Geometry worksheet with angle measurement problems and a riddle answer code.

Let’s solve each angle problem step by step. We’ll use basic geometry rules:

- In any triangle, the three angles add up to 180°.
- A straight line = 180°.
- Vertical angles (opposite when two lines cross) are equal.
- Adjacent angles on a straight line add to 180°.
- Right angle = 90°.

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Problem I: Triangle ABC

Given: ∠A = 40°, ∠C = 83°
Find: ∠B

→ Sum of angles in triangle = 180°
→ ∠B = 180° - 40° - 83° = 57°

✔ Answer for I: 57°

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Problem G: Triangle LKJ

Given: ∠L = 29°, ∠K = 115°
Find: ∠J

→ ∠J = 180° - 29° - 115° = 36°

✔ Answer for G: 36°

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Problem S: Angles around point O on line UX

We see:
- ∠UOV = 72°
- ∠VOW is marked with a right angle symbol → so it’s 90°
- Line UX is straight → total = 180°
So, ∠WOX = 180° - 72° - 90° = 18°

✔ Answer for S: 18°

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Problem A and E: Intersecting lines at Q

Lines PR and TS intersect at Q.
Given: One angle is 67° (∠RQS or similar — looks like ∠SQR = 67°)

Actually, looking at diagram: angle between R and S is labeled 67° → that’s ∠RQS = 67°

Then:
- ∠PQR is vertical to ∠SQT? Wait — let's label carefully.

Actually, from diagram:
Angle between rays QR and QS is 67° → so ∠RQS = 67°

Then:
- ∠PQR is adjacent to ∠RQS on straight line PS? No — better to think:

When two lines intersect, vertical angles are equal, and adjacent angles sum to 180°.

Assume the 67° is ∠RQS. Then:

→ ∠PQT is vertical to ∠RQS → so also 67°

→ ∠PQR is adjacent to ∠RQS → so 180° - 67° = 113°

Wait — check labels:

The diagram shows:

Line PT and RS crossing at Q.

Angle between R and S is 67° → that’s ∠RQS = 67°

Then:

- ∠PQR is next to it → same side → should be 180° - 67° = 113°? But wait — actually, if you look, ∠PQR and ∠RQS are adjacent along line PS? Not necessarily.

Better approach: The four angles around point Q:

If one angle is 67°, then its vertical opposite is also 67°, and the other two are each 180° - 67° = 113°.

Looking at the diagram labeling:

It says “67°” near angle between R and S → likely ∠RQS = 67°

Then:

- ∠PQT is vertical to ∠RQS → so 67°
- ∠PQR is adjacent → so 113°

But the problems ask:

(A) m∠PQR = ? → that’s the big angle on top left → yes, 113°
(E) m∠PQT = ? → that’s the bottom left angle → which is vertical to ∠RQS → 67°

Wait — let me double-check positions.

Actually, standard notation: ∠PQR means angle at Q between points P, Q, R.

In diagram: Points are arranged as:

Top: P and R
Bottom: T and S
Crossing at Q.

So ray QP goes up-left, QR goes up-right, QT goes down-left, QS goes down-right.

Given angle between QR and QS is 67° → that’s ∠RQS = 67°

Then:

- ∠PQT is vertical to ∠RQS → so also 67° → that’s answer for E
- ∠PQR is between QP and QR → which is adjacent to ∠RQS → since QP and QS are opposite? Actually, QP and QS are not necessarily opposite.

Wait — better: The straight line is PTS? Or PRS?

Actually, line PR is one line, line TS is another, crossing at Q.

So angles:

At intersection:

Vertical pairs:

∠PQT and ∠RQS are vertical → both 67°
∠PQR and ∠TQS are vertical → both 180° - 67° = 113°

Yes.

So:

(A) m∠PQR = 113°
(E) m∠PQT = 67°

✔ Answers:
A: 113°
E: 67°

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Problem N and O: At point A, with perpendicular BA and line AD

Diagram: Point A, with ray AB going up (perpendicular to AD), and ray AC going up-right at 56° from AD.

Given: ∠CAD = 56°, and BA ⊥ AD → so ∠BAD = 90°

Then:

(N) m∠DAB = angle between DA and AB → that’s the right angle → 90°

(O) m∠DAC = given as 56° → but wait, the question is m∠DAC — that’s the same as ∠CAD → 56°

Wait — no: ∠DAC is angle at A between D, A, C → which is exactly the 56° shown.

But let’s read:

(N) m∠DAB → D to A to B → that’s the full right angle → 90°
(O) m∠DAC → D to A to C → that’s the 56° part

Yes.

✔ Answers:
N: 90°
O: 56°

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Problem C and U: Triangle XYZ with exterior angle at Z

Given: ∠X = 79°, and exterior angle at Z is 138°

Exterior angle = sum of two remote interior angles.

So, exterior ∠ at Z = ∠X + Y
→ 138° = 79° + ∠Y
→ ∠Y = 138° - 79° = 59°

Also, (C) m∠XZY = that’s the interior angle at Z.

Since exterior is 138°, interior is 180° - 138° = 42°

Check: 79° + 59° + 42° = 180° → yes.

✔ Answers:
C: 42°
U: 59°

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Problem F and K: Triangle MLN, right-angled at L

Given: Right angle at L → ∠L = 90°
Also, at point N, there’s an angle of 37° outside — that’s the exterior angle.

So, interior angle at N: since exterior is 37°, interior ∠MNL = 180° - 37° = 143°? Wait — that can’t be, because triangle already has 90° at L.

Wait — look again.

Diagram: Triangle MLN, right angle at L. Line LN extended to the right, and angle between extension and MN is 37° → so that’s the exterior angle at N.

Therefore, interior angle at N (∠MNL) = 180° - 37° = 143°? But then triangle angles would be 90° + 143° + something > 180° — impossible.

Mistake!

Actually, the 37° is the angle between the extension of LN and the side MN — so that is the exterior angle, meaning the interior angle at N is supplementary to it only if it’s on a straight line.

Standard rule: Exterior angle = sum of two non-adjacent interior angles.

In triangle MLN:

Right angle at L → ∠L = 90°

Exterior angle at N = 37° → this equals ∠L + ∠M

Because exterior angle at N is formed by extending LN, so it equals sum of remote interiors: ∠L and ∠M.

So:

37° = ∠L + ∠M = 90° + ∠M → that gives negative — impossible.

I think I have the diagram wrong.

Re-examining: The 37° is drawn at point N, between the extension of LN and the side MN — so yes, that’s the exterior angle.

But in a triangle, exterior angle must be greater than either remote interior angle. Here, if ∠L is 90°, exterior angle can't be 37° — contradiction.

Unless... perhaps the 37° is the interior angle? But it's drawn outside.

Wait — maybe it's the angle between MN and the extension, but on the other side.

Another possibility: The 37° is the angle inside the triangle at N? But it's labeled outside.

Look back at image description: "at N, 37°" and it's outside the triangle, adjacent to angle MNL.

Perhaps it's the vertical angle or something.

Alternative interpretation: Maybe the 37° is the measure of ∠MNL itself? But the diagram shows it outside.

Let me think differently.

In many such diagrams, when they show an angle outside like that, it's often the alternate interior or corresponding, but here it's probably meant to be the exterior angle.

But mathematically, if ∠L = 90°, and exterior at N is 37°, then 37° = ∠L + ∠M = 90° + ∠M → ∠M = -53° — impossible.

So likely, the 37° is the interior angle at N.

Perhaps the label is misplaced, or I'm misreading.

Another idea: The 37° is the angle between MN and the horizontal, but since LN is horizontal, and MN is going down, the angle inside the triangle at N might be 37°.

Let me assume that ∠MNL = 37° — that makes sense.

Then, in triangle MLN:

∠L = 90°, ∠N = 37°, so ∠M = 180° - 90° - 37° = 53°

And (F) m∠MNL = 37°
(K) m∠M = 53°

That works.

Perhaps the 37° is labeled as the exterior, but in context, it's likely the interior angle.

To confirm: If exterior were 37°, interior would be 143°, too big. So probably, the 37° is the interior angle at N.

I'll go with that.

✔ Answers:
F: 37°
K: 53°

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Problem Q and B: Triangle DEF with transversal

Diagram: Line DF horizontal, triangle DEF above it. At D, angle between ED and DF is 41°. At F, exterior angle is 104°.

First, (Q) m∠EFD = that's the interior angle at F.

Exterior angle at F is 104°, so interior ∠EFD = 180° - 104° = 76°

Now, in triangle DEF:

∠D = 41° (given)
∠F = 76°
So ∠E = 180° - 41° - 76° = 63°

(B) m∠E = 63°

✔ Answers:
Q: 76°
B: 63°

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Problem M and R: Three lines intersecting at O

Lines: AD horizontal, BC and another line crossing.

Given: ∠AOD is straight line. Angles given: ∠AOB = 27°, ∠COD = 54°? Wait.

Diagram: Point O, with rays OA, OB, OC, OD.

OA and OD are opposite (straight line). OB and OC are other rays.

Given: ∠AOB = 27°, ∠DOC = 54°? The label says "54°" near D and C, so likely ∠DOC = 54°.

But we need ∠AOB and ∠BOC.

First, (M) m∠AOB = given as 27° — directly labeled.

Is that it? Yes, it's written as 27° between A and B.

So (M) = 27°

Now (R) m∠BOC = ?

Note that A-O-D is straight line, so ∠AOD = 180°

Angles around point O: ∠AOB + ∠BOC + ∠COD = 180°? Only if B and C are on the same side.

From diagram: Rays are OA (left), OD (right), OB up-left, OC up-right.

So angles: from OA to OB is 27°, from OC to OD is 54°, and between OB and OC is ∠BOC.

Since OA to OD is 180°, then:

∠AOB + ∠BOC + ∠COD = 180°

27° + ∠BOC + 54° = 180°

∠BOC = 180° - 27° - 54° = 99°

✔ Answers:
M: 27°
R: 99°

---

Now, let's list all answers with their letters:

I: 57°
G: 36°
S: 18°
A: 113°
E: 67°
N: 90°
O: 56°
C: 42°
U: 59°
F: 37°
K: 53°
Q: 76°
B: 63°
M: 27°
R: 99°

Now, recall the code at the top:

For joke 1: "Cross two ducks with a match?"
Answer code: 37° 57° 99° 67° 104° 76° 59° 113° 42° 53° 67° 99° 18°

We need to find which letter corresponds to each degree, and write the letter above the degree in the code.

Similarly for joke 2.

But the instruction is: "Look for each answer in the code. Each time the answer appears, write the letter of the exercise above it."

So for the first code sequence: 37° 57° 99° 67° 104° 76° 59° 113° 42° 53° 67° 99° 18°

We match:

37° → F
57° → I
99° → R
67° → E
104° → ? Not in our answers — skip? But it's in the code.
76° → Q
59° → U
113° → A
42° → C
53° → K
67° → E (again)
99° → R (again)
18° → S

So writing letters above:

Above 37°: F
Above 57°: I
Above 99°: R
Above 67°: E
Above 104°: ? (no match, leave blank or ignore)
Above 76°: Q
Above 59°: U
Above 113°: A
Above 42°: C
Above 53°: K
Above 67°: E
Above 99°: R
Above 18°: S

So the word formed is: F I R E _ Q U A C K E R S → "FIRE QUACKERS"? But 104° has no letter.

Perhaps 104° is not used, or I missed an answer.

In our solutions, we don't have 104°. But in the code it's there. Maybe it's a distractor.

Similarly, for the second code: 113° 68° 63° 34° 34° 54° 38° 54° 67° 99° 57° 90° 36° 59° 67°

Match:

113° → A
68° → ? not in our answers
63° → B
34° → ? not found
34° → ?
54° → ? not found
38° → ?
54° → ?
67° → E
99° → R
57° → I
90° → N
36° → G
59° → U
67° → E

So: A _ B _ _ _ _ _ E R I N G U E → doesn't make sense.

Perhaps I made a mistake in some answers.

Let me double-check problem F and K.

Earlier I assumed ∠MNL = 37°, but maybe it's the exterior.

Another interpretation: In triangle MLN, right-angled at L, and the 37° is the angle between MN and the extension of LN, so it's the exterior angle at N.

Then, exterior angle = sum of two remote interior angles = ∠L + ∠M

So 37° = 90° + ∠M → impossible.

Unless the right angle is not at L? But it's marked with square at L.

Perhaps the 37° is at M or something.

Let's calculate what it should be.

Suppose in triangle MLN, ∠L = 90°, and let ∠M = x, ∠N = y, x+y=90°.

The exterior angle at N is 180° - y.

If that is given as 37°, then 180° - y = 37° → y = 143°, impossible.

So the only logical conclusion is that the 37° is the interior angle at N.

Perhaps the diagram has the 37° as the angle of depression or something, but I think for school level, it's likely that ∠MNL = 37°.

Maybe "37°" is the measure of ∠LMN or something else.

Another idea: The 37° is the angle between MN and the horizontal, and since LN is horizontal, and ML is vertical, then in triangle, at N, the angle is between MN and LN, which is 37°, so yes, ∠MNL = 37°.

I think it's correct.

For the code, perhaps 104° is not needed, or it's a typo.

In the first code, after "FIRE" then "QUACKERS", but there's a gap at 104°.

Perhaps 104° corresponds to an answer I missed.

Let's list all our answers again:

From problems:

I: 57
G: 36
S: 18
A: 113
E: 67
N: 90
O: 56
C: 42
U: 59
F: 37
K: 53
Q: 76
B: 63
M: 27
R: 99

No 104°.

In problem Q and B, we have exterior angle 104°, but we didn't use it as an answer; we used the interior 76°.

The question asks for m∠EFD and m∠E, not the exterior.

So 104° is not an answer to any problem.

Similarly, in the code, it might be that for the joke, we only care about the degrees that appear in our answers.

For the first code: positions with our answers:

Position 1: 37° -> F
2: 57° -> I
3: 99° -> R
4: 67° -> E
5: 104° -> no letter
6: 76° -> Q
7: 59° -> U
8: 113° -> A
9: 42° -> C
10: 53° -> K
11: 67° -> E
12: 99° -> R
13: 18° -> S

So the letters are: F,I,R,E, ,Q,U,A,C,K,E,R,S

Reading as words: "FIRE" then "QUACKERS" — but "QUACKERS" has 8 letters, here we have Q,U,A,C,K,E,R,S — that's 8, but position 5 is missing.

Perhaps the 104° is ignored, and we read "FIREQUACKERS" but that's not a word.

"FIRE CRACKERS" — oh! Probably "FIRE CRACKERS".

But we have Q instead of C for position 6.

Position 6 is 76° -> Q, but for "CRACKERS", it should be C for the third letter.

Let's see the sequence: after FIRE, next is 104° (blank), then 76°->Q, 59°->U, 113°->A, 42°->C, 53°->K, 67°->E, 99°->R, 18°->S

So: _ Q U A C K E R S

If we ignore the blank, it's QUACKERS, but should be CRACKERS.

Perhaps I have a mistake in assignment.

Another possibility: for problem C, m∠XZY = 42°, but perhaps it's assigned to a different letter.

Or perhaps in the code, the degrees are to be matched, and for "CRACKERS", C is 42°, R is 99°, A is 113°, C is 42° again, K is 53°, E is 67°, R is 99°, S is 18°.

In the code, position 9 is 42° -> C, position 10 is 53° -> K, etc.

For "CRACKERS", the letters are C,R,A,C,K,E,R,S

In the code sequence, starting from position 6: 76°->Q, but should be C for "C" in crackers.

Unless position 6 is not part of it.

Perhaps the first four are "FIRE", then the next eight are "CRACKERS", but position 5 is 104°, which might be a separator or error.

Maybe 104° corresponds to an answer I have wrong.

Let's check problem Q and B again.

In triangle DEF, at F, exterior angle is 104°, so interior ∠EFD = 180° - 104° = 76°, which is Q.

But perhaps the question is asking for the exterior angle? No, it says m∠EFD, which is the interior angle at F.

Similarly, for other problems.

Another thought: in the first code, "104°" might be a red herring, or perhaps it's for a different problem.

Let's look at the second code for the second joke.

Second code: 113° 68° 63° 34° 34° 54° 38° 54° 67° 99° 57° 90° 36° 59° 67°

Match:

113° -> A
68° -> ? not in answers
63° -> B
34° -> ?
34° -> ?
54° -> ?
38° -> ?
54° -> ?
67° -> E
99° -> R
57° -> I
90° -> N
36° -> G
59° -> U
67° -> E

So: A, _, B, _, _, _, _, _, E, R, I, N, G, U, E

Reading: A_B____ERINGUE — not good.

"ABRINGUE" not a word.

Perhaps "A BRING UE" no.

Another idea: perhaps for problem O, m∠DAC = 56°, but 56° is not in the codes, so ok.

Let's list all unique answers we have: 18,27,36,37,42,53,56,57,59,63,67,76,90,99,113

In first code: 37,57,99,67,104,76,59,113,42,53,67,99,18 — all except 104 are in our answers.

In second code: 113,68,63,34,34,54,38,54,67,99,57,90,36,59,67 — 68,34,54,38 are not in our answers.

So for the jokes, we only fill in the letters for the degrees that match our answers.

For first joke: "Cross two ducks with a match?" -> answer should be "FIRE CRACKERS" but we have "FIRE" then "QUACKERS" with Q instead of C.

Unless for position 6, 76° is not Q, but what else?

Perhaps I misidentified problem Q.

Problem Q is m∠EFD = 76°, letter Q.

But in "CRACKERS", the first C is for 42°, which is problem C.

In the code, position 9 is 42° -> C, which would be the fourth letter of "CRACKERS" if we start from there.

Let's map the code to the word "FIRE CRACKERS".

"FIRE" : F,I,R,E -> 37,57,99,67 — matches positions 1,2,3,4.

Then space or something, then "CRACKERS": C,R,A,C,K,E,R,S

C=42°, R=99°, A=113°, C=42°, K=53°, E=67°, R=99°, S=18°

In the code, after position 4 (67°), position 5 is 104° (not used), position 6 is 76° (Q), position 7 is 59° (U), position 8 is 113° (A), position 9 is 42° (C), position 10 is 53° (K), position 11 is 67° (E), position 12 is 99° (R), position 13 is 18° (S)

So for "CRACKERS", we need C at position 6, but we have Q (76°).

Unless the 76° is not for Q, but for another problem.

Perhaps in problem Q, m∠EFD is not 76°.

Let's recalculate problem Q and B.

Triangle DEF: points D, E, F.

Line DF horizontal. At D, angle between ED and DF is 41° — so ∠EDF = 41°.

At F, the exterior angle is 104° — this is the angle between EF and the extension of DF beyond F.

So, the interior angle at F, ∠EFD, is adjacent to the exterior angle, so ∠EFD = 180° - 104° = 76° — that seems correct.

Then in triangle, ∠D = 41°, ∠F = 76°, so ∠E = 180° - 41° - 76° = 63° — B.

Seems correct.

Perhaps the letter for 76° is not Q, but the problem is labeled Q for m∠EFD, so it should be Q.

Another possibility: in the code, the degrees are to be matched, and for "CRACKERS", the C is 42°, which is at position 9, R is 99° at position 12, etc, but that doesn't form the word in order.

Perhaps the answer is "FIREWORKS" or something else.

Let's try to see what word is formed by the letters in order for the first code.

Positions with letters:

1: F (37°)
2: I (57°)
3: R (99°)
4: E (67°)
6: Q (76°)
7: U (59°)
8: A (113°)
9: C (42°)
10: K (53°)
11: E (67°)
12: R (99°)
13: S (18°)

So the sequence is F,I,R,E,Q,U,A,C,K,E,R,S

Read as "FIRE QUACKERS" — and "quackers" is a play on "crackers", since ducks quack.

Oh! "Cross two ducks with a match" -> "FIRE QUACKERS" — because ducks quack, so "quackers" instead of "crackers".

Yes! That makes sense for a pun.

Similarly, for the second joke: "Cross a stick of dynamite with a lemon pie?" -> probably "DYNAMITE PIE" or something, but let's see.

Second code: 113° 68° 63° 34° 34° 54° 38° 54° 67° 99° 57° 90° 36° 59° 67°

Letters:

1: A (113°)
3: B (63°)
9: E (67°)
10: R (99°)
11: I (57°)
12: N (90°)
13: G (36°)
14: U (59°)
15: E (67°)

So positions: 1:A, 3:B, 9:E, 10:R, 11:I, 12:N, 13:G, 14:U, 15:E

Sequence: A, _, B, _, _, _, _, _, E, R, I, N, G, U, E

Reading the letters in order: A,B,E,R,I,N,G,U,E — "ABERINGUE" not good.

Perhaps "A BRING UE" no.

"BRING" is there: B,R,I,N,G at positions 3,10,11,12,13

But before that A, after U,E.

"A BRING UE" not meaningful.

Perhaps "DYNAMITE" but D is not in our answers.

Another idea: perhaps for problem O, m∠DAC = 56°, but 56° is not in the code, so ok.

Let's list the letters we have for the second code: A,B,E,R,I,N,G,U,E

Perhaps it's "A BURGER IN E" no.

"BRING" and "UE" — "BRING UE" not good.

Perhaps I missed an answer.

What about problem G: 36° -> G, which is in position 13.

Problem S: 18° not in second code.

Perhaps the word is "DYNAMITE PIE" but we don't have D,Y,M,T,P.

Another thought: in the second code, position 2 is 68°, which might be for a problem I have wrong.

Let's check problem N and O again.

At point A, BA perpendicular to AD, so ∠BAD = 90°.

Ray AC such that ∠CAD = 56°, so ∠BAC = 90° - 56° = 34°.

Oh! I think I made a mistake here.

In problem N and O:

(N) m∠DAB = angle between D,A,B — which is the right angle, 90° — that's correct.

(O) m∠DAC = angle between D,A,C — which is given as 56° — but is that correct?

The diagram shows 56° between CA and DA, so yes, ∠CAD = 56°, so ∠DAC = 56°.

But perhaps the question is m∠BAC or something.

Let's read the problems:

(N) m∠DAB = ? — that's D to A to B, which is 90°.

(O) m∠DAC = ? — D to A to C, which is 56°.

But in the code, 34° appears twice, and 34° is not in our answers yet.

Notice that ∠BAC = ∠BAD - CAD = 90° - 56° = 34°.

But the problem doesn't ask for ∠BAC; it asks for ∠DAB and ∠DAC.

However, perhaps in the diagram, the 56° is labeled as ∠BAC or something.

Let's assume that the 56° is the angle between BA and CA, not between CA and DA.

In many diagrams, they might label the angle at A between B and C as 56°.

Let me re-read the user's image description.

In the initial request, for problem N and O: "at A, with perpendicular BA and line AD, and ray AC with 56°"

It says: "56°" near the angle between BA and CA or between CA and DA?

In the text: "56°" is written, and from context, it's likely between CA and DA, as I had.

But to get 34°, perhaps it's between BA and CA.

Suppose that the 56° is ∠BAC = 56°.

Then, since BA ⊥ AD, ∠BAD = 90°, so ∠CAD = 90° - 56° = 34°.

Then:

(N) m∠DAB = 90° (still)
(O) m∠DAC = 34° (if DAC is the same as CAD)

Then we have 34° for O.

Similarly, in the code, 34° appears twice, so perhaps for two problems.

But in our current setup, we have only one 34°.

Perhaps for problem O, it's 34°.

Let's try that.

Assume that the 56° is ∠BAC, not ∠CAD.

Then:

∠BAC = 56°
∠BAD = 90° (since BA ⊥ AD)
So ∠CAD = ∠BAD - ∠BAC = 90° - 56° = 34°

Then:

(N) m∠DAB = 90°
(O) m∠DAC = 34° (angle between D,A,C)

Yes.

Then we have 34° for O.

In the second code, 34° appears twice, so perhaps another problem has 34°.

What could it be? Perhaps in other problems.

For example, in problem G, we have 36°, close but not 34.

Or in F and K, we have 37° and 53°.

Perhaps for problem S or others.

Another possibility: in the intersecting lines, but we have 67° and 113°.

Perhaps for problem C and U, but we have 42° and 59°.

Let's see if there's another 34°.

Perhaps in the dynamite joke, but let's proceed with this change.

So revise:

For N and O:

(N) m∠DAB = 90°
(O) m∠DAC = 34° (assuming the 56° is ∠BAC)

Then our answers include 34° for O.

Now, in the second code: 113° 68° 63° 34° 34° 54° 38° 54° 67° 99° 57° 90° 36° 59° 67°

Match:

113° -> A
68° -> ? still not
63° -> B
34° -> O (first occurrence)
34° -> O (second occurrence) — but only one O, so perhaps another problem has 34°.

Maybe for problem G or others.

Perhaps the 68° is for a problem.

Let's calculate if there's a 34° elsewhere.

In problem F and K, if we have different values.

Another idea: in the first triangle, but no.

Perhaps for problem S, but we have 18°.

Let's list all answers with this change:

I: 57
G: 36
S: 18
A: 113
E: 67
N: 90
O: 34 (changed)
C: 42
U: 59
F: 37
K: 53
Q: 76
B: 63
M: 27
R: 99

Still no second 34°.

Unless problem O is listed twice, but it's only once.

Perhaps in the code, the two 34° are for the same letter, but usually each degree corresponds to a letter.

Another possibility: for problem "m∠DAC" and perhaps "m∠BAC" but the problem doesn't ask for m∠BAC.

The problems are only those listed.

Perhaps for the dynamite joke, "lemon pie" suggests something sweet, but let's try to form the word.

With O: 34°, then in second code:

Position 1: 113° -> A
2: 68° -> ?
3: 63° -> B
4: 34° -> O
5: 34° -> O (again)
6: 54° -> ?
7: 38° -> ?
8: 54° -> ?
9: 67° -> E
10: 99° -> R
11: 57° -> I
12: 90° -> N
13: 36° -> G
14: 59° -> U
15: 67° -> E

So letters: A, _, B, O, O, _, _, _, E, R, I, N, G, U, E

Reading: A_B OO ___ ERINGUE — not good.

"ABOO" not a word.

Perhaps "A BOO" but then "ERINGUE".

Another idea: perhaps the 68° is for problem G, but we have 36° for G.

Unless I miscalculated problem G.

Problem G: triangle LKJ, ∠L=29°, ∠K=115°, so ∠J=180-29-115=36° — correct.

Perhaps for problem S, but 18°.

Let's consider that in the second code, "68°" might be a typo, or perhaps it's 36°.

But it's written as 68°.

Perhaps for problem B, m∠E = 63°, but 63° is there.

Let's try to see what word is intended.

"Cross a stick of dynamite with a lemon pie?" -> probably "DYNAMITE PIE" or "EXPLOSION" or "BOMB" but let's think of puns.

"Dynamite" and "lemon pie" — perhaps "LEMONADE" or "PIE" but not matching.

Another common pun: "dynamite" and "pie" might give "DYNAMITE PIE" but how to spell with letters.

Perhaps "A BOMB" but not.

Let's look at the letters we have: A,B,O,O,E,R,I,N,G,U,E

Perhaps "A BORING UE" no.

"BRING" is there, and "OU" or something.

Perhaps "YOU BRING" but Y not there.

Another thought: in problem M and R, we have 27° and 99°, but 27° is not in second code.

Perhaps for the second joke, the answer is "A BURGER" but not.

Let's calculate if there's a 34° in other places.

In problem C and U, if we have different.

Or in F and K.

Suppose in triangle MLN, if ∠MNL = 37°, ∠L = 90°, ∠M = 53°, as before.

But 34° not there.

Perhaps the 37° is for a different problem.

Let's assume that for problem O, m∠DAC = 34°, and for another problem, say, if we have m∠BAC = 56°, but the problem doesn't ask for it.

Perhaps the problem has a typo, or in some interpretations.

Another idea: in the intersect