Practice worksheet for finding unknown angles in triangles using the angle sum theorem.

Let’s solve each problem one by one using the Angle Sum Theorem — which says:
> In any triangle, the sum of all three interior angles is always 180°.

We’ll also use other basic angle rules when needed (like straight lines = 180°, vertical angles are equal, etc.).

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Problem 1

Triangle ABC with point E on BC. Angles given:
- ∠BAE = 35°
- ∠CAE = 56° → So ∠BAC = 35° + 56° = 91°
- ∠ABC = 47°
- ∠ACB = 42°

Wait — let’s check if that adds up: 91° + 47° + 42° = 180° ✔ Good.

Now we need to find:
- ∠AEB and ∠AEC

Look at triangle ABE:
Angles: ∠ABE = 47°, ∠BAE = 35°, so ∠AEB = 180° - 47° - 35° = 98°

Then ∠AEC is next to it on a straight line → ∠AEB + ∠AEC = 180°
So ∠AEC = 180° - 98° = 82°

✔ Check with triangle AEC:
∠CAE = 56°, ∠ACE = 42°, so ∠AEC should be 180 - 56 - 42 = 82° ✔ Matches!

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Problem 2

Right triangle PQS? Wait — look carefully.

Point Q has right angle symbol → ∠PQR = 90°? Actually, in diagram, square at Q means ∠PQS = 90°? Let me re-read.

Actually, looking again: Triangle PQR? No — points are P, Q, R, S.

From diagram:

In triangle PQR? Wait — better to label what we see.

There’s a big triangle PQS? Or two triangles?

Actually, from diagram:

- Point Q has right angle → so ∠PQS = 90°? But then there's point R on QS.

Wait — perhaps triangle PQR and PRS?

Given:
- ∠QPR = 24°
- ∠RPS = 35° → so total ∠QPS = 24° + 35° = 59°
- Right angle at Q → so in triangle PQS, ∠PQS = 90°, ∠QPS = 59° → then ∠PSQ = 180 - 90 - 59 = 31°

But wait — they ask for ∠PRQ and ∠PSR.

Let’s focus on triangle PQR first.

In triangle PQR:
- ∠PQR = 90° (right angle)
- ∠QPR = 24°
→ So ∠PRQ = 180 - 90 - 24 = 66°

Now, ∠PRS is part of triangle PRS.

In triangle PRS:
- ∠RPS = 35°
- We just found ∠PRQ = 66°, but that’s adjacent to ∠PRS? Wait — actually, since Q-R-S is a straight line, then ∠PRQ and ∠PRS are supplementary? No — because R is between Q and S, so ∠PRQ and ∠PRS form a straight line? Not exactly — unless P-R is drawn.

Actually, looking at diagram: Points Q, R, S are colinear? Yes — since it’s a base line.

So ∠PRQ and ∠PRS are adjacent angles forming a straight line → so ∠PRS = 180° - ∠PRQ = 180 - 66 = 114°

Wait — but they ask for ∠PSR, not ∠PRS.

In triangle PRS:
We have:
- ∠RPS = 35°
- ∠PRS = ? — wait, no — if Q-R-S is straight, and we have triangle PRS, then angle at R inside triangle PRS is ∠PRS.

But earlier I said ∠PRQ = 66°, and since Q-R-S is straight, then ∠PRS = 180° - 66° = 114°? That can’t be right because then triangle PRS would have angles 35° + 114° + ∠PSR = 180 → ∠PSR = 31°, which matches our earlier calculation.

Yes! So:

∠PRQ = 66° (from triangle PQR)

∠PSR = 31° (from triangle PRS: 180 - 35 - 114 = 31°) OR from big triangle PQS: 180 - 90 - 59 = 31° ✔

So answers:
∠PRQ = 66°
∠PSR = 31°

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Problem 3

Two triangles sharing vertex S: triangle PRS and triangle QSR.

Given:
In triangle PRS:
- ∠RPS = 66°
- ∠PRS = 72°
→ So ∠RSP = 180 - 66 - 72 = 42°

In triangle QSR:
- ∠SQR = 30°
- ∠SRQ = 2°? Wait — looks like ∠QRS = 2°? That seems too small. Maybe typo? Or maybe it’s 20°? Wait — let me check original image description.

User wrote: “y° 30°” and “2°” near R in second triangle.

Assuming it’s correct: ∠QRS = 2°, ∠SQR = 30° → then ∠QSR = 180 - 30 - 2 = 148°

But also, note that ∠RSP and ∠QSR are vertical angles? No — they share side SR, but are on opposite sides.

Actually, points P-S-Q and R-S-T? Wait — diagram shows two triangles crossing at S.

Actually, ∠RSP and ∠QSR are vertically opposite? Let’s think.

If lines PR and QR cross at S? No — probably lines PS and QS meet at S, and RS is common.

Actually, likely ∠RSP and ∠QSR are adjacent or vertical?

Looking at standard such diagrams — often ∠RSP and ∠QSR are vertical angles? But here, if you draw it, point S is where two lines cross: one is P-S-Q, other is R-S-something.

Wait — in diagram, it’s labeled as triangle PRS and triangle QSR, sharing side SR? Or crossing?

Actually, from labels: Points P, R, S form one triangle; Q, S, R form another — so they share side SR.

Therefore, angles at S: ∠RSP and ∠RSQ are different.

But in triangle PRS, we calculated ∠RSP = 42°

In triangle QSR, angles: ∠SQR=30°, ∠SRQ=2° → ∠QSR=148°

But now, notice that ∠RSP and ∠QSR are on a straight line? If P-S-Q is straight, then yes!

Check: If P-S-Q is a straight line, then ∠RSP + ∠RSQ = 180°? But we have 42° + 148° = 190° — too much. Contradiction.

Ah — mistake! Probably ∠SRQ is not 2° — maybe it’s 20°? Or perhaps misread.

Wait — user wrote: “y° 30°” and “2°” — perhaps “2°” is ∠QRT or something else? Let me re-express.

Alternative approach: Perhaps the 2° is ∠QSR? No.

Another idea: Maybe the “2°” is a typo and should be 20°? Because 30+20=50, 180-50=130, and 42+130=172≠180.

Wait — perhaps the two triangles share the angle at S, and ∠RSP and ∠QSR are vertical angles? Then they should be equal.

But in triangle PRS: 66+72=138, so ∠RSP=42°

If vertical, then ∠QSR=42°, then in triangle QSR: 30° + x + 42° = 180 → x=108°, but they gave y° and 2° — doesn't match.

Perhaps the "2°" is ∠QSR? Then in triangle QSR: 30° + y° + 2° = 180 → y=148°, and ∠RSP=42°, and if P-S-Q is straight, then 42+148=190≠180 — still bad.

I think there might be a misinterpretation of the diagram.

Let me try this: In many such problems, when two triangles share a vertex and lines cross, the vertical angles are equal.

Assume that lines PR and QS intersect at S. Then ∠RSP and ∠QSR are vertical angles? No — vertical angles would be ∠RSP and ∠QST or something.

Standard notation: If two lines intersect, vertical angles are opposite.

Suppose line PSQ and line RST intersect at S. Then vertical angles are ∠PSR and ∠QST, and ∠PSQ and ∠RST.

But in the diagram, it's labeled as triangle PRS and triangle QSR — so likely points are arranged as P--S--Q and R--S--T, with S intersection.

Then in triangle PRS: angles at P=66°, R=72°, so at S: 42° — this is ∠PSR.

In triangle QSR: angles at Q=30°, R=2°? — but R is shared? Confusing.

Perhaps the "2°" is ∠QSR? Let's assume that.

Maybe the diagram has ∠QRS = 2°, but that seems unrealistic for a school problem. Likely a typo, and it's 20°.

Let me calculate with 20°: then in triangle QSR: 30° + 20° + ∠QSR = 180 → ∠QSR = 130°

Then if P-S-Q is straight, ∠PSR + ∠QSR = 42° + 130° = 172° ≠ 180 — still off.

Unless the 2° is not an angle in the triangle.

Another possibility: the "2°" is ∠QRT or external.

Perhaps it's ∠QSR = 2°, then y° = 180 - 30 - 2 = 148°, and ∠RSP = 42°, and they are not on a straight line.

But then why give both? The question asks for ∠RSP and ∠QSR, so perhaps they are separate.

And in that case, no contradiction if they are not adjacent.

So let's go with:

In triangle PRS: ∠RSP = 180 - 66 - 72 = 42°

In triangle QSR: ∠QSR = 180 - 30 - 2 = 148° (assuming the 2° is ∠QRS)

Even though 42 + 148 = 190, if they are not on a straight line, it's ok. Perhaps the diagram has them not collinear.

For the sake of proceeding, I'll take it as is.

So:
∠RSP = 42°
∠QSR = 148°

But let's double-check the user input: "y° 30°" and "2°" — perhaps "2°" is ∠QSR, and y° is ∠QRS.

The question asks for ∠RSP and ∠QSR, so if ∠QSR is given as 2°, then answer is 2°, but that seems odd.

Perhaps "2°" is a label for ∠QSR, so ∠QSR = 2°, then in triangle QSR, y° = 180 - 30 - 2 = 148°, and ∠RSP = 42° from other triangle.

And they are not required to add to 180 if not on straight line.

I think that's the only way.

So:
∠RSP = 42°
∠QSR = 2° (if that's what is labeled)

But the user wrote "y° 30°" and "2°", and asks for ∠QSR, so likely ∠QSR is the 2°.

To confirm, in some diagrams, the number is placed at the angle.

So I'll go with:
∠RSP = 42°
∠QSR = 2°

But let's see problem 7 has similar setup, and there angles are reasonable.

Perhaps for problem 3, the "2°" is a mistake, and it's 20° or 28°.

Another thought: perhaps the "2°" is ∠QRT, but the triangle is QSR, so unlikely.

I recall that in some worksheets, they have small angles, but 2° is very small.

Let's calculate the sum: if in triangle QSR, angles are 30°, y°, and 2°, sum 32+y=180, y=148, and in other triangle 66+72+42=180, so mathematically correct, even if unusual.

So I'll proceed.

Answers:
∠RSP = 42°
∠QSR = 2°

But the question asks for ∠RSP and ∠QSR, so that's it.

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Problem 4

Quadrilateral or triangle? Points P, Q, R, S.

Diagram: P-Q-R is a line? With right angle at P, and S below.

Specifically:
- ∠SPQ = 90° (right angle at P)
- ∠PSQ = 46°
- Need to find ∠RQS = x°

First, in triangle PQS:
- ∠P = 90°
- ∠S = 46°
- So ∠PQS = 180 - 90 - 46 = 44°

Now, P-Q-R is a straight line, so ∠PQS + ∠RQS = 180° (since they form a straight line at Q)

Thus, ∠RQS = 180 - 44 = 136°

So x = 136°

Answer: ∠RQS = 136°

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Problem 5

Points O, P, Q, R, S, T.

Given:
- ∠OPQ = 65°? Wait — at P, angle is 65°, and right angle at O.

So in triangle OPQ:
- ∠O = 90°
- ∠P = 65°
- So ∠OQP = 180 - 90 - 65 = 25°

Now, points O-Q-T are colinear? Probably, since it's a base.

Also, triangle SQT or something.

Given in triangle SRT or STR: ∠S = 50°, ∠T = 80°, so ∠SRT = 180 - 50 - 80 = 50°

But they ask for ∠QRT = x°

Note that ∠OQP and ∠RQT are vertical angles? Or same angle?

Actually, at point Q, we have angle from triangle OPQ: ∠OQP = 25°

Since O-Q-T is straight, and R is above, then ∠RQT is the angle between RQ and QT.

But ∠OQP is between OQ and QP.

If P-Q-R is a line? Not necessarily.

From diagram, likely that P-Q-R is not straight; instead, we have line O-Q-T, and point R connected to Q and T.

In triangle QRT, we need ∠QRT.

We know in triangle SRT: ∠S=50°, ∠T=80°, so ∠SRT=50°

But ∠SRT is the same as ∠QRT if S-R-Q is straight? Probably not.

Perhaps R is on ST or something.

Another approach: at point R, angles around.

Note that ∠OQP = 25°, and if we consider vertical angles or corresponding.

Perhaps ∠PQR is related.

Let's list knowns:

- In triangle OPQ: ∠O=90°, ∠P=65°, ∠OQP=25°

- In triangle SRT: ∠S=50°, ∠T=80°, ∠SRT=50°

Now, ∠OQP and ∠RQT are vertically opposite if lines cross, but likely not.

Perhaps points are arranged as O-Q-T straight, P above left, S above right, R intersection of PS and QT or something.

Common configuration: line OT, with Q on it, P connected to O and Q, S connected to T and R, and R is intersection of PS and QT? But QT is part of OT.

Assume that line PS intersects line OT at R.

Then at R, we have vertical angles.

In triangle OPQ, we have ∠OQP = 25°, which is at Q.

At R, in triangle SRT, ∠SRT = 50°.

But ∠SRT is the angle at R in triangle SRT, which is between S-R and T-R.

If R is on OT, then T-R is along OT, so ∠SRT is the angle between SR and RT.

Similarly, for triangle QRT, if Q and T are on the line, and R above, then ∠QRT is the angle at R between Q-R and T-R.

But if R is on OT, and Q is also on OT, then Q-R-T are colinear, so triangle QRT degenerate — impossible.

So likely, R is not on OT; rather, R is the intersection point of PS and another line.

From standard problems, often R is the intersection of PS and QT, but QT is part of OT.

Perhaps line PS crosses line OT at R.

Then, at R, we have angles.

In triangle OPQ, we have angle at Q: 25°.

This angle ∠OQP is between OQ and QP.

When line PS is drawn, it may create vertical angles.

Perhaps ∠OQP and ∠RQT are the same if P-Q-R is straight, but not specified.

Another idea: perhaps ∠PQR is the angle we need, but they ask for ∠QRT.

Let's look at triangle QRT.

We need more information.

Note that in triangle SRT, ∠SRT = 50°, and if we can find other angles.

Perhaps ∠OQP and ∠SRT are related through parallel lines or something, but no indication.

Let's calculate the angle at R for the whole figure.

Perhaps the key is that ∠OQP and the angle in triangle QRT are supplementary or something.

Recall that in such diagrams, often the angle at Q in triangle OPQ is equal to the angle at R in triangle QRT if there are parallel lines, but here no.

Let's try to find ∠PQS or something.

Perhaps point R is on PS, and we have triangle QRT with points Q, R, T.

Given that in triangle SRT, angles are 50° at S, 80° at T, so 50° at R.

Now, if S-R-P is a straight line, then ∠SRT and ∠PRQ are adjacent or vertical.

Assume that S-R-P is straight, so line SP.

Then at R, on line SP, we have angle from triangle SRT: ∠SRT = 50°, which is between S-R and T-R.

Then the adjacent angle on the other side, ∠PRT, would be 180° - 50° = 130°, since S-R-P straight.

Then in triangle QRT, we have points Q, R, T.

We know ∠ at T is 80° (same as in triangle SRT, since same angle).

∠ at R is ∠QRT, which is part of ∠PRT.

If Q is on the other side, then ∠QRT might be the same as ∠PRT if Q is on PR, but not.

Perhaps in triangle QRT, angle at R is the same as ∠PRT if Q is on the extension, but complicated.

Another approach: use the fact that the sum of angles around point R is 360°, but we don't have enough.

Let's consider triangle PQR or something.

Perhaps the right angle at O and the 65° suggest that we can find other angles.

Let's calculate the slope or use trigonometry, but that's overkill.

Simple way: in triangle OPQ, ∠OQP = 25°.

This angle is at Q, between OQ and QP.

Now, if we consider line QT, which is extension of OQ, so O-Q-T straight.

Then the angle between QP and QT is 180° - 25° = 155°, because OQ and QT are straight line.

So ∠PQT = 155°.

Now, if we have point R on PS, and we draw QR, then in triangle PQR or QRT.

But we need more.

Perhaps R is the intersection, and we have triangle QRT with known angles at T and at R from other triangle.

Let's assume that in triangle QRT, we know ∠ at T is 80°, and if we can find ∠ at Q.

At Q, the angle in triangle QRT is ∠RQT.

From earlier, ∠PQT = 155°, which is the angle between QP and QT.

If R is on QP, then ∠RQT = ∠PQT = 155°, but then triangle QRT would have angles 155° at Q, 80° at T, sum already 235>180, impossible.

So R is not on QP.

Perhaps R is on the other side.

Another idea: perhaps the 65° is at P for triangle OPS or something.

Let's read the diagram description: "P 65°" and "O right angle", so in triangle OPQ, yes.

"S 50°", "T 80°", so in triangle SRT.

And R is common.

Perhaps lines are: O-P, P-S, S-T, T-O, with Q on OT, R on PS, and QR drawn.

Then at R, in triangle SRT, ∠SRT = 50°.

In triangle OPQ, ∠OQP = 25°.

Now, if we consider quadrilateral or the whole, but for triangle QRT, we need its angles.

Note that ∠OQP and ∠RQT are vertically opposite if lines cross at Q, but at Q, we have lines OQ, QP, and QR.

Unless QR is drawn, but in the diagram, likely QR is a line.

Perhaps ∠OQP and the angle in triangle QRT at Q are the same if R is on QP, but we saw it's not.

Let's calculate the angle between QP and QS or something.

Perhaps use the fact that the sum of angles in polygon.

I recall that in such problems, the angle x can be found by noting that the two triangles share the angle at R or something.

Let's try this: in triangle SRT, ∠SRT = 50°.

This angle is at R, between S-R and T-R.

In triangle QRT, the angle at R is between Q-R and T-R.

If S-R-Q is a straight line, then ∠SRT and ∠QRT are adjacent and sum to 180°.

So if S-R-Q straight, then ∠QRT = 180° - 50° = 130°.

Then in triangle QRT, angles: at R 130°, at T 80°, sum 210>180, impossible.

So not straight.

If S-R-Q is not straight, then no direct relation.

Perhaps the 50° at S is for triangle PST or something.

Another thought: perhaps "S 50°" means ∠PST = 50°, but in the diagram, it's at S for triangle SRT.

Let's look for online or standard solution, but since I can't, let's assume that at point R, the angle for triangle QRT is the same as for SRT if Q and S are symmetric, but not.

Perhaps the right angle and 65° allow us to find that ∠OQP = 25°, and this is equal to ∠QRT because of alternate interior angles if lines are parallel, but no indication of parallel lines.

Let's calculate the difference.

Perhaps in triangle QRT, we can find angle at Q.

At Q, the angle between OQ and QP is 25°, and if QR is another line, but we don't know.

Perhaps the line QR is such that it makes equal angles, but not specified.

Let's try to use the fact that the sum of angles around point R is 360°.

But we have only one angle.

Perhaps from the diagram, triangle OPQ and triangle SRT are on opposite sides, and R is connected.

I think I need to make an assumption.

In many textbooks, for such a diagram, the angle x is found by: in triangle OPQ, ∠OQP = 25°, and this is equal to the angle at R in triangle QRT because they are corresponding or something, but let's see the values.

Suppose that ∠QRT = 25°, then in triangle QRT, angles: at R 25°, at T 80°, so at Q 75°.

Is that possible? Then no conflict.

But why 25°?

Perhaps because of vertical angles.

Another idea: when line PS is drawn, and it intersects OT at R, then at R, the vertical angle to ∠OQP might be equal.

In triangle OPQ, ∠OQP = 25°, which is at Q.

If line PS crosses OT at R, then at R, the angle between PR and OR might be equal to ∠OQP if parallel, but not.

Perhaps using triangle similarity, but no.

Let's calculate the angle at P for the whole.

In triangle OPQ, ∠OPQ = 65°.

If we consider triangle OPS or something.

Perhaps the 65° is for angle at P in triangle OPS, but the diagram shows it in triangle OPQ.

I think I found a way: in triangle OPQ, ∠OQP = 25°.

This angle is the same as the angle between QP and the line OT.

Now, if we consider triangle QRT, and if we assume that QR is parallel to OP or something, but not.

Perhaps the key is that the angle at R for triangle QRT is equal to the angle at P for triangle OPQ, which is 65°, but then in triangle QRT, 65° + 80° = 145°, so x = 35°, but not justified.

Let's look at problem 7 for clue.

In problem 7, it's similar: two triangles sharing a vertex, and angles given, and we use vertical angles.

In problem 7: triangle ABC and CDE, with C common, and ∠D=43°, ∠E=67°, so ∠DCE = 180-43-67=70°, and since vertical angles, ∠ACB = 70°, then in triangle ABC, if ∠A=x°, ∠B=x°, so 2x + 70 = 180, x=55°.

So for problem 5, perhaps similar.

In problem 5, at point R, the angle in triangle SRT is 50°, and if it is vertical to the angle in triangle QRT, then ∠QRT = 50°, but then in triangle QRT, 50° + 80° = 130°, so angle at Q is 50°, but we have no information.

But in the diagram, likely that the angle at R for the two triangles are vertical angles.

So assume that ∠SRT and ∠QRP are vertical angles, so equal, 50°.

Then in triangle QRT, if ∠QRT = 50°, and ∠QTR = 80°, then ∠RQT = 180 - 50 - 80 = 50°.

So x = 50°.

And it makes sense.

Moreover, in triangle OPQ, we have 25°, which is not used, but perhaps for another purpose, or maybe to verify.

So I'll go with that.

Answer: ∠QRT = 50°

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Problem 6

Points U, P, Q, R, S, T.

Given:
- ∠PQU = x+74°
- ∠QRS = 38+x°
- ∠UST = 33° (at S)

Need to find ∠PQU and ∠QRS.

Also, likely that lines are crossing at S or Q.

From diagram, probably lines UT and PR intersect at S, and Q is on UT or something.

Specifically, ∠UST = 33°, which is at S in triangle UST or something.

Assume that points are: line U-Q-R, and line P-S-T, intersecting at S.

Then at S, vertical angles.

In triangle UST or QST.

Given ∠UST = 33°, which is angle at S between U-S and T-S.

Then the vertical angle would be ∠PSR or something.

Also, ∠PQU = x+74°, which is at Q, between P-Q and U-Q.

∠QRS = 38+x°, at R, between Q-R and S-R.

Probably, in triangle QRS or something.

Note that ∠PQU and ∠RQS are vertical angles if lines cross at Q, but at Q, we have lines U-Q-R and P-Q- something.

Assume that line PR and line UT intersect at S.

Then at S, vertical angles are equal.

In triangle UST, if we have points U,S,T, with ∠UST = 33°.

But we need more.

Perhaps triangle QRS has angles involving x.

Another idea: perhaps ∠PQU and ∠RQS are the same if P-Q-R is straight, but not.

Let's consider that at point Q, the angle ∠PQU = x+74°, and if U-Q-R is straight, then the adjacent angle ∠PQR = 180 - (x+74) = 106 - x°.

Then in triangle QRS, we have angle at Q: ∠RQS = 106 - x° (if P-Q-R straight), angle at R: ∠QRS = 38 + x°, and angle at S: which is part of the 33° or related.

At S, if lines intersect, the angle in triangle QRS at S might be vertical to ∠UST or something.

Assume that ∠UST = 33° is the angle at S in triangle UST, and if U-S-P is straight, then the vertical angle to ∠UST is ∠PSR = 33°.

Then in triangle QRS, angle at S is ∠QSR = 33° (vertical angle).

Then in triangle QRS, sum of angles: ∠ at Q + ∠ at R + ∠ at S = 180°

So (106 - x) + (38 + x) + 33 = 180

Simplify: 106 - x + 38 + x + 33 = 180

The x cancels: 106 + 38 + 33 = 177 = 180? 177 ≠ 180, close but not exact.

177 vs 180, difference of 3°, so perhaps my assumption is wrong.

Maybe ∠QSR is not 33°.

Perhaps ∠UST = 33° is the angle, and in triangle QRS, the angle at S is the same as ∠UST if S is common, but not.

Another possibility: the 33° is ∠QST or something.

Perhaps in triangle QST, but not specified.

Let's try this: suppose that at S, the angle between U-S and T-S is 33°, and since U-Q-R is a line, and P-S-T is a line, intersecting at S, then the vertical angle to ∠UST is ∠PSR = 33°.

Then in triangle QRS, if Q, R, S are vertices, and if Q and R are on the lines, then angle at S in triangle QRS is ∠QSR, which may be different.

If Q is on US, R on PS, then in triangle QRS, angle at S is between Q-S and R-S, which is the same as between U-S and P-S, so if U-S-P is straight, then ∠QSR = 180° - ∠UST - ∠PSR or something.

If U-S-P is straight, and T-S- something, then at S, the ray SU and SP are opposite, so angle between SU and ST is 33°, then angle between SP and ST is 180° - 33° = 147°, since U-S-P straight.

Then in triangle QRS, if Q is on SU, R on SP, then angle at S is the angle between SQ and SR, which is the same as between SU and SP, which is 180°, but that can't be for a triangle.

So not.

Perhaps Q is not on SU.

Let's assume that line UT and line PR intersect at S.

Then at S, vertical angles are equal.

Let ∠USP = a, etc.

But we have ∠UST = 33°, which is part of it.

Perhaps ∠UST = 33° is the angle in triangle UST, but for our purpose, in triangle QRS, the angle at S is vertical to another angle.

Notice that in the expression, when we add, the x canceled, giving 177, close to 180, so perhaps the 33° is not correct, or there's a mistake in the problem.

Maybe the 33° is ∠QST or ∠RST.

Another idea: perhaps ∠UST = 33° is the angle at S for triangle UST, but in the context, it might be the angle between U-S and T-S, and for triangle QRS, the angle at S is the vertical angle, which is equal to 33° if it's the same type.

In my earlier calculation, I had 106 - x + 38 + x + 33 = 177, but 177 < 180, so perhaps the angle at S is not 33°, but 36° or something.

Perhaps the 33° is for a different angle.

Let's read: "33°" at S, and "x+74" at Q, "38+x" at R.

Perhaps in triangle QRS, the angles are: at Q: the supplement of x+74, at R: 38+x, at S: 33°, and sum to 180.

So (180 - (x+74)) + (38 + x) + 33 = 180

Calculate: 180 - x - 74 + 38 + x + 33 = 180

Simplify: (180 - 74 + 38 + 33) + (-x + x) = 177 + 0 = 177 = 180? Still 177.

180 - 74 = 106, 106 + 38 = 144, 144 + 33 = 177, yes.

So 177 = 180, which is not true, so error.

Unless the 33° is not the angle in the triangle.

Perhaps the 33° is ∠QST, and in triangle QRS, the angle at S is different.

Another possibility: the 33° is the angle at S for the intersection, and it is vertical to the angle in triangle QRS, so same 33°.

But still 177.

Perhaps the angle at Q is not the supplement.

Let's think differently.

Suppose that at point Q, the angle ∠PQU = x+74°, and this is an exterior angle or something.

Perhaps for triangle QRS, the angle at Q is ∠RQS, which is adjacent to ∠PQU.

If P-Q-R is a straight line, then ∠PQU + ∠RQU = 180°, so ∠RQU = 180 - (x+74) = 106 - x°.

Then in triangle QRS, angles: at Q: 106 - x°, at R: 38 + x°, at S: let's call it y°.

Sum: 106 - x + 38 + x + y = 144 + y = 180, so y = 36°.

But we have 33° given, so perhaps y = 33°, but 144 + 33 = 177 ≠ 180.

Unless the 33° is not y.

Perhaps the 33° is the angle at S for the other triangle, and it is equal to y by vertical angles, so y = 33°, but then 144 + 33 = 177 < 180, so not.

Perhaps there is a typo, and it's 36° instead of 33°.

Or perhaps the 33° is for a different angle.

Another idea: perhaps "33°" is ∠UST, and in the diagram, it is the angle between U-S and T-S, and for triangle QRS, the angle at S is the vertical angle, which is equal to 33°, but then we have discrepancy.

Perhaps the line is not straight.

Let's assume that the sum is 180, so 106 - x + 38 + x + s = 180, so 144 + s = 180, s = 36°.

So if the given 33° is a mistake, and it's 36°, then it works for any x, but that can't be.

In the equation, x cancels, so s must be 36°, so perhaps the 33° is incorrect, or it's for something else.

Perhaps the 33° is ∠QST, and it is part of the angle at S.

Let's look at the answer choices or think.

Perhaps in triangle UST, but we don't have enough.

Another approach: perhaps the 33° is the angle at S in triangle UST, and if we can find other angles, but not given.

Perhaps for point S, the angle between the lines is 33°, and it is used in both triangles.

Let's try to set up the equation as is.

From earlier, in triangle QRS, if we assume angle at S is 33°, then (180 - (x+74)) + (38 + x) + 33 = 180

As before, 106 - x + 38 + x + 33 = 177 = 180, which is never true.

So impossible.

Unless the angle at Q is not 180 - (x+74).

Perhaps ∠PQU = x+74° is the angle in the triangle, not the adjacent.

In some diagrams, ∠PQU might be the angle at Q for triangle PQU, but then for triangle QRS, it's different.

Perhaps Q is the vertex, and ∠PQU is the angle between P-Q and U-Q, and for triangle QRS, the angle at Q is between R-Q and S-Q, which may be different.

So without knowing the configuration, it's hard.

Perhaps from the diagram, U-Q-R is a straight line, and P-S-T is a straight line, intersecting at S, and Q is on U-R, R on U-R, S on P-T.

Then at S, the angle between U-S and T-S is 33°.

Then the vertical angle is between P-S and R-S or something.

Then in triangle QRS, points Q on U-S, R on P-S, S intersection.

Then angle at S in triangle QRS is the angle between Q-S and R-S, which is the same as between U-S and P-S.

If U-S-P is not straight, then it's the angle at S between the two lines.

If the two lines intersect at S, the angle between them is say θ, then in triangle QRS, angle at S is θ.

But we have ∠UST = 33°, which is the angle between U-S and T-S.

If T-S is the same as P-S, then ∠UST = angle between U-S and P-S = 33°.

So in triangle QRS, angle at S is 33°.

Then as before.

But then sum is 177, so perhaps the 33° is 36°, or the numbers are different.

Perhaps "33°" is a typo, and it's 36°.

Or perhaps in the problem, it's 36°.

Maybe the 38+x is for a different angle.

Another possibility: perhaps ∠QRS = 38+x° is not the angle in the triangle, but the name suggests it is.

Let's calculate what it should be.

From 144 + s = 180, s=36°, so if the given is 33°, perhaps it's for a different purpose.

Perhaps the 33° is ∠QST, and it is equal to the angle at S in triangle QRS by vertical angles, so s=33°, but then 144+33=177, so the sum is 177, which means the triangle is not Euclidean, impossible.

So likely a typo in the problem or in my reasoning.

Perhaps the angle at Q is not 106 - x.

Let's assume that ∠PQU = x+74° is the angle at Q for triangle PQU, and for triangle QRS, the angle at Q is the same if R is on PU, but not.

Perhaps in the diagram, the angle at Q for triangle QRS is ∠RQS = x+74°, but the label is ∠PQU, which might be the same if P and R are on the same line.

Suppose that P-Q-R is a straight line, then ∠PQU and ∠RQU are adjacent, so if ∠PQU = x+74°, then ∠RQU = 180 - (x+74) = 106 - x°, as before.

I think I have to accept that with the given, if we force the sum, s=36°, so perhaps the 33° is 36°, or vice versa.

Perhaps "33°" is ∠UST, and it is not the angle in the triangle, but the angle between, and for triangle QRS, the angle at S is 180 - 33 = 147° or something.

Let's try that.

Suppose that in triangle QRS, angle at S is 180 - 33 = 147° (if it's the supplementary).

Then sum: (106 - x) + (38 + x) + 147 = 106+38+147 = 291 > 180, impossible.

If angle at S is 33°, and we have 177, perhaps the missing 3° is from elsewhere.

Another idea: perhaps the 33° is for triangle UST, and it is used to find another angle, but not given.

Perhaps for point S, the angle is 33°, and it is vertical to the angle in triangle QRS, so same, and then the sum is 177, so maybe the problem has a mistake, or in some systems, but for school, likely a typo.

Perhaps "33°" is 36°.

Or perhaps the 38+x is 35+x or something.

Let's solve for x assuming the sum is 180 with s=33°.

From 106 - x + 38 + x + 33 = 177 = 180, which is never true, so no solution.

So must be that the angle at S is not 33° for the triangle.

Perhaps the 33° is ∠QST, and in triangle QRS, the angle at S is ∠QSR, which is different.

Let's assume that at S, the angle between Q-S and T-S is 33°, and for triangle QRS, the angle at S is between Q-S and R-S, which may be the same if R and T are on the same line, but not.

I think for the sake of time, I'll assume that the angle at S in triangle QRS is 36°, as per calculation, and proceed, or perhaps in the diagram, it's 36°.

Maybe "33°" is a misread, and it's 36°.

So let's set s = 36°.

Then from earlier, 106 - x + 38 + x + 36 = 180, which is 180 = 180, true for all x, so x can be anything, but that can't be.

In the equation, x cancels, so it's identity, so no constraint, but we have to find specific values.

So that means my assumption about the angle at Q is wrong.

Perhaps ∠PQU = x+74° is the angle in triangle QRS at Q.

So in triangle QRS, angle at Q is x+74°, angle at R is 38+x°, angle at S is 33°.

Then sum: (x+74) + (38+x) + 33 = 2x + 145 = 180

So 2x = 35, x = 17.5

Then ∠PQU = 17.5 + 74 = 91.5°
∠QRS = 38 + 17.5 = 55.5°

And angle at S 33°, sum 91.5+55.5+33=180, good.

And the 33° is given, so perhaps that's it.

In the diagram, ∠PQU might be the angle at Q for the triangle, not the adjacent.

So I'll go with that.

So x = 17.5

Then ∠PQU = 17.5 + 74 = 91.5°
∠QRS = 38 + 17.5 = 55.5°

But usually angles are integer, but possible.

Perhaps x is integer, so maybe not.

Another possibility: perhaps "x+74" and "38+x" are for different things.

Or perhaps the 33° is not in the triangle.

But with this, it works.

So for problem 6:
∠PQU = 91.5°
∠QRS = 55.5°

But let's keep as fractions: x=35/2=17.5, so 91.5 and 55.5.

Or 183/2 and 111/2, but decimal is fine.

---

Problem 7

Triangle ABC and CDE, with C common, and lines ACB and DCE intersecting at C.

Given:
- ∠CDE = 43°
- ∠CED = 67°
- In triangle ABC, ∠BAC = x°, ∠ABC = x° (isosceles)

First, in triangle CDE:
Sum of angles: ∠CDE + ∠CED + ∠DCE = 180°
43° + 67° + ∠DCE = 180°
110° + ∠DCE = 180°
So ∠DCE = 70°

Now, ∠DCE and ∠ACB are vertical angles (since lines AD and BE intersect at C, assuming A-C-D and B-C-E are straight lines).

So ∠ACB = ∠DCE = 70° (vertical angles are equal)

In triangle ABC:
∠BAC + ∠ABC + ∠ACB = 180°
x° + x° + 70° = 180°
2x + 70 = 180
2x = 110
x = 55

So ∠BAC = 55°, ∠ABC = 55°

Answer: ∠ABC = 55°, ∠BAC = 55°

---

Now, let's compile all answers.

Problem 1:
∠AEB = 98°
∠AEC = 82°

Problem 2:
∠PRQ = 66°
∠PSR = 31°

Problem 3:
∠RSP = 42°
∠QSR = 2° (assuming the 2° is given as ∠QSR)

But to be precise, from earlier, in triangle PRS: 66+72+42=180, so ∠RSP=42°
In triangle QSR: if ∠SQR=30°, ∠QRS=2°, then ∠QSR=148°, but the question asks for ∠QSR, so if it's labeled as 2°, then 2°, but that doesn't make sense.

Perhaps in the diagram, the "2°" is ∠QSR, so ∠QSR = 2°, and y° = 148° for ∠QRS.

So I'll put:
∠RSP = 42°
∠QSR = 2°

Problem 4:
∠RQS = 136°

Problem 5:
∠QRT = 50° (assumed vertical angle)

Problem 6:
∠PQU = 91.5°
∠QRS = 55.5°

Problem 7:
∠ABC = 55°
∠BAC = 55°

For problem 3, to match the request, and since the user has "y° 30°" and "2°", and asks for ∠QSR, likely ∠QSR is the 2°, so we'll go with that.

For problem 6, with x=17.5, angles 91.5° and 55.5°.

Now, let's write the final answers.

Final Answer:
1. ∠AEB = 98°, ∠AEC = 82°
2. ∠PRQ = 66°, ∠PSR = 31°
3. ∠RSP = 42°, ∠QSR = 2°
4. ∠RQS = 136°
5. ∠QRT = 50°
6. ∠PQU = 91.5°, ∠QRS = 55.5°
7. ∠ABC = 55°, ∠BAC = 55°

For problem 3, if ∠QSR is 2°, but in the triangle it should be large, but based on the label, we'll keep it.

Perhaps in problem 3, the "2°" is ∠QSR, so answer is 2°.

Similarly for others.

So boxed answers.

Since the instruction is to provide final answer after reasoning, and for each part.

So for the response, I'll list them.

Note: For problem 6, 91.5 and 55.5 can be written as 183/2 and 111/2, but decimal is acceptable.

Or perhaps leave as fractions, but in context, decimal is fine.

So here is the final answer section.