Exterior Angle Theorem Notes & Worksheets - Lindsay Bowden - Free Printable

Let’s solve each problem using the Exterior Angle Theorem, which states:

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

Also, remember that the sum of the interior angles of any triangle is always 180°.

---

Problem 1

We have an isosceles triangle (marked with tick marks on two sides), so the base angles are equal. One base angle is 62°, so the other base angle is also 62°.

The exterior angle is labeled `x°`. It is adjacent to the third interior angle (the vertex angle).

First, find the vertex angle:

> Vertex angle = 180° - 62° - 62° = 56°

Then, since the exterior angle `x°` is adjacent to the vertex angle, they form a linear pair:

> x° + 56° = 180° → x = 124

✔ Answer: x = 124

*(Missing angles: both base angles are 62°, vertex angle is 56°)*

---

Problem 2

We have an exterior angle of 140°. The two remote interior angles are 80° and `(2x)°`.

By Exterior Angle Theorem:

> 80 + 2x = 140
> 2x = 60
> x = 30

✔ Answer: x = 30

*(Missing angle: 2x = 60°; third interior angle = 180 - 80 - 60 = 40°)*

---

Problem 3

This is a right triangle (marked with square). One acute angle is 59°, another is `5x°`, and the right angle is 90°.

Sum of interior angles:

> 5x + 59 + 90 = 180
> 5x + 149 = 180
> 5x = 31
> x = 6.2

✔ Answer: x = 6.2

*(Missing angle: 5x = 31°)*

---

Problem 4

Exterior angle = 144°. Remote interior angles are 36° and `(x + 32)°`.

Apply Exterior Angle Theorem:

> 36 + (x + 32) = 144
> x + 68 = 144
> x = 76

Now check the other exterior angle given: `(2x + 1)°`

Plug in x = 76: 2(76) + 1 = 152 + 1 = 153°

Check: The interior angle adjacent to 153° would be 180 - 153 = 27°.

Sum of triangle angles: 36° + (x+32)=76+32=108° + 27° = 171° — wait, that doesn’t add to 180!

Wait — let’s re-analyze.

Actually, we have two exterior angles shown: one is 144°, the other is `(2x + 1)°`. But they are at different vertices.

Let’s use the 144° exterior angle first.

Remote interior angles: 36° and `(x + 32)°`

So:

> 36 + (x + 32) = 144 → x + 68 = 144 → x = 76

Now, let’s verify with the other exterior angle `(2x + 1)° = 2(76)+1 = 153°`

That exterior angle should equal the sum of the two remote interior angles: 36° and the angle opposite it.

What’s the third interior angle? Let’s compute all interior angles.

Interior angles:

- One is 36°
- Another is x + 32 = 76 + 32 = 108°
- Third angle = 180 - 36 - 108 = 36°

Wait — now we have two 36° angles? That’s fine.

Now, the exterior angle at the vertex with 36° (the one not adjacent to 144°) should be 180 - 36 = 144° — but we already used that.

Actually, the exterior angle `(2x + 1)°` is at the vertex where interior angle is 108°? No — exterior angle is adjacent to its interior angle.

If the interior angle is 108°, then the exterior angle should be 180 - 108 = 72°, but we got 153° — inconsistency?

Wait — perhaps I misassigned.

Let me draw mentally:

Triangle has angles:

- A: 36°
- B: x + 32 = 108°
- C: 180 - 36 - 108 = 36°

Exterior angle at A: 180 - 36 = 144° — matches given.

Exterior angle at B: 180 - 108 = 72° — but problem says `(2x + 1)° = 153°` — that doesn’t match.

There’s a mistake.

Wait — maybe the exterior angle `(2x + 1)°` is NOT at vertex B. Maybe it’s at vertex C?

Vertex C has interior angle 36°, so exterior angle = 144° — again, same as before.

But we have two exterior angles labeled: 144° and `(2x+1)°`.

Ah! Perhaps the 144° is at one vertex, and `(2x+1)°` is at another — meaning we can set up two equations?

But that would be over-constrained.

Alternative approach: Use the fact that the exterior angle equals sum of remote interiors.

We used 144° = 36 + (x+32) → x=76.

Now plug into `(2x + 1)` = 153°.

Is 153° an exterior angle? Then its remote interior angles should be 36° and 108°? 36 + 108 = 144 ≠ 153 — contradiction.

Wait — perhaps the diagram shows the exterior angle `(2x+1)°` as being adjacent to the angle `(x+32)°`? Let’s reinterpret.

Actually, looking back: In Problem 4, the exterior angle labeled `(2x+1)°` is at the bottom-right vertex. The interior angle there is unknown. The other two interior angles are 36° and `(x+32)°`.

So the exterior angle `(2x+1)°` should equal the sum of the two remote interior angles: 36° and `(x+32)°`.

So:

> 2x + 1 = 36 + (x + 32)
> 2x + 1 = x + 68
> 2x - x = 68 - 1
> x = 67

Now check:

x = 67

Then:

- Interior angle: x + 32 = 67 + 32 = 99°
- Other interior angle: 36°
- Third interior angle = 180 - 99 - 36 = 45°

Exterior angle `(2x + 1)` = 2(67) + 1 = 134 + 1 = 135°

Check: Is 135° equal to sum of remote interiors? 36° + 99° = 135° — YES!

Also, the other exterior angle shown is 144° — that must be adjacent to the 36° angle? 180 - 36 = 144 — YES!

Perfect.

✔ Answer: x = 67

*(Missing angles: interior angles are 36°, 99°, 45°; exterior angles are 144° and 135°)*

---

Problem 5

Exterior angle = `(3x + 5)°`

Remote interior angles: 58° and 32°

Apply theorem:

> 3x + 5 = 58 + 32
> 3x + 5 = 90
> 3x = 85
> x = 85/3 ≈ 28.333...

But let’s keep as fraction: x = 85/3

✔ Answer: x = 85/3

*(Missing angle: exterior angle = 3(85/3) + 5 = 85 + 5 = 90°; third interior angle = 180 - 58 - 32 = 90° — makes sense, right triangle!)*

---

Problem 6

Exterior angle = `(12x + 6)°`

Remote interior angles: `(5x - 2)°` and `64°`

Apply theorem:

> 12x + 6 = (5x - 2) + 64
> 12x + 6 = 5x + 62
> 12x - 5x = 62 - 6
> 7x = 56
> x = 8

✔ Answer: x = 8

*(Missing angles: 5x - 2 = 40 - 2 = 38°; 64°; third interior angle = 180 - 38 - 64 = 78°; exterior angle = 12(8)+6 = 96+6=102° — check: 38+64=102 — correct!)*

---

Problem 7

Triangle ABC: ∠ABC = 100°, ∠BCA = 26°

Segment extends from triangle creating ACD — so point D is on extension of AC, making ∠BCD the exterior angle at C.

We need to find m∠BCD.

By Exterior Angle Theorem, ∠BCD = sum of remote interior angles: ∠ABC + ∠BAC

We don’t have ∠BAC yet.

First, find ∠BAC:

> ∠BAC = 180 - ∠ABC - ∠BCA = 180 - 100 - 26 = 54°

Now, ∠BCD (exterior at C) = ∠ABC + ∠BAC = 100 + 54 = 154°

✔ Answer: ∠BCD = 154°

*(Note: no variable x here — just find the angle.)*

---

Problem 8

Triangle DEF: m∠DEF = (2x + 3)°, m∠EFD = 51°

Segment extends from triangle creating DG — so G is on extension of EF, making ∠GDF the exterior angle at D.

Given: m∠GDF = 100°

We need to solve for x.

By Exterior Angle Theorem:

> ∠GDF = sum of remote interior angles = ∠DEF + ∠EFD

So:

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

✔ Answer: x = 23

*(Check: ∠DEF = 2(23)+3 = 46+3 = 49°; 49 + 51 = 100 — correct!)*

---

## ✔ Final Answers:

1. x = 124
2. x = 30
3. x = 6.2
4. x = 67
5. x = 85/3 (or approx 28.33)
6. x = 8
7. ∠BCD = 154°
8. x = 23

All solved using the Exterior Angle Theorem and/or triangle angle sum.