Free Angle Pair Relationships Worksheets For Teaching - Free Printable

Problem: Find the measure of each indicated angle.

#### Part 1:
We are given the following diagram and need to find the measures of the angles:


- ∠FGE is a right angle, so \( m\angle FGE = 90^\circ \).
- ∠DGE is marked as \( 40^\circ \).
- ∠AGE is marked as \( 45^\circ \).

We need to find:
1. \( m\angle DGE \)
2. \( m\angle AGF \)
3. \( m\angle CGE \)
4. \( m\angle CGB \)
5. \( m\angle AGB \)

##### Step-by-Step Solution:

1. \( m\angle DGE \):
- Given directly in the diagram.
- \( m\angle DGE = 40^\circ \).

2. \( m\angle AGF \):
- Since \( \angle FGE \) is a right angle (\( 90^\circ \)) and \( \angle AGE = 45^\circ \):
\[
m\angle AGF = m\angle FGE - m\angle AGE = 90^\circ - 45^\circ = 45^\circ
\]

3. \( m\angle CGE \):
- \( \angle CGE \) is a straight angle minus \( \angle DGE \):
\[
m\angle CGE = 180^\circ - m\angle DGE = 180^\circ - 40^\circ = 140^\circ
\]

4. \( m\angle CGB \):
- \( \angle CGB \) is part of the straight line at point \( G \). Since \( \angle DGE = 40^\circ \) and \( \angle AGE = 45^\circ \):
\[
m\angle CGB = 180^\circ - (m\angle DGE + m\angle AGE) = 180^\circ - (40^\circ + 45^\circ) = 180^\circ - 85^\circ = 95^\circ
\]

5. \( m\angle AGB \):
- \( \angle AGB \) is a straight angle:
\[
m\angle AGB = 180^\circ
\]

##### Final Answers for Part 1:
\[
\boxed{
\begin{aligned}
i) & \quad m\angle DGE = 40^\circ \\
ii) & \quad m\angle AGF = 45^\circ \\
iii) & \quad m\angle CGE = 140^\circ \\
iv) & \quad m\angle CGB = 95^\circ \\
v) & \quad m\angle AGB = 180^\circ
\end{aligned}
}
\]

---

#### Part 2:
We are given the following diagram and need to find the measures of the angles:


- \( \angle PXS \) is marked as \( 79^\circ \).
- Lines \( VT \) and \( PS \) are parallel.

We need to find:
1. \( m\angle VWU \)
2. \( m\angle PXR \)
3. \( m\angle PXW \)
4. \( m\angle UWT \)
5. \( m\angle WXS \)

##### Step-by-Step Solution:

1. \( m\angle VWU \):
- Since \( VT \) is a straight line, \( \angle VWU \) is a straight angle:
\[
m\angle VWU = 180^\circ
\]

2. \( m\angle PXR \):
- \( \angle PXR \) is a vertical angle to \( \angle PXS \):
\[
m\angle PXR = m\angle PXS = 79^\circ
\]

3. \( m\angle PXW \):
- \( \angle PXW \) is a linear pair with \( \angle PXS \):
\[
m\angle PXW = 180^\circ - m\angle PXS = 180^\circ - 79^\circ = 101^\circ
\]

4. \( m\angle UWT \):
- Since \( VT \) and \( PS \) are parallel, \( \angle UWT \) is corresponding to \( \angle PXW \):
\[
m\angle UWT = m\angle PXW = 101^\circ
\]

5. \( m\angle WXS \):
- \( \angle WXS \) is a linear pair with \( \angle PXW \):
\[
m\angle WXS = 180^\circ - m\angle PXW = 180^\circ - 101^\circ = 79^\circ
\]

##### Final Answers for Part 2:
\[
\boxed{
\begin{aligned}
i) & \quad m\angle VWU = 180^\circ \\
ii) & \quad m\angle PXR = 79^\circ \\
iii) & \quad m\angle PXW = 101^\circ \\
iv) & \quad m\angle UWT = 101^\circ \\
v) & \quad m\angle WXS = 79^\circ
\end{aligned}
}
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
\text{Part 1:} & \quad \begin{aligned}
i) & \quad 40^\circ \\
ii) & \quad 45^\circ \\
iii) & \quad 140^\circ \\
iv) & \quad 95^\circ \\
v) & \quad 180^\circ
\end{aligned} \\
\text{Part 2:} & \quad \begin{aligned}
i) & \quad 180^\circ \\
ii) & \quad 79^\circ \\
iii) & \quad 101^\circ \\
iv) & \quad 101^\circ \\
v) & \quad 79^\circ
\end{aligned}
\end{aligned}
}
\]