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### **Problem 21**
- **Given**: In the diagram, \( \angle B = 35^\circ \), \( \angle C = 75^\circ \).
- **To Find**: The degree measure of \( \angle A \).

**Solution**:
The sum of the angles in any triangle is \( 180^\circ \). Therefore, we have:

\[
\angle A + \angle B + \angle C = 180^\circ
\]

Substituting the given values:

\[
\angle A + 35^\circ + 75^\circ = 180^\circ
\]

\[
\angle A + 110^\circ = 180^\circ
\]

\[
\angle A = 180^\circ - 110^\circ = 70^\circ
\]

**Answer**: \( \angle A = 70^\circ \)

---

### **Problem 22**
- **Given**: \( \overline{AB} \) and \( \overline{CD} \) are two lines intersecting at point \( O \). \( \angle AOD = 72^\circ \).
- **To Find**: The measure of \( \angle BOC \).

**Solution**:
In a pair of intersecting lines, opposite angles (also called vertically opposite angles) are equal. Thus:

\[
\angle BOC = \angle AOD = 72^\circ
\]

**Answer**: \( \angle BOC = 72^\circ \)

---

### **Problem 23**
- **Given**: A diagram of a pentagon (5-sided polygon) where all sides and angles are not equal, with one angle labeled as \( 145^\circ \), and the other internal angles to be found.
- **To Find**: The sum of the interior angles of the pentagon.

**Solution**:
The formula for finding the sum of the interior angles of a polygon is:

\[
\text{Sum of interior angles} = (n-2) \times 180^\circ
\]

For a pentagon (\( n = 5 \)):

\[
\text{Sum of interior angles} = (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ
\]

**Answer**: The sum of the interior angles of the pentagon is \( 540^\circ \).

---

### **Problem 24**
- **Given**: \( \overline{AB} \parallel \overline{CD} \) and \( \overline{EF} \) is a transversal cutting them at points \( G \) and \( H \) respectively. \( \angle AGH = 110^\circ \) and \( \angle GHE = x^\circ \).
- **To Find**: The measure of \( \angle GHE \).

**Solution**:
Since \( \overline{AB} \parallel \overline{CD} \) and \( \overline{EF} \) is a transversal, the angles \( \angle AGH \) and \( \angle GHE \) are supplementary because they form a pair of co-interior angles.

\[
\angle AGH + \angle GHE = 180^\circ
\]

Substituting the given value:

\[
110^\circ + x^\circ = 180^\circ
\]

\[
x^\circ = 180^\circ - 110^\circ = 70^\circ
\]

**Answer**: \( x = 70^\circ \)

---

### **Problem 25**
- **Given**: The ratio of the angles in a quadrilateral is \( 2:3:5:8 \).
- **To Find**: The degree measure of each angle.

**Solution**:
The sum of the angles in a quadrilateral is \( 360^\circ \).

Let the angles be \( 2x \), \( 3x \), \( 5x \), and \( 8x \).

\[
2x + 3x + 5x + 8x = 360^\circ
\]

\[
18x = 360^\circ
\]

\[
x = \frac{360^\circ}{18} = 20^\circ
\]

Thus, the angles are:
- \( 2x = 2 \times 20^\circ = 40^\circ \)
- \( 3x = 3 \times 20^\circ = 60^\circ \)
- \( 5x = 5 \times 20^\circ = 100^\circ \)
- \( 8x = 8 \times 20^\circ =

This page provides detailed solutions to problems involving the sum of angles in triangles, the measure of vertically opposite angles, the interior angles of polygons, and the ratio of angles in a quadrilateral. Perfect for students in Grades 7-9 looking to reinforce their understanding of geometry.

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