Complementary, Supplementary, and Vertical Angles Practice ... - Free Printable

Let’s solve each problem step by step. We’ll go one at a time, carefully checking our work.

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Problem 1:
We see two angles on a straight line: 120° and x°.
Angles on a straight line add up to 180° → they are supplementary.
So:
x = 180 - 120 = 60

→ Type: supplementary, x = 60

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Problem 2:
Two lines cross — vertical angles are equal.
One angle is 119°, the opposite (vertical) angle is x°.
Vertical angles are always equal → x = 119

But wait — let’s check: adjacent angles would be supplementary. The question asks to classify as complementary, supplementary, or adjacent. These two angles are vertical, but that’s not one of the options. However, since they’re formed by intersecting lines, and we’re told to pick from those three, note: vertical angles are NOT adjacent, not necessarily complementary or supplementary unless specified.

Wait — actually, looking again: the diagram shows two intersecting lines, with one angle labeled 119° and the vertically opposite angle labeled x°. Since vertical angles are equal, x = 119. But what classification? The pair of angles shown are vertical, which isn’t listed. But perhaps the question means: for the pair given, are they complementary, supplementary, or adjacent?

Actually, in this case, the two angles marked (119° and x°) are vertical angles, so they are not adjacent, and their sum is 238° — not 90 or 180. So maybe the instruction is to classify the relationship between the two angles shown? That doesn’t fit any of the three categories perfectly.

Wait — perhaps I misread. Let me re-read the instructions: “Write if angles are complementary, supplementary, or adjacent.” It might mean: for the figure, identify the type of angle pair shown, then find x.

In Problem 2, the two angles shown are vertical — but vertical is not an option. Alternatively, maybe it's implying the relationship between the two angles that form a linear pair? No — the diagram clearly marks the vertical angle.

Hold on — perhaps the question expects us to say “adjacent” only if they share a side and vertex without overlapping. In this case, 119° and x° do NOT share a side — they are opposite. So not adjacent. Not complementary (sum ≠ 90). Not supplementary (sum ≠ 180). Hmm.

Wait — maybe I made a mistake. Let’s look at other problems first and come back.

Actually, let’s think differently: sometimes in these worksheets, when two lines intersect, and you’re given one angle and asked for the vertical angle, you just use vertical angles theorem. And for classification — perhaps they want “neither”, but that’s not an option. Or maybe the question is poorly worded.

Alternatively — perhaps in Problem 2, the two angles shown are meant to be considered as a pair — but they are vertical, so not adjacent, not comp/supp. This is confusing.

Wait — let’s check Problem 8: same setup — two intersecting lines, one angle 59°, other x° — likely vertical again. So probably for both 2 and 8, x equals the given angle because they are vertical.

And for classification — maybe the worksheet considers vertical angles under “adjacent”? No, that’s incorrect. Adjacent angles share a common side and vertex, and don’t overlap. Vertical angles do not share a side.

Perhaps the instruction is to classify the *relationship* between the two angles that are drawn together — like in Problem 1, they are on a straight line → supplementary. In Problem 3, they form a right angle → complementary.

For Problem 2: the two angles shown are vertical — but since “vertical” isn’t an option, and they are not adjacent, not comp, not supp — this is a problem.

Wait — maybe I should look at the actual diagrams mentally:

Problem 2: two lines crossing, forming four angles. One is labeled 119°, the one directly opposite is labeled x°. So yes, vertical angles → x = 119. For classification — perhaps the question allows “none of the above”, but it doesn’t say that. Or maybe it’s a trick.

Another idea: perhaps “classify angles” refers to the type of angle pair that includes the unknown x and the given angle. In this case, they are vertical, but since vertical isn't listed, and they are not adjacent, not comp, not supp — this is odd.

Let’s skip and come back. Maybe after doing others, it will make sense.

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Problem 3:
Right angle symbol (square corner), so total 90°. Two angles: 55° and x°. They are adjacent and form a right angle → complementary.
x = 90 - 55 = 35

→ Type: complementary, x = 35

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Problem 4:
Straight line → 180°. Angles: 40° and x° → supplementary.
x = 180 - 40 = 140

→ Type: supplementary, x = 140

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Problem 5:
Three rays from a point. One angle is 80°, another is x°, and there’s a right angle symbol between them? Wait — looking at description: it says "80°" and "x°", and there’s a square symbol — meaning the angle between the two outer rays is 90°? Or is the 80° part of it?

Actually, typical diagram: two angles sharing a ray, and together they form a larger angle. If there’s a right angle mark covering both 80° and x°, then 80 + x = 90 → x = 10. And they are adjacent and complementary.

Yes — that makes sense. The right angle symbol spans both angles, so they add to 90° → complementary.
x = 90 - 80 = 10

→ Type: complementary, x = 10

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Problem 6:
Straight line → 180°. Angles: 98° and x° → supplementary.
x = 180 - 98 = 82

→ Type: supplementary, x = 82

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Problem 7:
Right angle symbol (90° total). One angle is 22°, other is x° → complementary.
x = 90 - 22 = 68

→ Type: complementary, x = 68

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Problem 8:
Intersecting lines. One angle 59°, opposite angle x° → vertical angles → equal.
x = 59

Classification: again, vertical angles — not adjacent, not comp, not supp. Same issue as Problem 2.

Perhaps the worksheet intends for us to say “adjacent” only if they are next to each other, but here they are opposite. Maybe for Problems 2 and 8, since they are vertical, and vertical angles are not classified under those three, but the problem forces us to choose — perhaps it’s a mistake, or perhaps in some curricula they consider them under “adjacent” incorrectly.

Wait — let’s read the instruction again: “Write if angles are complementary, supplementary, or adjacent.” It doesn’t say “only one of these”, but implies to pick the correct term.

In geometry, vertical angles are a separate category. But since it’s not listed, and for Problems 2 and 8, the angles are not adjacent (they don’t share a side), not complementary (sum >90), not supplementary (sum <180 for 59+59=118, 119+119=238 — neither is 180).

This is problematic. Perhaps for Problem 2, the 119° and x° are not the vertical pair? Maybe the diagram shows 119° and x° as adjacent? But the user said “two lines crossing, with 119° and x° on opposite sides”.

I think there might be a misunderstanding in the diagram interpretation. Let me assume based on standard problems:

In many such worksheets, when two lines intersect and you're given one angle and asked for the vertically opposite, you use vertical angles theorem, and for classification, if it's not fitting, perhaps leave blank or write "vertical", but the instruction says to choose from those three.

Another possibility: in Problem 2, the two angles shown are actually adjacent? But that wouldn't make sense with the values.

Let’s calculate: if 119° and x° are adjacent on a straight line, then x = 180 - 119 = 61, and they would be supplementary. But the diagram likely shows them as vertical.

Given the context, and to move forward, I'll assume that for Problems 2 and 8, since they are vertical angles, and the worksheet might have a typo, but logically, x is equal to the given angle, and for classification, perhaps they expect "none", but since it's not an option, maybe for those, we can say "adjacent" is incorrect, so perhaps the intended answer is to recognize they are vertical, but since not listed, maybe skip classification? But the problem requires it.

Let’s look at Problem 9.

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Problem 9:
Straight line → 180°. Angles: x° and 6° → supplementary.
x = 180 - 6 = 174

→ Type: supplementary, x = 174

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Now back to Problem 2 and 8.

Perhaps in Problem 2, the 119° and x° are not the vertical pair, but rather, the diagram shows two angles that are adjacent and form a linear pair? But 119 + x = 180 would give x=61, and they would be supplementary.

But the user described it as "two lines crossing, with 119° and x° on opposite sides", which suggests vertical.

However, in many online worksheets, sometimes the labeling is such that x is the adjacent angle. Let me think differently.

Perhaps for Problem 2, the angle labeled 119° and the angle labeled x° are actually adjacent and on a straight line? But that would be unusual labeling.

Another idea: in some diagrams, when two lines intersect, they label one angle and the vertically opposite, but ask for the adjacent one. But here it's labeled x° on the opposite.

I recall that in some systems, "classify the angles" might refer to the type of the angle itself (acute, obtuse, etc.), but the instruction says "complementary, supplementary, or adjacent", which are relationships between pairs.

Given the time, and to provide answers, I'll assume that for Problems 2 and 8, since they are vertical angles, and vertical angles are not complementary, supplementary, or adjacent, but the problem might have a mistake, or perhaps in this context, they want us to say "adjacent" for all intersecting line angles, which is wrong.

Let’s check Problem 10 and 11 for clues.

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Problem 10:
Lines intersecting. Given angle 43°. Find angles 1,2,3.

Typically, angle 1 is vertical to 43° → so angle 1 = 43°.

Angle 2 is adjacent to 43° on a straight line → supplementary → 180 - 43 = 137°.

Angle 3 is vertical to angle 2 → so angle 3 = 137°.

Or depending on labeling: usually, if 43° is given, and angles are numbered around the intersection.

Assume: the 43° angle is between two lines. Angle 1 is opposite to it → vertical → 43°.

Angle 2 is adjacent to 43° → supplementary → 137°.

Angle 3 is opposite to angle 2 → vertical → 137°.

Reasoning: vertical angles are equal; adjacent angles on a straight line are supplementary.

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Problem 11:
Diagram with perpendicular lines and a diagonal. Angles labeled a,b,c,d,e.

From the description: there is a right angle symbol, so some angles are 90°.

Specifically, angle a has a right angle symbol, so a = 90°.

Also, the horizontal and vertical lines are perpendicular, so the angles at the intersection are 90° each, but then a diagonal cuts through.

Angles: a, b, c, d, e.

Typically, in such diagrams:

- Angle a is the right angle between vertical and horizontal, so a = 90°.

- Then the diagonal creates angles b, c, d, e.

Usually, b and e are vertical, c and d are vertical, etc.

Part a) Vertical angles: pairs that are opposite. Likely, b and e are vertical, c and d are vertical. Also, a is 90°, and its vertical counterpart might be the angle below, but not labeled. Probably, vertical pairs are: b and e, c and d.

Part b) Complementary: ∠c and ? Complementary means sum to 90°. Since a is 90°, and if c is part of it, but typically, if the diagonal is cutting the right angle, then b and c might be complementary if they form the 90° angle.

Assume: the vertical and horizontal lines intersect at O. Diagonal line passes through O. Angle a is the top-left quadrant, marked 90°. Then angle b is between vertical and diagonal in top-right, angle c between diagonal and horizontal in top-right, so b + c = 90° → complementary.

Similarly, in bottom-left, e and d might be complementary.

Part c) Supplementary: ∠c and ? Sum to 180°. Likely, c and d are on a straight line? Or c and the angle adjacent on the straight line.

If the horizontal line is straight, then angle c and angle d are on opposite sides, but not necessarily supplementary unless they are adjacent on a straight line.

Actually, angle c and angle e might be supplementary if they are on a straight line, but let's think.

Standard: on a straight line, adjacent angles are supplementary.

For example, angle c and angle d: if they are on the horizontal line, but c is above, d is below, so not adjacent.

Perhaps angle c and the angle that is adjacent to it on the straight line. For instance, if we consider the horizontal line, the angles on one side: from left to right, angle e (bottom-left), then the diagonal, then angle c (top-right)? Messy.

Better to assign:

Assume the intersection point O.

- Vertical line: up and down.

- Horizontal line: left and right.

- Diagonal line: say, from bottom-left to top-right.

Then:

- Angle a: between up-vertical and left-horizontal → 90° (given by symbol).

- Angle b: between up-vertical and diagonal (in top-right quadrant).

- Angle c: between diagonal and right-horizontal (in top-right quadrant).

- Angle d: between right-horizontal and down-vertical (bottom-right quadrant).

- Angle e: between down-vertical and diagonal (bottom-left quadrant).

Then:

- Angles b and c are adjacent and together make the top-right 90° angle? No, the top-right quadrant is between up-vertical and right-horizontal, which is 90°, and it's split into b and c by the diagonal. So b + c = 90° → complementary.

Similarly, in bottom-left, angle e and the angle between diagonal and left-horizontal — but that's not labeled. Angle e is between down-vertical and diagonal, and the angle between diagonal and left-horizontal is not labeled, but perhaps it's part of e or something.

Actually, the bottom-left quadrant is between left-horizontal and down-vertical, which is 90°, and it's split by the diagonal into two angles: one is e (between down-vertical and diagonal), and the other is between diagonal and left-horizontal — let's call it f, but not labeled. So e + f = 90°.

But in the diagram, only a,b,c,d,e are labeled, so probably d is in bottom-right, e in bottom-left.

Angle d: between right-horizontal and down-vertical — that's the bottom-right quadrant, which is 90°, but if the diagonal is there, it might be split, but in this case, the diagonal is from bottom-left to top-right, so in bottom-right, it's not cut; angle d is the entire bottom-right quadrant? But that would be 90°, and similarly for others.

I think I have it: the diagonal line divides the plane, and the angles are defined as follows:

- At the intersection, the four quadrants are divided, but with the diagonal, we have six angles? No, typically five labels for the regions.

Perhaps a is the right angle at top-left, so between positive y-axis and negative x-axis.

Then b is between positive y-axis and the diagonal (which is in first quadrant).

c is between the diagonal and positive x-axis.

d is between positive x-axis and negative y-axis (fourth quadrant).

e is between negative y-axis and the diagonal (third quadrant).

Then:

- Angle a = 90° (given).

- Angles b and c are in the first quadrant, and since the first quadrant is 90°, b + c = 90°.

- Similarly, in the third quadrant, angle e and the angle between diagonal and negative x-axis — but that's not labeled. The angle between negative x-axis and negative y-axis is 90°, and it's split by the diagonal into e and another angle, say f, so e + f = 90°.

But f is not labeled, so perhaps for the purpose of this problem, we consider only the labeled angles.

Part a) Vertical angles: angles that are opposite each other.

- Angle b and angle e: are they vertical? Angle b is in first quadrant between y-axis and diagonal, angle e is in third quadrant between negative y-axis and diagonal. If the diagonal is a straight line, then the ray for b and the ray for e are opposite, so yes, b and e are vertical angles.

Similarly, angle c and angle d: angle c is between diagonal and positive x-axis, angle d is between positive x-axis and negative y-axis — not opposite.

Angle c and the angle in the third quadrant between diagonal and negative x-axis — but that's not labeled.

Actually, the vertical angle to c would be the angle in the third quadrant between the diagonal and the negative x-axis, which is not labeled; it's part of e or something.

Perhaps d and the angle in the second quadrant — but a is already there.

This is messy. Let's assume standard labeling:

In many textbooks, for such a diagram:

- Vertical angles are: b and e, and also c and the angle opposite to it, but since d is labeled, perhaps c and d are not vertical.

Another way: the vertical angle to b is the angle directly opposite, which would be in the third quadrant, between the diagonal and the negative y-axis — that's angle e. Yes, so b and e are vertical.

Similarly, the vertical angle to c is the angle in the third quadrant between the diagonal and the negative x-axis — but that's not labeled; it's adjacent to e. Since e is between negative y-axis and diagonal, the angle between diagonal and negative x-axis is different.

Perhaps angle d is in the fourth quadrant, and its vertical angle is in the second quadrant, which is part of a or something.

To simplify, commonly in such problems:

- Vertical angles: b and e, and also c and the angle that is not labeled, but since d is there, perhaps d and the angle in the second quadrant.

But angle a is 90°, and it's composed of the second quadrant, which is between negative x-axis and positive y-axis, and if the diagonal is not in it, then a is alone.

I think for this problem, the vertical pairs are: b and e, and also c and d? Let's see the positions.

If the diagonal is from bottom-left to top-right, then:

- Ray for b: from O along diagonal to top-right.

- Ray for e: from O along diagonal to bottom-left — so same line, opposite directions.

Then the angle between positive y-axis and diagonal (b) and the angle between negative y-axis and diagonal (e) — these are not vertical; vertical angles are formed by two intersecting lines, and are opposite.

The two lines are: the vertical line (y-axis), the horizontal line (x-axis), and the diagonal line. But there are three lines, so multiple intersections, but all pass through O.

The vertical angles are determined by pairs of lines.

For example, the vertical line and the diagonal line intersect, forming two pairs of vertical angles: one pair is the angles between y-axis and diagonal in the first and third quadrants, which are b and e — yes, so b and e are vertical angles.

Similarly, the horizontal line and the diagonal line intersect, forming vertical angles: the angle between x-axis and diagonal in the first quadrant (c) and the angle between negative x-axis and diagonal in the third quadrant — let's call it f, but f is not labeled; it's part of the third quadrant.

In the third quadrant, the angle between negative x-axis and diagonal is not labeled; angle e is between negative y-axis and diagonal, so the angle between negative x-axis and diagonal is adjacent to e, and together with e make the 90° of the third quadrant.

So perhaps f = 90° - e, but not labeled.

Then for the horizontal and vertical lines, they form vertical angles of 90° each, but a is one of them.

So for labeled angles, the only clear vertical pair is b and e.

Then c and d: c is in first quadrant between diagonal and x-axis, d is in fourth quadrant between x-axis and negative y-axis — not vertical.

The vertical angle to c would be in the third quadrant between diagonal and negative x-axis, which is not labeled.

So perhaps only b and e are vertical among the labeled angles.

But the problem asks for "Vertical: _____ and _____", implying a pair, so likely b and e.

Part b) Complementary: ∠c and ? As above, b + c = 90°, since they make the first quadrant 90°. So ∠c and ∠b are complementary.

Part c) Supplementary: ∠c and ? Sum to 180°. On the straight line of the x-axis, the angles on one side: from left to right, the angle on the upper side is c (in first quadrant), and on the lower side, in the fourth quadrant, angle d is between x-axis and negative y-axis, but that's 90° if no diagonal, but with diagonal, d is part of it.

Actually, the straight line is the x-axis. The angles adjacent on the x-axis: for example, the angle between negative x-axis and the diagonal in the third quadrant, and the angle between diagonal and positive x-axis in the first quadrant — but those are not both labeled.

Angle c is between diagonal and positive x-axis. The adjacent angle on the straight line would be the angle between positive x-axis and the next ray, which is the negative y-axis for d, but d is not on the same side.

Perhaps angle c and angle e are supplementary? Let's calculate later.

Part d) All adjacent: angles that share a common side and vertex.

For example, a and b share the y-axis ray, a and e share the y-axis ray? a is between neg x and pos y, b is between pos y and diagonal, so they share the pos y-ray, so adjacent.

Similarly, b and c share the diagonal ray, c and d share the pos x-ray, d and e share the neg y-ray, e and a share the neg x-ray? e is between neg y and diagonal, a is between neg x and pos y, so they share the origin but not a common ray unless the diagonal is involved.

Adjacent angles must share a common ray.

So:

- a and b: share the positive y-axis ray → adjacent.

- b and c: share the diagonal ray → adjacent.

- c and d: share the positive x-axis ray → adjacent.

- d and e: share the negative y-axis ray → adjacent.

- e and a: do they share a ray? e is between neg y and diagonal, a is between neg x and pos y — no common ray, so not adjacent.

- Also, a and c: no common ray.

So adjacent pairs are: a-b, b-c, c-d, d-e.

But the question says "All adjacent: ________________" — probably list all pairs or all angles that are adjacent to each other, but it's ambiguous. Perhaps list the angles that have adjacent partners, or list the pairs.

Typically, it might mean list all pairs of adjacent angles.

Part e) If angle c is 30°, find all angles.

Given c = 30°.

Since b + c = 90° (first quadrant), b = 60°.

a = 90° (given).

Now, angle d: in the fourth quadrant, between pos x-axis and neg y-axis. This is 90°, and it's not cut by any line, so d = 90°? But the diagonal is in the first and third quadrants, so in fourth quadrant, no diagonal, so d should be 90°.

Similarly, in third quadrant, between neg x-axis and neg y-axis, 90°, split by diagonal into e and f, where f is between neg x-axis and diagonal.

Since the diagonal is a straight line, the angle between pos x-axis and diagonal is c = 30°, so the angle between neg x-axis and diagonal is 180° - 30° = 150°? No.

The diagonal line makes an angle with the x-axis. If in the first quadrant, it makes 30° with the x-axis (since c = 30°), then in the third quadrant, it makes 30° with the negative x-axis, because vertical angles or corresponding.

Actually, the angle between the diagonal and the positive x-axis is c = 30°, so the angle between the diagonal and the negative x-axis is 180° - 30° = 150°, but that's the reflex angle; the acute angle is 30° on the other side.

When two lines intersect, the vertical angles are equal, so the angle between the diagonal and the negative x-axis in the third quadrant is equal to the angle between the diagonal and the positive x-axis in the first quadrant, which is c = 30°, because they are vertical angles? Let's see.

The line of the diagonal and the x-axis intersect at O. The angle in the first quadrant between diag and pos x-axis is c = 30°. The vertical angle to that is the angle in the third quadrant between diag and neg x-axis, which should also be 30°.

Yes! Because vertical angles are equal. So the angle between the diagonal and the negative x-axis in the third quadrant is 30°.

Then, in the third quadrant, the total angle is 90° between neg x-axis and neg y-axis. This is split by the diagonal into two parts: one is the angle between neg x-axis and diagonal = 30°, and the other is between diagonal and neg y-axis = e.

So e = 90° - 30° = 60°.

Then, angle d: in the fourth quadrant, between pos x-axis and neg y-axis. This is 90°, and since no line cuts it, d = 90°.

But is that correct? The diagonal is only in first and third quadrants, so yes, fourth quadrant is untouched, so d = 90°.

Similarly, a = 90°.

Now, verify: b = 60°, c = 30°, a = 90°, d = 90°, e = 60°.

Check vertical angles: b and e are both 60°, and they are vertical — good.

c and the angle in third quadrant between diag and neg x-axis is 30°, which is not labeled, but if it were, it would be vertical to c.

d = 90°, and its vertical angle would be a = 90°, since a is in second quadrant, d in fourth, and they are vertical if the lines are perpendicular, which they are.

Yes, a and d are vertical angles, both 90°.

So vertical pairs: a and d, b and e, and c and the unlabeled 30° in third quadrant.

But for labeled angles, vertical pairs are a-d, b-e.

Complementary: b and c: 60+30=90, good. Also, e and the unlabeled 30°: 60+30=90, but unlabeled.

Supplementary: for example, c and the angle adjacent on the straight line. On the x-axis, the angle c (30°) and the angle on the other side of the x-axis in the fourth quadrant — but d is 90°, not adjacent.

The straight line is the x-axis. The angles on the upper half-plane: from neg x to pos x, passing through upper, the angles are: in second quadrant a=90°, but a is between neg x and pos y, not on the x-axis directly.

Perhaps angle c and angle e are not supplementary.

Angle c = 30°, angle e = 60°, sum 90, not 180.

Angle c and angle d: 30+90=120, not 180.

Angle c and angle a: 30+90=120.

Perhaps angle c and the angle that is 150° somewhere.

On the straight line of the diagonal, but it's complicated.

For supplementary, likely c and the angle that is adjacent on the straight line of the x-axis. For example, the angle between pos x-axis and the next ray in the clockwise direction, which is d, but d is 90°, and c is 30°, but they are not on a straight line together; they are on different sides.

Actually, the angle c and the angle in the fourth quadrant between pos x-axis and the diagonal — but there is no such angle; the diagonal is not in fourth quadrant.

Perhaps for angle c, the supplementary angle is the one that forms a linear pair with it. Since c is between diag and pos x-axis, the adjacent angle on the straight line of the pos x-axis would be the angle between pos x-axis and the next ray, which is the neg y-axis for d, but d is 90°, and c is 30°, and they are adjacent if they share the pos x-axis ray, which they do, and together they make the angle from diag to neg y-axis, which is not 180°.

The sum c + d = 30 + 90 = 120°, not 180.

The only way to get 180 is if we take c and the angle on the other side of the x-axis. For example, the angle between pos x-axis and the diagonal in the fourth quadrant, but there is no such angle; the diagonal is in first and third.

So perhaps for angle c, the supplementary angle is not among the labeled ones, or perhaps it's angle e plus something.

Let's calculate the angle between the diagonal and the neg y-axis in the fourth quadrant — but it's not there.

I think for this problem, when they say "supplementary: ∠c and _____", they mean which labeled angle is supplementary to c, but none may be, or perhaps d is not 90°.

Earlier I assumed d = 90°, but is that correct? In the fourth quadrant, between pos x-axis and neg y-axis, if no line cuts it, yes, 90°. But perhaps the diagonal affects it? No, the diagonal is in first and third, so fourth is clear.

Perhaps angle d is the angle between pos x-axis and the diagonal, but in the fourth quadrant, but the diagonal is not in fourth quadrant; it's in first and third.

Unless the diagonal is full line, so in fourth quadrant, the ray is the extension, but the angle is measured from pos x-axis to the diagonal ray in the fourth quadrant, which would be the same as in the first quadrant but on the other side.

When two lines intersect, they form four angles. Here, we have three lines, so more.

For the x-axis and the diagonal line, they intersect at O, forming four angles:

- In first quadrant: c = 30° (between pos x and diag)

- In second quadrant: between neg x and diag — this is 180° - 30° = 150°? No, the angle between neg x and diag in the second quadrant is the supplement if on a straight line, but let's think.

The line of the diagonal makes an angle θ with the x-axis. If in first quadrant it's 30°, then the angle in the second quadrant between neg x-axis and the diagonal is 180° - 30° = 150°, but that's the larger angle; the smaller angle is 30° on the other side, but in terms of the region, the angle inside the second quadrant between neg x-axis and the diagonal is 180° - 30° = 150° if we measure the smaller one, but usually we take the smaller angle.

In the context of the diagram, the angle in the second quadrant between neg x-axis and the diagonal is not labeled; it's part of a or something.

Perhaps for angle c = 30°, the supplementary angle is the one that is 150°, which might be angle a plus something, but a is 90°.

I recall that in such problems, often angle c and angle e are not supplementary, but angle c and the angle that is vertical to it or something.

Let's use the fact that on a straight line, adjacent angles sum to 180°.

For example, on the x-axis, the angles on the upper side: from neg x to pos x, the angle is 180°, composed of the angle in second quadrant and first quadrant.

In second quadrant, the angle between neg x-axis and pos y-axis is a = 90°, but that's not on the x-axis; the angle on the x-axis would be the angle between the rays.

Perhaps the angle between the neg x-axis and the pos x-axis along the upper path is 180°, and it is split by the pos y-axis and the diagonal.

From neg x-axis to pos y-axis: angle a = 90°.

From pos y-axis to pos x-axis: this is split into b and c, with b+ c = 90°.

So total from neg x to pos x via upper: a + b + c = 90 + b + c = 90 + 90 = 180°, good.

Similarly, on the lower side, from neg x to pos x via lower: angle between neg x and neg y is 90° (part of third quadrant), then from neg y to pos x is d = 90°, so 90 + 90 = 180°.

Now, for angle c = 30°, which is between diag and pos x-axis.

The adjacent angle on the straight line of the pos x-axis would be the angle between pos x-axis and the next ray in the clockwise direction, which is the neg y-axis for d, but d is 90°, and they share the pos x-axis ray, so they are adjacent, and their sum is c + d = 30 + 90 = 120°, not 180, so not supplementary.

The supplementary angle to c would be the angle that is on the other side of the pos x-axis, but in the lower half, between pos x-axis and the diagonal in the fourth quadrant. But since the diagonal is in the first and third, in the fourth quadrant, the ray of the diagonal is the extension, so the angle between pos x-axis and the diagonal ray in the fourth quadrant is the same as in the first quadrant but measured clockwise, so it would be 30° below the x-axis, so the angle is 30°, but then c and this angle are vertical or something.

Actually, the angle between the pos x-axis and the diagonal in the fourth quadrant is the vertical angle to the angle in the second quadrant between neg x-axis and the diagonal, which is not labeled.

Perhaps for this problem, when they say "supplementary: ∠c and _____", they mean ∠c and the angle that is 150°, which might be angle a + b or something, but a+ b = 90 + 60 = 150°, and c = 30°, sum 180°, and they are not adjacent, but supplementary means sum to 180, not necessarily adjacent.

Oh! I forgot that supplementary angles don't have to be adjacent; they just need to sum to 180°.

So for ∠c = 30°, any angle that is 150° is supplementary to it.

In the diagram, what angle is 150°? From earlier, a + b = 90° + 60° = 150°, but a and b are separate angles.

The angle from neg x-axis to the diagonal in the second quadrant: from neg x-axis to pos y-axis is a = 90°, then from pos y-axis to diagonal is b = 60°, so from neg x-axis to diagonal is a + b = 150°.

And this angle is not labeled as a single angle; it's composed of a and b.

But perhaps in the diagram, there is an angle that is 150°, or perhaps they consider the combination.

Maybe angle e is 60°, not 150.

Another possibility: the angle between the neg x-axis and the diagonal in the third quadrant is 30°, as we said, so not 150.

Perhaps for angle c, the supplementary angle is angle d plus something.

Let's calculate the angle between the diagonal and the neg y-axis in the fourth quadrant — but it's not there.

I think for this problem, with c = 30°, then:

- b = 60° (since b+c=90)

- a = 90°

- e = 60° (since in third quadrant, angle between neg x and diag is 30°, so e = 90° - 30° = 60°)

- d = 90° (fourth quadrant)

Now, is there an angle of 150°? The angle from pos x-axis to neg x-axis via the lower path is 180°, composed of d and the angle in third quadrant between neg y and neg x, which is 90°, so d + 90° = 180°, but 90° is not labeled.

The angle from pos x-axis to the diagonal in the fourth quadrant: since the diagonal makes 30° with pos x-axis in first quadrant, in fourth quadrant, it makes 30° below, so the angle between pos x-axis and the diagonal ray in fourth quadrant is 30°, but then the angle between that and the neg y-axis would be 60°, etc.

Perhaps the supplementary angle to c is the angle that is 150°, which is the reflex angle or something, but unlikely.

Another idea: angle c and angle e are not supplementary, but angle c and the angle that is vertical to the complement or something.

Let's list the measures:

- a = 90°

- b = 60°

- c = 30°

- d = 90°

- e = 60°

Sum of all around point: 90+60+30+90+60 = 330°, but should be 360°, so missing 30°. Ah, yes, the angle in the third quadrant between neg x-axis and the diagonal is 30°, which is not labeled. Let's call it f = 30°.

Then total: a+b+c+d+e+f = 90+60+30+90+60+30 = 360°, good.

So f = 30°, vertical to c.

Now, for supplementary to c=30°, we need 150°. What angles sum to 150°? a+b = 90+60=150°, or d+e = 90+60=150°, or a+f = 90+30=120, etc.

So a+b = 150°, and c = 30°, sum 180°, so c and (a+b) are supplementary, but a and b are separate.

Perhaps the problem considers the angle from neg x-axis to the diagonal in the second quadrant as a single angle, but it's not labeled.

Maybe for "supplementary: ∠c and _____", they mean ∠c and d, but 30+90=120≠180.

Or ∠c and ∠e: 30+60=90.

Perhaps ∠c and the angle that is 150°, which is not labeled, so maybe it's ∠a and ∠b together, but the blank is for one angle.

Another thought: in some interpretations, angle d is not 90°; perhaps the diagonal affects it, but no.

Perhaps for angle c, the supplementary angle is the one adjacent on the straight line of the diagonal.

On the diagonal line, the angles on one side: from the ray in first quadrant to the ray in third quadrant, the angle is 180°.

On the upper side, from diag in first to diag in third, passing through pos y-axis, the angles are b and a and the angle in second quadrant between pos y and neg x, which is part of a, but a is already 90° for the whole second quadrant.

From the diagonal ray in first quadrant to the diagonal ray in third quadrant, the shorter arc is 180° since it's a straight line.

The angles around: from diag in first to pos y-axis: b = 60°

From pos y-axis to neg x-axis: a = 90°

From neg x-axis to diag in third: f = 30°

Sum: 60+90+30 = 180°, good.

So on this path, the angles are b, a, f.

For angle c = 30°, which is on the other side, between diag and pos x-axis.

The supplementary angle would be the angle on the other side of the pos x-axis, but in the lower half.

Perhaps it's easier to accept that for ∠c = 30°, the supplementary angle is 150°, and in the diagram, the angle that is 150° is the combination of a and b, but since the blank is for one angle, and d is 90°, e is 60°, etc, perhaps they mean ∠c and ∠ (a+b) , but not single.

Maybe in the diagram, angle d is the angle from pos x-axis to neg y-axis, which is 90°, but if we consider the angle from pos x-axis to the diagonal in the fourth quadrant, it is 30° (since vertical to the 30° in second quadrant or something), but then the angle between that and neg y-axis is 60°, so d would be split, but in the diagram, d is labeled as one angle, so probably not.

I think for the sake of time, and since this is taking too long, I'll assume that for Problem 11e, with c=30°, then:

- a = 90°

- b = 60° (since b+c=90)

- c = 30°

- d = 90° (assumed)

- e = 60° (since in third quadrant, angle between neg y and diag is e, and between neg x and diag is 30°, so e = 90° - 30° = 60°)