2-The Angle Addition Postulate | PDF | Geometry | Space - Free Printable

Let’s solve each problem one by one. These are all about angles around a point or on a straight line — remember:

- Angles on a straight line add up to 180°.
- Angles around a point add up to 360°.

We’ll use these rules for every problem.

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Problem 2:
Find m∠FED if m∠FDB = 105° and m∠BDE = 34°.

Look at the diagram (even though we can’t see it, from context): Points F, D, E are likely on a straight line with B somewhere in between? Wait — actually, since ∠FDB and ∠BDE are given, and we’re to find ∠FED… Hmm, maybe points F, D, E form a triangle? Or perhaps D is the vertex?

Wait — let’s think differently. If you have angles at point D: ∠FDB and ∠BDE, then together they make ∠FDE.

So:
m∠FDE = m∠FDB + m∠BDE = 105° + 34° = 139°

But the question asks for m∠FED — that’s angle at E in triangle FED? But we don’t have enough info unless it’s a straight line.

Wait — maybe points F, D, E are colinear? No, because then angle at E wouldn’t make sense.

Actually, looking at standard problems like this — often, when two angles are given at a point along a path, and you’re asked for another angle, it’s about angles on a straight line or around a point.

Wait — perhaps ∠FDB and ∠BDE are adjacent angles forming ∠FDE, and then ∠FED is part of a triangle? But without more info, that doesn’t work.

Hold on — maybe it's simpler. Let me re-read: “Find m∠FED” — so vertex is E. That suggests triangle FED or something.

But here’s a better idea: In many such worksheets, when you see names like ∠FDB and ∠BDE, and then ask for ∠FED, it might be that points F, D, E are connected, and B is on DE or FD.

Alternatively — perhaps D is the common vertex, and rays go to F, B, E. So ∠FDB and ∠BDE are adjacent, making ∠FDE = 105° + 34° = 139°. Then if F, D, E are not colinear, but we need ∠FED — still stuck.

Wait — maybe it’s a typo or mislabel? Or perhaps it’s supposed to be ∠FDE? But the problem says ∠FED.

Another thought: Maybe triangle FDE, and we know two angles at D? No.

Let me try to assume that points F, D, E form a triangle, and B is on side DE. Then ∠FDB is an exterior angle? Not sure.

This is confusing. Let’s skip and come back — or look at pattern.

Looking at Problem 3:
m∠GBC = 105°, m∠CBH = 104°, find m∠GBH.

That makes sense: G-B-H with C in between? So ∠GBH = ∠GBC + ∠CBH = 105° + 104° = 209°? But that’s more than 180° — possible if it’s reflex, but usually in basic geometry, we take smaller angle.

Wait — 105 + 104 = 209, which is greater than 180, so maybe it’s the other way? Or perhaps points are arranged so that ∠GBH is the sum.

But 209° is valid as a reflex angle, but typically in such problems, if it’s over 180, they might expect the smaller one, but 360 - 209 = 151, which isn't matching.

Perhaps I’m misunderstanding the configuration.

Let me try a different approach. In many of these, the angles are around a point or on a line.

For Problem 2: Suppose D is the vertex, and we have ray DF, DB, DE. If ∠FDB = 105° and ∠BDE = 34°, then if B is between F and E, then ∠FDE = 105° + 34° = 139°. But the question is ∠FED — which is at E.

Unless... perhaps it's a triangle, and we need to find angle at E, but we only have angles at D? Not possible.

Wait — maybe "m∠FED" is a mistake, and it should be "m∠FDE"? That would make sense. Because then it's just adding the two angles.

In many textbooks, they do that. Let me check Problem 4: Find m∠WVZ if m∠ZVT = 62° and m∠WTZ = 35° — wait, WTZ? That has T, not V. Probably typo.

This is messy. Perhaps I should assume that for Problem 2, ∠FED is meant to be ∠FDE, so answer is 105 + 34 = 139°.

Similarly, Problem 3: m∠GBH = m∠GBC + m∠CBH = 105 + 104 = 209°, but that seems large. Or perhaps it's the difference? 105 - 104 = 1? Unlikely.

Another idea: Perhaps the points are on a straight line. For example, in Problem 3, if G, B, H are colinear, and C is not on the line, then ∠GBC and ∠CBH are adjacent angles that make ∠GBH, but if G-B-H is straight, then ∠GBH should be 180°, but 105+104=209>180, so not straight.

Perhaps B is the vertex, and rays to G, C, H. So ∠GBC and ∠CBH are adjacent, so total ∠GBH = 105 + 104 = 209°. And since it's around a point, it could be, but usually we report the smaller angle, so 360 - 209 = 151°. But the problem doesn't specify.

I recall that in some systems, they allow reflex angles, but for middle school, probably not.

Let's look at Problem 5: m∠JKO = 59°, m∠LKP = 26°, find m∠JKP.

Again, K is vertex. Likely, J-K-P with L in between or something. If ∠JKO and ∠LKP are given, but O and P are different points.

This is tricky without diagrams.

Perhaps for all these, the angles are adjacent and we add them.

Let me try Problem 6: Find m∠NDC if m∠EDC = 145° and m∠EDN = 61°.

If D is vertex, and rays to E, N, C. If ∠EDC = 145°, and ∠EDN = 61°, then if N is between E and C, then ∠NDC = ∠EDC - ∠EDN = 145 - 61 = 84°.

That makes sense! So subtraction when one angle is part of another.

Similarly, for Problem 2: If ∠FDB = 105°, ∠BDE = 34°, and if B is between F and E, then ∠FDE = 105 + 34 = 139°, but the question is ∠FED — still not matching.

Unless in Problem 2, it's ∠FDE, and "E" and "D" are swapped in the question.

Perhaps "m∠FED" is a typo, and it should be "m∠FDE". I think that's likely, because otherwise it doesn't make sense.

Similarly, in Problem 4: Find m∠WVZ if m∠ZVT = 62° and m∠WTZ = 35° — WTZ has T, not V, so probably typo, should be m∠WTV or something.

Assume that for Problem 2, it's m∠FDE = m∠FDB + m∠BDE = 105° + 34° = 139°.

For Problem 3: m∠GBH = m∠GBC + m∠CBH = 105° + 104° = 209°, but since it's likely the smaller angle, 360 - 209 = 151°, but let's see the numbers.

105 + 104 = 209, and 360 - 209 = 151, but perhaps it's not around a point.

Another possibility: In Problem 3, if G, B, H are on a straight line, then ∠GBH = 180°, but 105 + 104 = 209 > 180, so not.

Perhaps C is on the other side.

Let's calculate the difference: |105 - 104| = 1°, unlikely.

Or perhaps ∠GBH = |m∠GBC - m∠CBH| if they overlap, but 105 - 104 = 1, too small.

I think for consistency, in Problem 6, we did subtraction: 145 - 61 = 84°.

In Problem 5: m∠JKO = 59°, m∠LKP = 26°, find m∠JKP.

If K is vertex, and rays to J, O, L, P. If O and L are the same or something.

Perhaps ∠JKP = ∠JKO + ∠OKP, but we have ∠LKP.

Assume that L is on KO or something.

Suppose that from K, ray KJ, then KO, then KP, and L is on KO or KP.

If m∠JKO = 59°, and m∠LKP = 26°, and if L is on the extension or something.

Perhaps ∠JKP = m∠JKO + m∠OKP, but we don't have ∠OKP.

Another idea: Perhaps "m∠LKP" is meant to be "m∠OKP", and L is a typo for O.

Then m∠JKP = m∠JKO + m∠OKP = 59° + 26° = 85°.

That makes sense.

Similarly, for Problem 2, if "m∠FED" is "m∠FDE", then 105 + 34 = 139°.

For Problem 3, if m∠GBH = m∠GBC + m∠CBH = 105 + 104 = 209°, but since it's probably not, or perhaps it's 180 - something.

Let's look at Problem 7: Find m∠JKE if m∠IKE = 31° and m∠IKJ = 52°.

K is vertex. Likely, I-K-J and I-K-E, so if J and E are on opposite sides, then ∠JKE = ∠IKJ + ∠IKE = 52° + 31° = 83°.

Yes, that makes sense.

Similarly, Problem 8: Find m∠IHQ if m∠GHQ = 170° and m∠GHI = 159°.

H is vertex. If G-H-Q and G-H-I, then if I is between G and Q, then ∠IHQ = ∠GHQ - ∠GHI = 170° - 159° = 11°.

Yes.

Problem 9: Find m∠LEF if m∠LEM = 20° and m∠KEM = 110°.

E is vertex. Rays to L, M, K. If M is between L and K, then ∠LEK = ∠LEM + ∠MEK = 20° + 110° = 130°, but the question is ∠LEF — again, F instead of K? Probably typo, should be ∠LEK.

Assume that.

Problem 10: Find m∠JDA if m∠ADC = 81° and m∠JDC = 113°.

D is vertex. Rays to J, A, C. If A is between J and C, then ∠JDC = ∠JDA + ∠ADC, so 113° = m∠JDA + 81°, thus m∠JDA = 113° - 81° = 32°.

Perfect.

So general rule: When two angles share a common ray, and you're to find the whole or part, add or subtract accordingly.

Now back to Problem 2: Find m∠FED — but based on pattern, likely it's m∠FDE, and it's the sum: 105° + 34° = 139°.

Similarly, Problem 3: m∠GBH = m∠GBC + m∠CBH = 105° + 104° = 209°, but 209° is greater than 180°, and in most contexts, they might expect the smaller angle, but 360 - 209 = 151°, or perhaps it's correct as is.

But let's calculate: 105 + 104 = 209, and if it's around a point, it could be, but typically for such problems, if it's over 180, they might have intended the other combination.

Perhaps in Problem 3, the angles are on a straight line, but 105 + 104 = 209 > 180, so not.

Another possibility: Perhaps ∠GBC and ∠CBH are not adjacent in the way I think. Maybe B is the vertex, and G, C, H are points, but C is not between G and H.

For example, if ray BC is between BG and BH, then yes, sum.

But 209° is unusual.

Let's see the numbers: 105 and 104, sum 209, difference 1.

Perhaps it's |105 - 104| = 1°, but that seems too small.

Or perhaps for Problem 3, it's m∠GBH = 180° - something, but no.

I recall that in some cases, if the angles are on a straight line, but here it's not specified.

Let's assume that for Problem 3, since 105 + 104 = 209, and 360 - 209 = 151, and 151 is a nice number, perhaps that's it, but why would they give those numbers.

Perhaps the diagram shows that G, B, H are not colinear, and the angle at B for GBH is the sum.

But to be consistent with other problems, in Problem 6, we did subtraction: 145 - 61 = 84°.

In Problem 8: 170 - 159 = 11°.

In Problem 10: 113 - 81 = 32°.

In Problem 5: if we assume m∠JKP = m∠JKO + m∠OKP, and if m∠LKP is m∠OKP, then 59 + 26 = 85°.

In Problem 7: 52 + 31 = 83°.

In Problem 9: if m∠LEF is m∠LEK, then 20 + 110 = 130°.

For Problem 2: if m∠FED is m∠FDE, then 105 + 34 = 139°.

For Problem 3: m∠GBH = m∠GBC + m∠CBH = 105 + 104 = 209°, but perhaps it's correct, or perhaps it's a different configuration.

Let's calculate 180 - 105 = 75, 180 - 104 = 76, not helpful.

Another idea: Perhaps in Problem 3, the points are such that ∠GBC and ∠CBH are on a straight line, but then their sum should be 180, but 105+104=209≠180, so not.

Perhaps B is not the vertex for the final angle, but that doesn't make sense.

I think I have to go with the addition for Problem 3 as well, even if it's 209°, or perhaps it's 151° as the reflex, but let's see the answer choices or typical values.

Since this is a worksheet, and other answers are reasonable, perhaps for Problem 3, it's 105 + 104 = 209°, and they accept it.

But let's check Problem 4: Find m∠WVZ if m∠ZVT = 62° and m∠WTZ = 35°.

WTZ has T, not V, so likely typo. Probably m∠WTV = 35° or m∠WV T = 35°.

Assume that m∠WVT = 35°, and m∠ZVT = 62°, then if V is vertex, and T is common, then ∠WVZ = ∠WVT + ∠TVZ = 35° + 62° = 97°, if T is between W and Z.

Yes.

So for all, we add or subtract based on configuration.

For Problem 3, since no subtraction is indicated, and angles are given as adjacent, likely sum is 209°.

But to confirm, let's do Problem 1, but Problem 1 is not numbered; the first is 2.

In the image, it starts from 2, so perhaps 1 is missing.

Anyway, I'll proceed with the calculations as per the pattern.

So let's list them:

Problem 2: m∠FDE = m∠FDB + m∠BDE = 105° + 34° = 139° (assuming "FED" is typo for "FDE")

Problem 3: m∠GBH = m∠GBC + m∠CBH = 105° + 104° = 209°

But 209° is unusual; perhaps it's the smaller angle, 360 - 209 = 151°. Let me see if 151 makes sense with other numbers.

Perhaps in the diagram, the angles are on the other side.

Another thought: In some configurations, if C is on the opposite side, then ∠GBH = |m∠GBC - m∠CBH| = |105 - 104| = 1°, but that seems unlikely.

Or perhaps it's 180 - (105 + 104 - 180) or something, complicated.

Let's calculate the supplement.

I recall that in some problems, if two angles are given at a point on a line, but here it's not specified.

Perhaps for Problem 3, the angle GBH is the angle not including C, so if G-B-H is a straight line, then ∠GBH = 180°, but then why give 105 and 104.

Unless 105 and 104 are not both at B for the same line.

I think I have to go with 209° for now, or perhaps it's 151°.

Let's look online or think differently.

Perhaps "m∠GBH" means the angle at B between G and H, and if C is inside, then it's the sum, but 209>180, so in geometry, the angle is usually taken as the smaller one, so min(209, 360-209) = min(209,151) = 151°.

And 151 is a nice number, while 209 is not.

In Problem 6, 145 - 61 = 84, which is less than 180.

In Problem 8, 170 - 159 = 11<180.

In Problem 10, 113 - 81 = 32<180.

In Problem 5, 59+26=85<180.

In Problem 7, 52+31=83<180.

In Problem 9, 20+110=130<180.

In Problem 2, 105+34=139<180.

For Problem 3, 105+104=209>180, so likely they intend the smaller angle, 360-209=151°.

Moreover, 151 is close to 180, and 105 and 104 are both obtuse, so their sum is reflex, so the actual angle GBH is 151°.

I think that's reasonable.

So for Problem 3: m∠GBH = 360° - (105° + 104°) = 360° - 209° = 151°.

Yes.

Similarly, for other problems, if sum exceeds 180, we take the smaller one, but in this case, only Problem 3 has sum >180.

Problem 4: if m∠WVZ = m∠WVT + m∠TVZ = 35° + 62° = 97° <180, ok.

So let's finalize.

Problem 2: Assuming "m∠FED" is a typo and should be "m∠FDE", then 105° + 34° = 139°.

Problem 3: m∠GBH = 360° - (105° + 104°) = 360° - 209° = 151°.

Problem 4: Assuming "m∠WTZ" is "m∠WVT" or "m∠WTV", and m∠ZVT = 62°, so if V is vertex, and T is common, then m∠WVZ = m∠WVT + m∠TVZ. If m∠WVT = 35°, then 35° + 62° = 97°.

Problem 5: m∠JKP = m∠JKO + m∠OKP. Assuming "m∠LKP" is "m∠OKP", so 59° + 26° = 85°.

Problem 6: m∠NDC = m∠EDC - m∠EDN = 145° - 61° = 84° (since N is between E and C).

Problem 7: m∠JKE = m∠IKJ + m∠IKE = 52° + 31° = 83°.

Problem 8: m∠IHQ = m∠GHQ - m∠GHI = 170° - 159° = 11°.

Problem 9: Assuming "m∠LEF" is "m∠LEK", then m∠LEK = m∠LEM + m∠MEK = 20° + 110° = 130°.

Problem 10: m∠JDA = m∠JDC - m∠ADC = 113° - 81° = 32°.

Now, to write the answers.

But for Problem 2, if it's really ∠FED, we can't solve, so must be typo.

Similarly for others.

So final answers:

2) 139°
3) 151°
4) 97°
5) 85°
6) 84°
7) 83°
8) 11°
9) 130°
10) 32°

Let me double-check Problem 3: 105 + 104 = 209, 360 - 209 = 151, yes.

Problem 4: 35 + 62 = 97, yes.

All seem consistent.

So I'll go with that.

Final Answer:
2) 139°
3) 151°
4) 97°
5) 85°
6) 84°
7) 83°
8) 11°
9) 130°
10) 32°