXR, not ∠AXS.

Perhaps in the diagram, R is on the same line, but unlikely.

Another idea: perhaps "R" in ∠AXR refers to the point R, but the ray is from X to R, which may not be straight, but in geometry problems, sometimes it's assumed that we consider the angle formed by the lines.

To resolve, let's look at the answer format and common answers.

For (d) ∠AXR, and (e) ∠BYQ, and from symmetry, likely:

For (d): ∠AXR = ∠BYS = ∠CMR = ∠DNS or something.

Perhaps use the fact that at O, angles are equal.

Let's assume that ∠AXR is the angle at X between A and the line to R, but since R is on RQ, and X is on PS, then the line XR is not a transversal, so it's complicated.

Perhaps in the context, "XR" means the ray along the transversal that contains R, but from X, it's not defined.

I think there might be a mistake in my initial assumption.

Let's read the figure description again.

The figure has points: P, O, R on top; A,X,Y,B on middle line; C,M,N,D on bottom line; S, Q on bottom.

Transversals: PS from P to S, passing through O and X and M.

RQ from R to Q, passing through O and Y and N.

So at point X, only PS and AB intersect, so angles at X are between AB and PS.

Similarly at Y, between AB and RQ.

So for ∠AXR, if R is not on PS, then it must be that "R" is a misnomer, and it should be "S" or "P".

Perhaps "R" in ∠AXR is meant to be the direction, but in standard notation, it's the point.

Another possibility: in some notations, the third letter indicates the arm, so ∠AXR means the angle at X with arms XA and XR, and if XR is the ray towards R, but since R is not on the line from X, it might be the angle between XA and the line XR, but that would require defining the line.

I think for the sake of solving, and given that in many similar problems, for ∠AXR, it is intended to be the angle with the transversal, and since R is on RQ, but X is on PS, perhaps it's a different interpretation.

Let's look at part (c)(i) : ∠BYQ corresponds to ∠DNQ, as we said.

For (d) ∠AXR, perhaps it is the angle at X between A and the line to R, but since R is on the other side, maybe it's the angle that is alternate or something.

Perhaps ∠AXR is the same as ∠OXY or something.

Let's calculate the values later, but for now, let's assume that for (d), ∠AXR is meant to be ∠AXS, as it's common.

Or perhaps in the diagram, the ray from X to R is considered, but in reality, from X, the ray to R is not straight, so likely a typo.

Given that, and to proceed, I'll assume that for (d), ∠AXR is ∠AXS, and for (e), ∠BYQ is as is.

So for (d) ∠AXS = ∠BXP = ∠CMS = ∠DMP

For (e) ∠BYQ = ∠AYR = ∠DNQ = ∠CNR

But the problem has three blanks for each, so three angles besides itself.

So for (d): ∠AXR = ∠BXP = ∠CMS = ∠DMP

For (e): ∠BYQ = ∠AYR = ∠DNQ = ∠CNR

Now for (a) and (b).

(a) ∠POR = ?

As vertical angle, ∠POR = ∠SOQ

(b) ∠ROQ = ∠POS

(c) (i) ∠BYQ corresponds to ∠DNQ

(ii) ∠SMC corresponds to ∠AXS (as we said earlier)

But in (c)(ii), it's ∠SMC, which is at M, between S, M, C. S is down-left, C is left on CD, so between SM and MC.

Corresponding angle on AB is at X, between SX and XA, which is ∠AXS.

Yes.

So now for the final answers.

But let's confirm with the value problems.

Perhaps for (d) and (e), the angles are equal due to parallel lines and vertical angles, and we can list them.

So for (d) ∠AXR = ∠BXP = ∠CMS = ∠DMP (assuming XR means XS)

But to match the notation, perhaps in the diagram, R is used for the direction, but I think it's safe to go with that.

For (e) ∠BYQ = ∠AYR = ∠DNQ = ∠CNR

Now for Problem 4.

Problem 4(a):

AV || DJ

Lines AV and DJ are parallel, cut by transversal MH.

Angles given: at left, 7x - 12° ; at right, 4x - 6°

These are corresponding angles or alternate interior?

Looking at the figure: M-H is the transversal.

At left, angle between M-A-V is 7x-12°, which is on the upper left.

At right, angle between H-D-J is 4x-6°, on the upper right.

Since AV || DJ, and MH is transversal, then these two angles are corresponding angles if they are in the same relative position.

Both are on the "top" side of the transversal, and on the "left" for the first, "right" for the second — not the same side.

Actually, for corresponding angles, they should be on the same side of the transversal and same side of the parallels.

Here, 7x-12° is at A, between MA and AV, so if MA is the transversal coming from M, then this angle is between the transversal and the parallel line AV.

Similarly, 4x-6° is at D, between HD and DJ.

Since AV || DJ, and MH is transversal, then the angle at A and the angle at D are corresponding if they are both on the same side.

In this case, both angles are on the "upper" side of the parallels, and the transversal is crossing, so likely they are corresponding angles.

Specifically, ∠MAV and ∠HDJ are corresponding angles.

And since lines are parallel, corresponding angles are equal.

So 7x - 12 = 4x - 6

Solve:

7x - 4x = -6 + 12

3x = 6

x = 2

Check: 7*2 - 12 = 14-12=2°; 4*2-6=8-6=2° — equal, good.

So x=2 for (a)

Problem 4(b):

XY || PQ and MN || RS

Figure shows a quadrilateral-like shape with points X,N,S,Y on one side, P,M,R,Q on the other.

Angles given: at X, 11x+20° ; at Y, 12x+22°

Since XY || PQ, and MN || RS, likely this is a parallelogram or has properties.

The angles at X and Y are consecutive angles or something.

Probably, the figure is such that XY and PQ are parallel, MN and RS are parallel, so it might be a parallelogram if the sides are connected properly.

Points: X to N to S to Y, and P to M to R to Q, with XY || PQ, MN || RS.

Likely, the polygon is X-N-S-Y and P-M-R-Q, but with connections.

Probably, it's a quadrilateral with vertices X,P,Q,Y or something.

From the diagram description: points X,N,S,Y are on the bottom, P,M,R,Q on the top, with XY || PQ, and MN || RS.

Also, angles at X and Y are given: at X, angle between XN and XP or something.

Typically, in such figures, the angle at X is between the two sides meeting at X, which are XN and XP, but XP is not defined.

From the labels: at X, angle is 11x+20°, which is likely the interior angle of the polygon at X.

Similarly at Y, 12x+22°.

Since XY || PQ, and assuming that the figure is a trapezoid or parallelogram, but with MN || RS, it might be that the opposite sides are parallel, so it could be a parallelogram.

If XY || PQ and MN || RS, and if MN and RS are the other pair, then likely it's a parallelogram, so opposite angles are equal, or consecutive angles sum to 180°.

But here, angles at X and Y are given, which are adjacent if it's a quadrilateral X-Y-something.

Assume the quadrilateral is X-P-Q-Y or X-N-S-Y, but with diagonals.

Perhaps the vertices are X, P, Q, Y, with XP, PQ, QY, YX.

But then MN and RS are additional lines.

The condition is XY || PQ and MN || RS, so probably MN and RS are the other pair of sides.

So likely, the quadrilateral has sides XP, PQ, QY, YX, but then MN and RS are not sides.

Perhaps it's a hexagon or something.

Another possibility: points are arranged as X and Y on bottom, P and Q on top, with XY || PQ, and then M and N on left, R and S on right, with MN || RS, and the figure is bounded by X-M-P, M-N, N-S, S-Y, Y-Q, Q-R, etc., but that's messy.

From the angle labels: at X, angle 11x+20° is between the two lines meeting at X, which are likely XN and XP, but in the diagram, from X, there is line to N and to P or to M.

Typically, in such problems, the angle at X is the interior angle of the polygon formed.

Moreover, since XY || PQ, and if we consider the transversal XP or something.

Notice that the angles at X and Y are on the same side, and if the figure is convex, and XY is one side, then angles at X and Y are consecutive angles.

In a parallelogram, consecutive angles are supplementary.

Is this a parallelogram? With XY || PQ and MN || RS, if MN and RS are the other pair, then yes, it should be a parallelogram.

For example, if the quadrilateral is X-M-R-Y or something.

Assume that the quadrilateral has vertices X, M, R, Y, with XM, MR, RY, YX.

Then if XY || MR? But the condition is XY || PQ, and MN || RS.

Perhaps P and Q are on the top, so maybe the quadrilateral is X-P-Q-Y, with XP, PQ, QY, YX.

Then if XY || PQ, that's one pair of opposite sides parallel.

Then MN || RS — M and N are on XP and YQ or something.

This is ambiguous.

Perhaps "MN || RS" means that the lines MN and RS are parallel, and they are the other pair of sides.

In many such problems, when two pairs of opposite sides are parallel, it's a parallelogram, and opposite angles are equal, or consecutive angles sum to 180°.

Here, angles at X and Y are given, and if X and Y are adjacent vertices, then their angles should sum to 180° if it's a parallelogram.

If they are opposite, then equal.

In the diagram, likely X and Y are adjacent, since both are on the bottom.

So assume that in quadrilateral X-P-Q-Y, with XY || PQ, and if XP || YQ, then it's a parallelogram, but the condition is MN || RS, not XP || YQ.

Perhaps M and N are on XP and YQ, and R and S on the top, but it's complicated.

Another way: perhaps the angles at X and Y are corresponding or alternate due to the parallels.

Let's look at the values.

Suppose that since XY || PQ, and there is a transversal, say XP or YQ.

At X, the angle 11x+20° might be between XY and XP.

At Y, 12x+22° between YX and YQ.

If XP and YQ are the same line or something.

Perhaps the line from X to P and Y to Q are the transversals.

But with MN || RS, it might be that the figure has properties that make the angles equal or supplementary.

Perhaps in this configuration, the angles at X and Y are equal because of the parallels.

Or sum to 180°.

Let me try setting them equal: 11x+20 = 12x+22 => 20-22 = 12x-11x => -2 = x, negative, impossible.

Set sum to 180°: 11x+20 + 12x+22 = 180 => 23x +42 = 180 => 23x = 138 => x=6

Check: 11*6+20=66+20=86°; 12*6+22=72+22=94°; 86+94=180° — good.

And in a parallelogram, consecutive angles are supplementary, so if X and Y are adjacent vertices, this makes sense.

Moreover, with XY || PQ and if the other pair is parallel, then yes.

So likely x=6 for (b).

Now back to Problem 3.

For (a) ∠POR = ∠SOQ (vertical angles)

(b) ∠ROQ = ∠POS (vertical angles)

(c) (i) ∠BYQ corresponds to ∠DNQ

(ii) ∠SMC corresponds to ∠AXS

(d) ∠AXR = ∠BXP = ∠CMS = ∠DMP (assuming XR means XS)

But to match the notation, perhaps in the diagram, the ray is labeled, but I think for the answer, we can write the angles.

Since the problem has blanks, likely specific names.

For (d), perhaps ∠AXR = ∠OYR or something, but let's stick with the vertical and corresponding.

Another thought: in some books, ∠AXR might mean the angle at X in triangle or something, but I think for consistency, we'll use:

For (d): ∠AXR = ∠BXP = ∠CMS = ∠DMP

For (e): ∠BYQ = ∠AYR = ∠DNQ = ∠CNR

But in (e), ∠AYR is at Y, between A,Y,R, which is correct.

Now for the final answers.

So summarizing:

Problem 3:

(a) ∠POR = ∠SOQ

(b) ∠ROQ = ∠POS

(c) (i) ∠BYQ corresponds to ∠DNQ

(ii) ∠SMC corresponds to ∠AXS

(d) ∠AXR = ∠BXP = ∠CMS = ∠DMP

(e) ∠BYQ = ∠AYR = ∠DNQ = ∠CNR

Problem 4:

(a) x = 2

(b) x = 6

Now for the boxed answer, since the user asked for final answer, and there are multiple parts, probably list them.

But the instruction is to provide final answer after reasoning, and for multiple questions, perhaps box each or list.

Since it's a single response, I'll write the answers clearly.

For Problem 3, the answers are angles, for 4, numerical.

So for the final output, I'll state:

Final Answer for Problem 3:

(a) \angle SOQ

(b) \angle POS

(c) (i) \angle DNQ

(ii) \angle AXS

(d) \angle BXP = \angle CMS = \angle DMP

(e) \angle AYR = \angle DNQ = \angle CNR

For Problem 4:

(a) x = 2

(b) x = 6

But in (d) and (e), there are three angles, so perhaps list them as per the blanks.

In the problem, for (d) it's ∠AXR = ___ = ___ = ___ , so three angles equal to it.

Similarly for (e).

So in (d), the three angles are ∠BXP, ∠CMS, ∠DMP

In (e), ∠AYR, ∠DNQ, ∠CNR

Yes.

Now to write the final answer.

Also, for (c)(i) and (ii), it's "write in the angle", so just the angle name.

So I think this is complete.