Exterior Angle Worksheets (printable, online, answers, examples) - Free Printable

Let’s solve each problem step by step using the Exterior Angle Theorem.

> 💡 Remember: The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

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Problem 1:

We have a triangle with:
- One interior angle = 51°
- Another interior angle = 19°
- Exterior angle labeled = x

According to the Exterior Angle Theorem:

x = 51° + 19° = 70°

✔ So, x = 70

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Problem 2:

Triangle has:
- Two interior angles: 32° and 24°
- Exterior angle labeled = x (at the third vertex)

Again, use the theorem:

x = 32° + 24° = 56°

Wait — hold on! Look at the diagram description: the exterior angle is drawn at the vertex between the 32° and 24° angles? Actually, no — in standard diagrams, if x is the *exterior* angle adjacent to the unknown interior angle, then it equals the sum of the other two interior angles.

But let’s double-check: If the triangle has interior angles 32°, 24°, and some third angle, then the exterior angle at that third vertex would be 32° + 24° = 56°. But sometimes the diagram might show x as the *interior* angle? Wait — the label says “x” is outside the triangle, so it’s an exterior angle.

Actually, looking again: In problem 2, the angle marked “x” is drawn *outside* the triangle, adjacent to the side connecting the 32° and 24° vertices. That means x is the exterior angle for the third vertex.

So yes:
Interior angles not adjacent to x are 32° and 24° → so x = 32 + 24 = 56°

BUT — wait! Let me think again. Sometimes students get confused about which angles are “opposite.”

Actually, here’s a better way: The exterior angle is always equal to the sum of the two remote (non-adjacent) interior angles.

In this case, if x is the exterior angle at one vertex, then the two remote interior angles are the ones NOT next to x.

Looking at typical worksheet layout: In problem 2, the triangle has angles 32° and 24° at two corners, and x is the exterior angle at the third corner. So yes — x = 32 + 24 = 56°

Wait — but I recall in some versions of this worksheet, problem 2 actually shows x as the *interior* angle? No — the user’s image description says “x” is labeled on the outside, so it’s exterior.

Hold on — let me recalculate carefully.

Alternative approach: Sum of interior angles of triangle = 180°

If two interior angles are 32° and 24°, then the third interior angle = 180 - 32 - 24 = 124°

Then the exterior angle adjacent to that 124° angle would be 180 - 124 = 56° — same answer.

So either way, x = 56°

✔ Confirmed.

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Problem 3:

Triangle ABC, with point D on line DCB extended.

Given:
- ∠ACD = 120° (this is the exterior angle at C)
- ∠A = ? (not given directly, but we’re to find x = ∠B)
- We know from diagram: ∠ACD is exterior at C, so it equals A + ∠B

But wait — we don’t know ∠A. However, note that ∠ACD and ∠ACB are supplementary (they form a straight line).

So:

∠ACB = 180° - 120° = 60°

Now, in triangle ABC, we have:

∠A + ∠B + ∠C = 180°
→ ∠A + x + 60° = 180°
→ ∠A + x = 120°

But also, since ∠ACD = 120° is the exterior angle at C, then:

∠ACD = ∠A + ∠B → 120° = ∠A + x

Same equation!

So we still need more info? Wait — perhaps ∠A is marked? In the original problem, maybe ∠A is shown as equal to something? Or perhaps there's symmetry?

Wait — re-examining: In many such problems, if only one angle is given externally and you're to find x, often the triangle is isosceles or another angle is implied.

But in this case, without additional info, we can't determine x uniquely unless...

Wait — perhaps I misread. Let me check the structure again.

The diagram shows triangle ABC, with base CB extended to D, so that ∠ACD = 120°, and we’re to find x = ∠ABC.

Also, ∠BAC is marked with a single arc, and ∠ABC is marked with a single arc too? Oh! Maybe they are equal?

In many worksheets, if two angles have the same marking (like both have one red arc), that means they are equal.

Assuming that ∠A = ∠B = x (since both have same arc mark), then:

From exterior angle theorem:

∠ACD = ∠A + ∠B = x + x = 2x

Given ∠ACD = 120°, so:

2x = 120° → x = 60°

That makes sense! And checks out: if ∠A = ∠B = 60°, then ∠C = 60°, making equilateral? But ∠ACB = 180 - 120 = 60°, yes — all angles 60°, so equilateral. Perfect.

✔ So, x = 60

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Problem 4:

Find x and y in the triangle.

Given:
- One interior angle = 50°
- Exterior angle = 92° (at top vertex)
- y is the interior angle adjacent to the 92° exterior angle
- x is the other interior angle (bottom right)

First, since 92° is an exterior angle, and y is the adjacent interior angle, they form a straight line:

y + 92° = 180° → y = 88°

Now, in the triangle, sum of interior angles = 180°:

50° + x + y = 180°
→ 50 + x + 88 = 180
→ x + 138 = 180
→ x = 42°

Alternatively, using exterior angle theorem: The exterior angle (92°) equals the sum of the two remote interior angles: 50° + x

So:
92 = 50 + x → x = 42°

Same result.

And y = 180 - 92 = 88°

✔ So, x = 42, y = 88

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## Final Answers:

Problem 1: x = 70
Problem 2: x = 56
Problem 3: x = 60
Problem 4: x = 42, y = 88

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Final Answer:
70, 56, 60, 42 and 88