Triangle Sum Theorem and Exterior Angle Theorem Worksheet featuring problems to calculate unknown angles in triangles.

Let's solve each problem step by step.

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Problem 1:

We have triangle ABC with angles:
- ∠ABC = 98°
- ∠ACB = 24°
- ∠BAC = x°

We know that the sum of angles in a triangle is 180°.

So:

x + 98 + 24 = 180
x + 122 = 180
x = 180 - 122 = 58

✔ So, ∠BAC = 58°

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Problem 2:

Triangle AED has:
- ∠EAD = 29°
- ∠ADE = 44°
- ∠AED = x° (interior angle)

Sum of angles in triangle = 180°

x + 29 + 44 = 180
x + 73 = 180
x = 107° → This is ∠AED

But we are asked for ∠AEX, which is the exterior angle at E.

Exterior angle = 180° - interior angle
→ ∠AEX = 180 - 107 = 73°

Alternatively, using Exterior Angle Theorem:
Exterior angle = sum of two opposite interior angles
→ ∠AEX = 29° + 44° = 73°

✔ So, ∠AEX = 73°

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Problem 3:

Triangle DBC has:
- ∠DCB = 60°
- ∠DBC = 2x°
- Exterior angle at D = 84° (angle EDB)

By Exterior Angle Theorem:
Exterior angle = sum of two opposite interior angles
→ 84° = 60° + 2x°
→ 2x = 84 - 60 = 24
→ x = 12

Then ∠DBC = 2x = 24°

✔ So, ∠DBC = 24°

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Problem 4:

Triangle KLN has:
- ∠KLN = (2x - 3)°
- ∠KNL = 51°
- Exterior angle at K = 100° (angle MKL)

Exterior angle = sum of two opposite interior angles
→ 100 = (2x - 3) + 51
→ 100 = 2x + 48
→ 2x = 52
→ x = 26

Now find ∠KLN = 2x - 3 = 2(26) - 3 = 52 - 3 = 49°

✔ So, ∠KLN = 49°

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Problem 5:

Triangle ABD has:
- ∠DAB = (5x - 7)°
- ∠DBA = (100 - x)°
- ∠ADB = 2x°

Sum of angles = 180°

So:

(5x - 7) + (100 - x) + 2x = 180
Combine like terms:
5x - x + 2x = 6x
-7 + 100 = 93
→ 6x + 93 = 180
→ 6x = 87
→ x = 14.5

Now find ∠DBA = 100 - x = 100 - 14.5 = 85.5°

Also, ∠CAD is the exterior angle at A.

∠CAD = 180° - ∠DAB = 180 - (5x - 7)
First compute 5x - 7 = 5(14.5) - 7 = 72.5 - 7 = 65.5
→ ∠CAD = 180 - 65.5 = 114.5°

Alternatively, using Exterior Angle Theorem:
∠CAD = ∠DBA + ∠ADB = 85.5 + 29 = 114.5° ✔

✔ So, ∠DBA = 85.5°, ∠CAD = 114.5°

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Problem 6:

Triangle XYZ has:
- ∠YXZ = 60°
- ∠XWZ? Wait — actually, it’s triangle XYW? Let me re-read.

Actually, triangle X Y W? No — points are X, Y, W. But angle at W is 60°, angle at X is 60°, so triangle XYW?

Wait — label: Triangle is XYZ? No — vertices are X, Y, W? Actually, looking again:

Points: X, Y, W — and Z is on extension of YW.

Angle at X = 60°, angle at W = 60°, so triangle XYW? Actually, triangle is XYW? Or XYW?

Actually, triangle is XYW? Let me assume triangle is XYW with:

- ∠YXW = 60°
- ∠XWY = 60°
- Then ∠XYW = ?

Sum = 180 → ∠XYW = 180 - 60 - 60 = 60° → equilateral?

But then exterior angle at Y is ∠XYZ = x°

Exterior angle = 180 - interior angle = 180 - 60 = 120°

Or by Exterior Angle Theorem: exterior angle = sum of two opposite interior angles = 60 + 60 = 120°

✔ So, ∠XYZ = 120°

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Problem 7:

Triangle ABC has AB = AC (marked with double tick marks), so it’s isosceles.

∠BAC = 66°

Since AB = AC, base angles ∠ABC = ∠ACB

Sum of angles = 180 → ∠ABC + ∠ACB = 180 - 66 = 114°
Each = 57°

Now, point D is on extension of BC, so CD is extended, and AD is drawn.

We are to find ∠ADC = x°

Note: ∠ACB = 57°, so adjacent angle ∠ACD = 180 - 57 = 123° (linear pair)

In triangle ACD, we have:

- ∠ACD = 123°
- ∠CAD = ? Not given directly.

Wait — look at diagram: there’s a mark on AC and CD — both have double tick marks? Wait, no:

Actually, in the diagram:

- AB and AC have double ticks → AB = AC
- Also, AC and CD have double ticks? Wait — looking again:

Actually, in problem 7:

It shows triangle ABC with AB = AC (double ticks on AB and AC). Then from C, they extend to D, and mark CD with double ticks — meaning CD = AC? Or CD = something else?

Actually, looking carefully: the double tick on AC and on CD — so AC = CD.

So triangle ACD is isosceles with AC = CD.

Therefore, ∠CAD = ∠CDA = x°

And ∠ACD = 123° (as above)

Sum of angles in triangle ACD:

x + x + 123 = 180
2x = 57
x = 28.5

✔ So, ∠ADC = 28.5°

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Problem 8:

Triangle ACE is right-angled at C (∠ACE = 90°)

Exterior angle at A is 122° (angle between EA extended and AD)

So, interior angle ∠CAE = 180 - 122 = 58°

In triangle ACE:

∠ACE = 90°
∠CAE = 58°
So ∠AEC = 180 - 90 - 58 = 32°

Or by Exterior Angle Theorem:
Exterior angle at A = ∠ACE + ∠AEC
122 = 90 + x
x = 32°

✔ So, ∠AEC = 32°

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Final Answers:

1. ∠BAC = 58°
2. ∠AEX = 73°
3. ∠DBC = 24°
4. ∠KLN = 49°
5. ∠DBA = 85.5°, ∠CAD = 114.5°
6. ∠XYZ = 120°
7. ∠ADC = 28.5°
8. ∠AEC = 32°

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Final Answer:
∠BAC = 58°, ∠AEX = 73°, ∠DBC = 24°, ∠KLN = 49°, ∠DBA = 85.5°, ∠CAD = 114.5°, ∠XYZ = 120°, ∠ADC = 28.5°, ∠AEC = 32°