Worksheet featuring 12 geometry problems that require students to find the value of "x" using the triangle sum theorem and exterior angle theorem.

Let’s solve each problem one by one using the Triangle Sum Theorem (the three interior angles of a triangle add up to 180°) and the Exterior Angle Theorem (an exterior angle equals the sum of the two opposite interior angles).

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Problem 1:

Given: Two angles are 58° and 47°, find x.

→ 58 + 47 = 105
→ 180 - 105 = 75

✔ Already solved correctly in image: x = 75

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Problem 2:

Angles: 96°, 21°, and x°

→ 96 + 21 = 117
→ 180 - 117 = 63

✔ x = 63

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Problem 3:

Right triangle → one angle is 90°, another is 31°, third is (3x - 1)°

→ 90 + 31 = 121
→ 180 - 121 = 59 → so (3x - 1) = 59
→ 3x = 60
→ x = 20

✔ x = 20

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Problem 4:

This is an exterior angle situation. The exterior angle is x°, and the two remote interior angles are 21° and 34°.

→ Exterior angle = sum of two opposite interior angles
→ x = 21 + 34 = 55

✔ x = 55

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Problem 5:

Exterior angle is 100°, interior angles are (2x+3)° and 51°

→ 100 = (2x + 3) + 51
→ 100 = 2x + 54
→ 2x = 46
→ x = 23

✔ x = 23

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Problem 6:

Exterior angle is 94°, interior angles are 60° and 2x°

→ 94 = 60 + 2x
→ 2x = 34
→ x = 17

✔ x = 17

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Problem 7:

Interior angles: 52°, 43°, and x°

→ 52 + 43 = 95
→ 180 - 95 = 85

✔ x = 85

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Problem 8:

All three angles are expressed in terms of x: x°, 3x°, 2x°

→ x + 3x + 2x = 6x
→ 6x = 180
→ x = 30

✔ x = 30

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Problem 9:

Right triangle → one angle is 90°, others are (2x-2)° and (x+5)°

→ 90 + (2x - 2) + (x + 5) = 180
→ Combine like terms: 90 - 2 + 5 + 2x + x = 180
→ 93 + 3x = 180
→ 3x = 87
→ x = 29

✔ x = 29

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Problem 10:

This is a tricky one — it looks like a straight line with two small triangles or maybe just angles on a line? Wait — actually, this appears to be a triangle with two exterior angles given: 10° and 23°, and we’re to find the interior angle x°.

But wait — if you look at the diagram, it seems like the 10° and 23° are *exterior* angles adjacent to the base angles of the triangle, and x is the top interior angle.

Actually, let’s think differently: In such diagrams, sometimes the two bottom angles are supplementary to the given exterior angles.

So:

Left base angle = 180 - 10 = 170? That can’t be right — too big for a triangle.

Wait — perhaps the 10° and 23° are the *remote interior* angles for some exterior angle? But there’s no clear exterior angle marked.

Alternatively — maybe this is a triangle where the two base angles are 10° and 23°, and x is the vertex angle?

That would make sense! Let me check the drawing again mentally — yes, likely the 10° and 23° are the two base interior angles, and x is the top angle.

Then:

→ 10 + 23 = 33
→ 180 - 33 = 147

✔ x = 147

*(Note: If the 10° and 23° were exterior, the math wouldn't work for a triangle. So they must be interior base angles.)*

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Problem 11:

Equilateral-looking triangle — all three angles are x°

→ x + x + x = 3x = 180
→ x = 60

✔ x = 60

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Problem 12:

Three angles: (x+40)°, (2x-5)°, (3x-17)°

Add them up:

→ (x + 40) + (2x - 5) + (3x - 17) = 180
→ Combine: x + 2x + 3x = 6x; 40 - 5 - 17 = 18
→ 6x + 18 = 180
→ 6x = 162
→ x = 27

✔ x = 27

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Final Answer:
1) 75
2) 63
3) 20
4) 55
5) 23
6) 17
7) 85
8) 30
9) 29
10) 147
11) 60
12) 27