Triangle exterior angle worksheet for geometry practice.

Let’s solve each triangle problem one by one. We’ll use the Exterior Angle Theorem, which says:

> The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

Also, remember:
→ Angles on a straight line add up to 180°.
→ Interior angles in any triangle add up to 180°.

We’ll go step by step for each numbered problem.

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Problem 1)
Triangle ABC with exterior angle at C (angle DCX).
Given: ∠A = 64°, ∠B = 57°
Find: m∠DCX

Using Exterior Angle Theorem:
m∠DCX = ∠A + B = 64 + 57 = 121°

✔ Check: Interior angle at C = 180 - (64+57) = 59° → then exterior = 180 - 59 = 121° → same answer.

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Problem 2)
Triangle XYZ with exterior angle at Y (angle UYX).
Given: ∠Z = 30°, ∠X = 90°
Find: m∠UYX

Exterior angle at Y = ∠Z + X = 30 + 90 = 120°

✔ Check: Interior angle at Y = 180 - (30+90) = 60° → exterior = 180 - 60 = 120° → correct.

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Problem 3)
Triangle PQR with exterior angle at Q (angle SQT).
Given: ∠P = 38°, ∠R = 29°
Find: m∠SQT

Exterior angle at Q = ∠P + R = 38 + 29 = 67°

✔ Check: Interior angle at Q = 180 - (38+29) = 113° → exterior = 180 - 113 = 67° → correct.

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Problem 4)
Triangle KLM with exterior angle at L (angle MLN? Wait — diagram shows point N outside, so likely angle KLN or similar. But label says “m∠KLN” — let’s assume it's the exterior angle at L.)

Wait — looking again: Triangle KLM, points K, L, M. Point N is on extension of ML beyond L? So angle KLN is exterior at L.

Given: ∠K = 47°, ∠M = 70°
Find: m∠KLN

Exterior angle at L = ∠K + M = 47 + 70 = 117°

✔ Check: Interior angle at L = 180 - (47+70) = 63° → exterior = 180 - 63 = 117° → correct.

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Problem 5)
Triangle BCX? Wait — diagram shows triangle BCD? Actually, labels are: Points B, C, D, and X is outside? Label says “m∠BCX”

Looking carefully: Triangle BCD? Or triangle BXC? Let’s read given: ∠CBD = 38°, ∠CDB = 50°, find m∠BCX.

Assuming triangle CBD, with exterior angle at C (point X is on extension of DC beyond C?).

So exterior angle at C = ∠CBD + ∠CDB = 38 + 50 = 88°

✔ Check: Interior angle at C = 180 - (38+50) = 92° → exterior = 180 - 92 = 88° → correct.

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Problem 6)
Triangle ACX? Diagram: Points A, C, X. Given: ∠CAX = 47°, ∠ACX = ? Wait — no, we’re to find m∠ACX? That can’t be right — that’s an interior angle.

Wait — re-read: Problem 6 says “m∠ACX = ___” but in diagram, point X is outside? Actually, looking at standard layout — probably triangle ABC with point X on extension of BC, so angle ACX is exterior at C.

But given angles: At A: 47°, at B: 30°? Wait — diagram shows: In triangle ABC, ∠A = 47°, ∠B = 30°, and we need m∠ACX — which is exterior at C.

Yes! So exterior angle at C = ∠A + ∠B = 47 + 30 = 77°

✔ Check: Interior angle at C = 180 - (47+30) = 103° → exterior = 180 - 103 = 77° → correct.

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Problem 7)
Triangle DEF? Labels: Points D, E, F, and X outside. Find m∠DFX.

Given: ∠D = 50°, ∠E = 40°? Wait — diagram shows: In triangle DEF, ∠D = 50°, ∠FED = 40°? Actually, label says “∠DEF = 40°”, and “∠EDF = 50°”.

Then exterior angle at F (angle DFX) = ∠D + ∠E = 50 + 40 = 90°

✔ Check: Interior angle at F = 180 - (50+40) = 90° → wait, that would make exterior 90° too? No — if interior is 90°, exterior is 90° only if it’s a straight line — yes, 180 - 90 = 90°. So actually, this triangle has angles 50, 40, 90 — so exterior at F is adjacent to 90°, so 90°? That doesn't match theorem.

Wait — mistake! If interior angle at F is 90°, then exterior angle should be 90°? But according to theorem, exterior = sum of other two = 50 + 40 = 90° — yes, consistent.

So m∠DFX = 90°

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Problem 8)
Triangle ABX? Diagram: Points A, B, X. Given: ∠BAX = 30°, ∠ABX = 60°, find m∠ABX? Wait — no, label says “m∠ABX = ___” — but that’s given as 60°? That can’t be.

Wait — re-examining: Probably triangle ABX with point X such that we’re finding exterior angle at B? But label says “m∠ABX” — which is interior.

Wait — perhaps typo in my reading. Looking back at original image description — problem 8 says “m∠ABX = ___” but in context, likely it’s exterior angle at B.

Actually, standard notation: If X is outside, and we have triangle AB something, and angle ABX is formed by extending side CB or something.

Wait — better approach: Assume triangle ABC, with point X on extension of AB beyond B, so angle CBX is exterior at B.

But label says “m∠ABX” — that might be the exterior angle.

Given: In triangle ABC, ∠A = 30°, ∠C = 60°, find exterior angle at B.

Then exterior angle at B = ∠A + C = 30 + 60 = 90°

And since ∠ABX is likely that exterior angle, answer is 90°

✔ Check: Interior angle at B = 180 - (30+60) = 90° → exterior = 90°? Again, same issue — if interior is 90°, exterior is 90° — mathematically correct because 180 - 90 = 90.

So yes, m∠ABX = 90°

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Problem 9)
Triangle STV? Labels: Points S, T, V, and X outside. Find m∠STX.

Given: ∠S = 40°, ∠V = 30°? Diagram shows: In triangle STV, ∠TSV = 40°, ∠TVS = 30°, find exterior angle at T (angle STX).

Exterior angle at T = ∠S + ∠V = 40 + 30 = 70°

✔ Check: Interior angle at T = 180 - (40+30) = 110° → exterior = 180 - 110 = 70° → correct.

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Now, compiling all answers:

1) 121°
2) 120°
3) 67°
4) 117°
5) 88°
6) 77°
7) 90°
8) 90°
9) 70°

Final Answer:
1) 121°
2) 120°
3) 67°
4) 117°
5) 88°
6) 77°
7) 90°
8) 90°
9) 70°