Geometry worksheet focusing on arcs, central angles, and inscribed angles in circles, with diagrams and problems for calculating angle measures.

Problem Analysis:

The worksheet involves problems related to arcs, central angles, and inscribed angles in circles. We will solve each part step by step.

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#### Part 1: Finding Measures in Circle \( \odot P \)

Given:
- \( m\angle WPX = 28^\circ \)
- \( m\angle ZPY = 38^\circ \)
- \( WZ \) and \( XV \) are diameters.

We need to find the following measures:
1. \( \widehat{YZ} \)
2. \( \widehat{WX} \)
3. \( \angle VPZ \)
4. \( \widehat{VWX} \)
5. \( \angle XPY \)
6. \( \widehat{XY} \)
7. \( \widehat{XWY} \)
8. \( \widehat{WZX} \)

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

1. \( \widehat{YZ} \):
- Since \( WZ \) is a diameter, \( \widehat{WZ} = 180^\circ \).
- The central angle \( \angle ZPY = 38^\circ \), so the arc \( \widehat{ZY} \) subtended by this angle is also \( 38^\circ \).
- Therefore, \( \widehat{YZ} = 180^\circ - 38^\circ = 142^\circ \).

2. \( \widehat{WX} \):
- Similarly, since \( XV \) is a diameter, \( \widehat{WXV} = 180^\circ \).
- The central angle \( \angle WPX = 28^\circ \), so the arc \( \widehat{WX} \) subtended by this angle is \( 28^\circ \).

3. \( \angle VPZ \):
- The total angle around point \( P \) is \( 360^\circ \).
- We know \( \angle WPX = 28^\circ \) and \( \angle ZPY = 38^\circ \).
- The remaining angle \( \angle VPZ \) is:
\[
\angle VPZ = 360^\circ - (28^\circ + 38^\circ + 90^\circ + 90^\circ) = 360^\circ - 246^\circ = 114^\circ
\]
(Note: \( 90^\circ \) each for the right angles formed by the diameters \( WZ \) and \( XV \)).

4. \( \widehat{VWX} \):
- The arc \( \widehat{VWX} \) is the sum of arcs \( \widehat{VW} \) and \( \widehat{WX} \).
- Since \( XV \) is a diameter, \( \widehat{VW} = 180^\circ - \widehat{WX} \).
- From part 2, \( \widehat{WX} = 28^\circ \), so \( \widehat{VW} = 180^\circ - 28^\circ = 152^\circ \).
- Therefore, \( \widehat{VWX} = \widehat{VW} + \widehat{WX} = 152^\circ + 28^\circ = 180^\circ \).

5. \( \angle XPY \):
- The angle \( \angle XPY \) is the remaining central angle after subtracting \( \angle WPX \) and \( \angle ZPY \) from the semicircle formed by diameter \( XV \).
- Since \( XV \) is a diameter, the total angle is \( 180^\circ \).
- Therefore, \( \angle XPY = 180^\circ - 28^\circ - 38^\circ = 114^\circ \).

6. \( \widehat{XY} \):
- The arc \( \widehat{XY} \) is the remaining arc after subtracting \( \widehat{WX} \) and \( \widehat{YZ} \) from the semicircle formed by diameter \( WZ \).
- Since \( WZ \) is a diameter, the total arc is \( 180^\circ \).
- Therefore, \( \widehat{XY} = 180^\circ - 28^\circ - 142^\circ = 10^\circ \).

7. \( \widehat{XWY} \):
- The arc \( \widehat{XWY} \) is the sum of arcs \( \widehat{XW} \) and \( \widehat{WY} \).
- From part 2, \( \widehat{WX} = 28^\circ \).
- The arc \( \widehat{WY} \) is the remaining arc after \( \widehat{WZ} \) minus \( \widehat{ZY} \):
\[
\widehat{WY} = 180^\circ - 38^\circ = 142^\circ
\]
- Therefore, \( \widehat{XWY} = \widehat{WX} + \widehat{WY} = 28^\circ + 142^\circ = 170^\circ \).

8. \( \widehat{WZX} \):
- The arc \( \widehat{WZX} \) is the semicircle formed by diameter \( WZ \), which is \( 180^\circ \).

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#### Part 2: Calculating the Value of \( x \) in Each Figure

##### Problem 9:
- Given \( \angle O = 36^\circ \).
- The inscribed angle is half the central angle:
\[
x = \frac{1}{2} \times 36^\circ = 18^\circ
\]

##### Problem 10:
- Given \( \angle O = 146^\circ \).
- The inscribed angle is half the central angle:
\[
x = \frac{1}{2} \times 146^\circ = 73^\circ
\]

##### Problem 11:
- Given \( \angle O = 49^\circ \).
- The inscribed angle is half the central angle:
\[
x = \frac{1}{2} \times 49^\circ = 24.5^\circ
\]

##### Problem 12:
- Given \( \angle O = 62^\circ \).
- The inscribed angle is half the central angle:
\[
x = \frac{1}{2} \times 62^\circ = 31^\circ
\]

##### Problem 13:
- The triangle is isosceles with two equal sides as radii.
- The base angles are equal, and the sum of angles in a triangle is \( 180^\circ \):
\[
2x + 100^\circ = 180^\circ \implies 2x = 80^\circ \implies x = 40^\circ
\]

##### Problem 14:
- The triangle is isosceles with two equal sides as radii.
- The base angles are equal, and the sum of angles in a triangle is \( 180^\circ \):
\[
2x + 150^\circ = 180^\circ \implies 2x = 30^\circ \implies x = 15^\circ
\]

##### Problem 15:
- The inscribed angle is half the central angle:
\[
x = \frac{1}{2} \times 18^\circ = 9^\circ
\]

##### Problem 16:
- The inscribed angle is half the central angle:
\[
x = \frac{1}{2} \times 27^\circ = 13.5^\circ
\]

##### Problem 17:
- The inscribed angle is half the central angle:
\[
x = 2 \times 55^\circ = 110^\circ
\]

##### Problem 18:
- The inscribed angle is half the central angle:
\[
x = 2 \times 91^\circ = 182^\circ
\]

##### Problem 19:
- The inscribed angle is half the central angle:
\[
x = 2 \times 35^\circ = 70^\circ
\]

##### Problem 20:
- The inscribed angle is half the central angle:
\[
x = 2 \times 29^\circ = 58^\circ
\]

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Final Answers:

1. \( \widehat{YZ} = 142^\circ \)
2. \( \widehat{WX} = 28^\circ \)
3. \( \angle VPZ = 114^\circ \)
4. \( \widehat{VWX} = 180^\circ \)
5. \( \angle XPY = 114^\circ \)
6. \( \widehat{XY} = 10^\circ \)
7. \( \widehat{XWY} = 170^\circ \)
8. \( \widehat{WZX} = 180^\circ \)

For Part 2:
9. \( x = 18^\circ \)
10. \( x = 73^\circ \)
11. \( x = 24.5^\circ \)
12. \( x = 31^\circ \)
13. \( x = 40^\circ \)
14. \( x = 15^\circ \)
15. \( x = 9^\circ \)
16. \( x = 13.5^\circ \)
17. \( x = 110^\circ \)
18. \( x = 182^\circ \)
19. \( x = 70^\circ \)
20. \( x = 58^\circ \)

\boxed{142^\circ, 28^\circ, 114^\circ, 180^\circ, 114^\circ, 10^\circ, 170^\circ, 180^\circ, 18^\circ, 73^\circ, 24.5^\circ, 31^\circ, 40^\circ, 15^\circ, 9^\circ, 13.5^\circ, 110^\circ, 182^\circ, 70^\circ, 58^\circ} \]