Solved Unit: Angle RelationshipsHomework 4NameDate -EXTERIOR ... - Free Printable

Let's solve each problem step by step using exterior angle theorems and triangle angle properties.

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🔷 Key Concept:

- The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.
- The sum of angles in a triangle is always 180°.
- A straight line = 180°, so adjacent angles on a straight line are supplementary.

We’ll solve each question, then match answers to letters to decode the riddle:

> "WHAT DO YOU CALL A CRUSHED ANGLE?"

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## ✔ Problem 1: Find $ m\angle KMU $

Given:
- Triangle $ \triangle KLM $
- $ \angle K = 50^\circ $
- $ \angle L = 90^\circ $ (right angle)
- $ MU $ is an extension of $ LM $, so $ \angle KMU $ is an exterior angle at vertex $ M $

So,
$$
\angle K + \angle L = 50^\circ + 90^\circ = 140^\circ
$$
This equals the exterior angle at $ M $, so:
$$
m\angle KMU = 140^\circ
$$

✔ Answer: 140 → G

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## ✔ Problem 2: Find $ m\angle BDC $

Given:
- Triangle $ \triangle BCD $
- $ \angle C = 60^\circ $
- Exterior angle at $ B $ is $ 120^\circ $

Since $ \angle EBD = 120^\circ $, this is an exterior angle at $ B $. So it equals the sum of the two non-adjacent interior angles:
$$
\angle BDC + \angle C = 120^\circ
$$
$$
\angle BDC + 60^\circ = 120^\circ \Rightarrow \angle BDC = 60^\circ
$$

Wait — that can't be right because we already have $ \angle C = 60^\circ $, and if $ \angle BDC = 60^\circ $, then the third angle would be $ 60^\circ $, making it equilateral. But let's check.

Actually, the exterior angle at B is $ 120^\circ $, so the interior angle at B is:
$$
180^\circ - 120^\circ = 60^\circ
$$

Now, in triangle $ \triangle BCD $:
- $ \angle B = 60^\circ $
- $ \angle C = 60^\circ $
- So $ \angle D = 180^\circ - 60^\circ - 60^\circ = 60^\circ $

So $ m\angle BDC = 60^\circ $

✔ Answer: 60 → T

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## ✔ Problem 3: Find $ m\angle NPI $

Triangle $ \triangle NOP $
- $ \angle N = 65^\circ $
- $ \angle O = 78^\circ $
- $ \angle P $ is the vertex where the exterior angle $ \angle NPI $ is formed.

First, find $ \angle P $:
$$
\angle P = 180^\circ - 65^\circ - 78^\circ = 37^\circ
$$

Then, $ \angle NPI $ is the exterior angle at $ P $, so:
$$
\angle NPI = 180^\circ - \angle P = 180^\circ - 37^\circ = 143^\circ
$$

Alternatively, use exterior angle theorem:
$$
\angle NPI = \angle N + \angle O = 65^\circ + 78^\circ = 143^\circ
$$

✔ Answer: 143 → L

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## ✔ Problem 4: Find the value of $ x $

Triangle $ \triangle ABC $
- $ \angle A = (2x + 8)^\circ $
- $ \angle B = (3x + 5)^\circ $
- Exterior angle at $ C $ is $ 123^\circ $

By exterior angle theorem:
$$
\text{Exterior angle} = \angle A + \angle B
$$
$$
(2x + 8) + (3x + 5) = 123
$$
$$
5x + 13 = 123
$$
$$
5x = 110 \Rightarrow x = 22
$$

✔ Answer: 22 → C

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## ✔ Problem 5: Find the value of $ x $

Triangle $ \triangle QRO $
- $ \angle Q = 25^\circ $
- $ \angle R = (2x + 3)^\circ $
- Exterior angle at $ O $ is $ (3x - 17)^\circ $

Exterior angle at $ O $ = sum of opposite interior angles:
$$
\angle Q + \angle R = \text{Exterior at } O
$$
$$
25 + (2x + 3) = 3x - 17
$$
$$
2x + 28 = 3x - 17
$$
$$
28 + 17 = 3x - 2x \Rightarrow x = 45
$$

✔ Answer: 45 → E

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## ✔ Problem 6: Find the value of $ x $

Triangle $ \triangle HIJ $
- $ \angle I = (5x + 2)^\circ $
- $ \angle H = (6x + 4)^\circ $
- Exterior angle at $ J $ is $ (12x - 4)^\circ $

Exterior angle at $ J $ = sum of $ \angle I + \angle H $
$$
(5x + 2) + (6x + 4) = 12x - 4
$$
$$
11x + 6 = 12x - 4
$$
$$
6 + 4 = 12x - 11x \Rightarrow x = 10
$$

✔ Answer: 10 → N

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## ✔ Problem 7: Find $ m\angle CVU $

Triangle $ \triangle UVT $
- $ \angle U = 63^\circ $
- $ \angle T = (x + 13)^\circ $
- $ \angle V $ is split: one part is $ (3x)^\circ $, and the other is the exterior angle $ \angle CVU $?

Wait — look at diagram:
- $ C-V-T $ is a straight line
- $ \angle CVU $ is the exterior angle at $ V $
- So $ \angle CVU = \angle U + \angle T $

But we don’t know $ x $. We need to use triangle angle sum.

In triangle $ \triangle UVT $:
- $ \angle U = 63^\circ $
- $ \angle T = (x + 13)^\circ $
- $ \angle V = (3x)^\circ $

Sum of angles:
$$
63 + (x + 13) + 3x = 180
$$
$$
63 + x + 13 + 3x = 180
$$
$$
4x + 76 = 180
$$
$$
4x = 104 \Rightarrow x = 26
$$

Now, $ \angle CVU $ is the exterior angle at V, so:
$$
\angle CVU = \angle U + \angle T = 63 + (26 + 13) = 63 + 39 = 102^\circ
$$

Wait — but 102 is not in the answer key.

Wait! Let’s double-check.

Wait — the exterior angle $ \angle CVU $ is adjacent to $ \angle V $, which is $ 3x = 3(26) = 78^\circ $

So:
$$
\angle CVU = 180^\circ - 78^\circ = 102^\circ
$$

Still 102 — not in options.

Wait — maybe I made a mistake.

Wait — the exterior angle at $ V $ is $ \angle CVU $, and from the triangle:
$$
\angle CVU = \angle U + \angle T = 63 + (x+13) = 63 + 26 + 13 = 102^\circ
$$

But 102 is not in the table.

Wait — perhaps I misread.

Wait — the table has:
- G: 140
- A: 114
- E: 45
- O: 52
- E: 62
- T: 60
- C: 22
- D: 25
- M: 112
- L: 143
- R: 135
- N: 10

No 102. So maybe I messed up.

Wait — maybe $ \angle V $ is not $ 3x $? Let's look again.

From the diagram:
- At point $ V $, there's a line $ C-V-T $, with $ \angle CVU = (3x)^\circ $, and $ \angle UV T = (x + 13)^\circ $

Wait — no! It says:
- $ \angle CVU = (3x)^\circ $
- $ \angle UVT = (x + 13)^\circ $
- And $ \angle U = 63^\circ $

But $ \angle CVU $ and $ \angle UVT $ are adjacent angles forming a straight line? No — they are both at $ V $, but $ \angle CVU $ is the exterior angle, and $ \angle UVT $ is the interior angle at $ V $

So:
- $ \angle CVU = (3x)^\circ $
- $ \angle UVT = (x + 13)^\circ $
- They form a linear pair → sum to 180°

So:
$$
3x + (x + 13) = 180
\Rightarrow 4x + 13 = 180
\Rightarrow 4x = 167 \Rightarrow x = 41.75
$$

That doesn't make sense — probably not.

Wait — maybe $ \angle CVU $ is the exterior angle, so:
$$
\angle CVU = \angle U + \angle T
$$
But we don’t know $ \angle T $

Wait — label:
- $ \angle T = (x + 13)^\circ $
- $ \angle U = 63^\circ $
- $ \angle V = ? $

But the exterior angle at $ V $ is $ \angle CVU = (3x)^\circ $

So:
$$
\angle CVU = \angle U + \angle T = 63 + (x + 13) = x + 76
$$

But also $ \angle CVU = 3x $

So:
$$
3x = x + 76 \Rightarrow 2x = 76 \Rightarrow x = 38
$$

Now, $ \angle CVU = 3x = 3(38) = 114^\circ $

✔ Answer: 114 → A

Perfect!

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## ✔ Problem 8: Find $ m\angle YXZ $

Triangle $ \triangle XYZ $
- $ \angle X = (3x + 5)^\circ $
- $ \angle Y = (4x)^\circ $
- Exterior angle at $ Z $ is $ (8x - 14)^\circ $

Exterior angle at $ Z $ = sum of $ \angle X + \angle Y $
$$
(3x + 5) + (4x) = 8x - 14
\Rightarrow 7x + 5 = 8x - 14
\Rightarrow 5 + 14 = 8x - 7x \Rightarrow x = 19
$$

Now, $ \angle YXZ $ is $ \angle X = 3x + 5 = 3(19) + 5 = 57 + 5 = 62^\circ $

✔ Answer: 62 → E

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## ✔ Problem 9: Find $ m\angle MNI $

Triangle $ \triangle LMN $
- $ \angle M = (15x - 12)^\circ $
- $ \angle L = (6x)^\circ $
- Exterior angle at $ N $ is $ (19x + 2)^\circ $

Exterior angle at $ N $ = sum of $ \angle M + \angle L $
$$
(15x - 12) + (6x) = 19x + 2
\Rightarrow 21x - 12 = 19x + 2
\Rightarrow 21x - 19x = 2 + 12
\Rightarrow 2x = 14 \Rightarrow x = 7
$$

Now, $ \angle MNI $ is the exterior angle at $ N $, so:
$$
\angle MNI = 19x + 2 = 19(7) + 2 = 133 + 2 = 135^\circ
$$

✔ Answer: 135 → R

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## 📊 Summary of Answers:

| Question | Answer | Letter |
|--------|--------|--------|
| 1 | 140 | G |
| 2 | 60 | T |
| 3 | 143 | L |
| 4 | 22 | C |
| 5 | 45 | E |
| 6 | 10 | N |
| 7 | 114 | A |
| 8 | 62 | E |
| 9 | 135 | R |

Now plug into the riddle:

> WHAT DO YOU CALL A CRUSHED ANGLE?

Order given:
```
______ ______ ______ ______ ______ ______ ______ ______ ______
7 9 8 4 2 7 6 1 3 5
```

So:
- 7 → A (from #7: 114 → A)
- 9 → R (from #9: 135 → R)
- 8 → E (from #8: 62 → E)
- 4 → C (from #4: 22 → C)
- 2 → T (from #2: 60 → T)
- 7 → A (again)
- 6 → N (from #6: 10 → N)
- 1 → G (from #1: 140 → G)
- 3 → L (from #3: 143 → L)
- 5 → E (from #5: 45 → E)

So the sequence is:
- 7: A
- 9: R
- 8: E
- 4: C
- 2: T
- 7: A
- 6: N
- 1: G
- 3: L
- 5: E

So the word is:
A R E C T A N G L E

Wait — that’s “RECTANGLE”?

But the order is:
A R E C T A N G L E

Wait — that’s A RECTANGLE?

But "A RECTANGLE" has 10 letters.

But our letters are:
1. A
2. R
3. E
4. C
5. T
6. A
7. N
8. G
9. L
10. E

So: A R E C T A N G L E → A RECTANGLE

But the riddle is:
"What do you call a crushed angle?"

Answer: A rectangle 😄

Because a crushed angle sounds like a rectangle — a pun!

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✔ Final Answer:

A RECTANGLE

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🔤 Decoded Riddle:

> What do you call a crushed angle?
> A RECTANGLE

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✔ All Solutions:

| # | Answer | Letter |
|---|--------|--------|
| 1 | 140 | G |
| 2 | 60 | T |
| 3 | 143 | L |
| 4 | 22 | C |
| 5 | 45 | E |
| 6 | 10 | N |
| 7 | 114 | A |
| 8 | 62 | E |
| 9 | 135 | R |

Sequence:
7(A), 9(R), 8(E), 4(C), 2(T), 7(A), 6(N), 1(G), 3(L), 5(E) → A RECTANGLE

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✔ Final Answer: A RECTANGLE 🟩