Interior and Exterior Angles of a Triangle Worksheet for geometry practice.

To solve the problem, we need to use the properties of the interior and exterior angles of a triangle. Here are the key concepts:

1. Sum of Interior Angles of a Triangle: The sum of the interior angles of a triangle is always \(180^\circ\).
2. Exterior Angle Theorem: The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.

Let's solve each part step by step.

---

Problem 1

Triangle \( \triangle DEC \):
- Given: \( \angle DCE = 21^\circ \), \( \angle CED = 31^\circ \)
- Find: \( \angle ECD \)

Using the sum of interior angles:
\[
\angle ECD + \angle DCE + \angle CED = 180^\circ
\]
\[
\angle ECD + 21^\circ + 31^\circ = 180^\circ
\]
\[
\angle ECD + 52^\circ = 180^\circ
\]
\[
\angle ECD = 180^\circ - 52^\circ = 128^\circ
\]

Answer: \( \angle ECD = 128^\circ \)

---

Problem 2

Triangle \( \triangle TQR \):
- Given: \( \angle TQR = 55^\circ \), \( \angle QRT = 39^\circ \)
- Find: \( \angle TQR \) (already given, so we find the third angle \( \angle QTR \))

Using the sum of interior angles:
\[
\angle TQR + \angle QRT + \angle QTR = 180^\circ
\]
\[
55^\circ + 39^\circ + \angle QTR = 180^\circ
\]
\[
94^\circ + \angle QTR = 180^\circ
\]
\[
\angle QTR = 180^\circ - 94^\circ = 86^\circ
\]

Answer: \( \angle QTR = 86^\circ \)

---

Problem 3

Triangle \( \triangle NOP \):
- Given: \( \angle NOP = 77^\circ \), \( \angle OPN = 77^\circ \)
- Find: \( \angle MNP \)

Using the sum of interior angles:
\[
\angle NOP + \angle OPN + \angle PNO = 180^\circ
\]
\[
77^\circ + 77^\circ + \angle PNO = 180^\circ
\]
\[
154^\circ + \angle PNO = 180^\circ
\]
\[
\angle PNO = 180^\circ - 154^\circ = 26^\circ
\]

Answer: \( \angle MNP = 26^\circ \)

---

Problem 4

Triangle \( \triangle MOQ \):
- Given: \( \angle MOQ = 87^\circ \), \( \angle OQM = 63^\circ \)
- Find: \( \angle MOQ \) (already given, so we find the third angle \( \angle OMQ \))

Using the sum of interior angles:
\[
\angle MOQ + \angle OQM + \angle OMQ = 180^\circ
\]
\[
87^\circ + 63^\circ + \angle OMQ = 180^\circ
\]
\[
150^\circ + \angle OMQ = 180^\circ
\]
\[
\angle OMQ = 180^\circ - 150^\circ = 30^\circ
\]

Answer: \( \angle OMQ = 30^\circ \)

---

Problem 5

Triangle \( \triangle NPM \):
- Given: \( \angle NPM = 60^\circ \), \( \angle PMN = 60^\circ \)
- Find: \( \angle NPM \) (already given, so we find the third angle \( \angle MNP \))

Using the sum of interior angles:
\[
\angle NPM + \angle PMN + \angle MNP = 180^\circ
\]
\[
60^\circ + 60^\circ + \angle MNP = 180^\circ
\]
\[
120^\circ + \angle MNP = 180^\circ
\]
\[
\angle MNP = 180^\circ - 120^\circ = 60^\circ
\]

Answer: \( \angle MNP = 60^\circ \)

---

Problem 6

Triangle \( \triangle HFW \):
- Given: \( \angle HFW = 39^\circ \), \( \angle FWH = 26^\circ \)
- Find: \( \angle HFX \) (exterior angle at \( F \))

Using the exterior angle theorem:
\[
\angle HFX = \angle HFW + \angle FWH
\]
\[
\angle HFX = 39^\circ + 26^\circ = 65^\circ
\]

Answer: \( \angle HFX = 65^\circ \)

---

Problem 7

Triangle \( \triangle PQR \):
- Given: \( \angle QPR = 51^\circ \), \( \angle PRQ = 46^\circ \)
- Find: \( \angle RPQ \)

Using the sum of interior angles:
\[
\angle QPR + \angle PRQ + \angle RPQ = 180^\circ
\]
\[
51^\circ + 46^\circ + \angle RPQ = 180^\circ
\]
\[
97^\circ + \angle RPQ = 180^\circ
\]
\[
\angle RPQ = 180^\circ - 97^\circ = 83^\circ
\]

Answer: \( \angle RPQ = 83^\circ \)

---

Problem 8

Triangle \( \triangle ABC \):
- Given: \( \angle BAC = 60^\circ \), \( \angle ACB = 47^\circ \)
- Find: \( \angle XAC \) (exterior angle at \( A \))

Using the exterior angle theorem:
\[
\angle XAC = \angle ABC + \angle ACB
\]

First, find \( \angle ABC \):
\[
\angle BAC + \angle ABC + \angle ACB = 180^\circ
\]
\[
60^\circ + \angle ABC + 47^\circ = 180^\circ
\]
\[
107^\circ + \angle ABC = 180^\circ
\]
\[
\angle ABC = 180^\circ - 107^\circ = 73^\circ
\]

Now, find \( \angle XAC \):
\[
\angle XAC = \angle ABC + \angle ACB
\]
\[
\angle XAC = 73^\circ + 47^\circ = 120^\circ
\]

Answer: \( \angle XAC = 120^\circ \)

---

Final Answers

\[
\boxed{
\begin{aligned}
1. & \ \angle ECD = 128^\circ \\
2. & \ \angle TQR = 86^\circ \\
3. & \ \angle MNP = 26^\circ \\
4. & \ \angle MOQ = 30^\circ \\
5. & \ \angle NPM = 60^\circ \\
6. & \ \angle HFX = 65^\circ \\
7. & \ \angle RPQ = 83^\circ \\
8. & \ \angle XAC = 120^\circ \\
\end{aligned}
}
\]