Answer :
Given the mathematical properties and relationships within regular polygons, let's solve each part of the given problem:
### Problem 1: A Pentagon with angles in the ratio 2:3:4:5:6
#### Part 1: Degrees
A pentagon has a total interior angle sum of [tex]\( (5 - 2) \times 180^\circ = 540^\circ \)[/tex].
Given the ratio 2:3:4:5:6, let’s denote the angles as [tex]\(2x\)[/tex], [tex]\(3x\)[/tex], [tex]\(4x\)[/tex], [tex]\(5x\)[/tex], and [tex]\(6x\)[/tex]. The sum of these angles is equal to 540°:
[tex]\[ 2x + 3x + 4x + 5x + 6x = 540^\circ \][/tex]
[tex]\[ 20x = 540^\circ \][/tex]
[tex]\[ x = 27^\circ \][/tex]
Thus, the angles are:
- [tex]\(2x = 2 \times 27^\circ = 54^\circ\)[/tex]
- [tex]\(3x = 3 \times 27^\circ = 81^\circ\)[/tex]
- [tex]\(4x = 4 \times 27^\circ = 108^\circ\)[/tex]
- [tex]\(5x = 5 \times 27^\circ = 135^\circ\)[/tex]
- [tex]\(6x = 6 \times 27^\circ = 162^\circ\)[/tex]
#### Part 2: Grades
Conversion from degrees to grades (1 degree = 10/9 grades), we have:
- [tex]\(54^\circ \times \frac{10}{9} = 60 \text{ grades}\)[/tex]
- [tex]\(81^\circ \times \frac{10}{9} = 90 \text{ grades}\)[/tex]
- [tex]\(108^\circ \times \frac{10}{9} = 120 \text{ grades}\)[/tex]
- [tex]\(135^\circ \times \frac{10}{9} = 150 \text{ grades}\)[/tex]
- [tex]\(162^\circ \times \frac{10}{9} = 180 \text{ grades}\)[/tex]
### Problem 2: A Hexagon with angles in the ratio 1:2:3:4:5:5
#### Part 1: Degrees
A hexagon has a total interior angle sum of [tex]\( (6 - 2) \times 180^\circ = 720^\circ \)[/tex].
Given the ratio 1:2:3:4:5:5, let’s denote the angles as [tex]\(x\)[/tex], [tex]\(2x\)[/tex], [tex]\(3x\)[/tex], [tex]\(4x\)[/tex], [tex]\(5x\)[/tex], [tex]\(5x\)[/tex]. The sum of these angles is:
[tex]\[ x + 2x + 3x + 4x + 5x + 5x = 720^\circ \][/tex]
[tex]\[ 20x = 720^\circ \][/tex]
[tex]\[ x = 36^\circ \][/tex]
Thus, the angles are:
- [tex]\(x = 36^\circ\)[/tex]
- [tex]\(2x = 2 \times 36^\circ = 72^\circ\)[/tex]
- [tex]\(3x = 3 \times 36^\circ = 108^\circ\)[/tex]
- [tex]\(4x = 4 \times 36^\circ = 144^\circ\)[/tex]
- [tex]\(5x = 5 \times 36^\circ = 180^\circ\)[/tex]
#### Part 2: Grades
- [tex]\(36^\circ \times \frac{10}{9} = 40 \text{ grades}\)[/tex]
- [tex]\(72^\circ \times \frac{10}{9} = 80 \text{ grades}\)[/tex]
- [tex]\(108^\circ \times \frac{10}{9} = 120 \text{ grades}\)[/tex]
- [tex]\(144^\circ \times \frac{10}{9} = 160 \text{ grades}\)[/tex]
- [tex]\(180^\circ \times \frac{10}{9} = 200 \text{ grades}\)[/tex]
### Problem 3: An Octagon with angles in the ratio 1:2:5:6:5
Given ratio needs correcting: Typical ratios sum of octagon isn’t provided.
### Problem 4: Ratio of interior to exterior angles is 2:1
#### Part 1: Number of sides
Let the exterior angle be [tex]\(e\)[/tex]. Since the ratio of the interior to exterior angle is 2:1, the interior angle, [tex]\(i\)[/tex], is [tex]\(2e\)[/tex].
[tex]\[ i + e = 180^\circ \quad \text{(since interior and exterior angles are supplementary)} \][/tex]
Substituting [tex]\(i = 2e\)[/tex],
[tex]\[ 2e + e = 180^\circ \][/tex]
[tex]\[ 3e = 180^\circ \][/tex]
[tex]\[ e = 60^\circ \][/tex]
The number of sides is [tex]\( \frac{360^\circ}{60^\circ} = 6 \)[/tex]. Thus the polygon is a hexagon.
#### Part 2: Express the angles in Degrees and Radians
- Interior angle [tex]\(i = 2 \times 60^\circ = 120^\circ \)[/tex]
- [tex]\( 120^\circ \times \frac{\pi}{180} = \frac{2\pi}{3} \)[/tex] radians.
- Exterior angle [tex]\( e = 60^\circ \)[/tex]
- [tex]\( 60^\circ \times \frac{\pi}{180} = \frac{\pi}{3} \)[/tex] radians.
### Problem 5: Ratio of interior to exterior angles 7:2
#### Part 1: Number of sides
Let the exterior angle be [tex]\(e\)[/tex]. The interior angle [tex]\(i\)[/tex] is [tex]\(7e/2\)[/tex].
[tex]\[ i + e = 180^\circ \quad \text{(since interior and exterior angles are supplementary)} \][/tex]
Substituting [tex]\(i = 7e/2\)[/tex],
[tex]\[ \frac{7e}{2} + e = 180^\circ \][/tex]
[tex]\[ \frac{9e}{2} = 180^\circ \][/tex]
[tex]\[ 9e = 360^\circ \][/tex]
[tex]\[ e = 40^\circ \][/tex]
The number of sides is [tex]\( \frac{360^\circ}{40^\circ} = 9 \)[/tex]. Thus the polygon is a nonagon.
#### Part 2: Express the angles in Degrees and Radians
- Interior angle [tex]\(i = \frac{7}{2} \times 40^\circ = 140^\circ \)[/tex]
- [tex]\( 140^\circ \times \frac{\pi}{180} = \frac{7\pi}{9} \)[/tex] radians.
- Exterior angle [tex]\( e = 40^\circ \)[/tex]
- [tex]\( 40^\circ \times \frac{\pi}{180} = \frac{2\pi}{9} \)[/tex] radians.
### Problem 1: A Pentagon with angles in the ratio 2:3:4:5:6
#### Part 1: Degrees
A pentagon has a total interior angle sum of [tex]\( (5 - 2) \times 180^\circ = 540^\circ \)[/tex].
Given the ratio 2:3:4:5:6, let’s denote the angles as [tex]\(2x\)[/tex], [tex]\(3x\)[/tex], [tex]\(4x\)[/tex], [tex]\(5x\)[/tex], and [tex]\(6x\)[/tex]. The sum of these angles is equal to 540°:
[tex]\[ 2x + 3x + 4x + 5x + 6x = 540^\circ \][/tex]
[tex]\[ 20x = 540^\circ \][/tex]
[tex]\[ x = 27^\circ \][/tex]
Thus, the angles are:
- [tex]\(2x = 2 \times 27^\circ = 54^\circ\)[/tex]
- [tex]\(3x = 3 \times 27^\circ = 81^\circ\)[/tex]
- [tex]\(4x = 4 \times 27^\circ = 108^\circ\)[/tex]
- [tex]\(5x = 5 \times 27^\circ = 135^\circ\)[/tex]
- [tex]\(6x = 6 \times 27^\circ = 162^\circ\)[/tex]
#### Part 2: Grades
Conversion from degrees to grades (1 degree = 10/9 grades), we have:
- [tex]\(54^\circ \times \frac{10}{9} = 60 \text{ grades}\)[/tex]
- [tex]\(81^\circ \times \frac{10}{9} = 90 \text{ grades}\)[/tex]
- [tex]\(108^\circ \times \frac{10}{9} = 120 \text{ grades}\)[/tex]
- [tex]\(135^\circ \times \frac{10}{9} = 150 \text{ grades}\)[/tex]
- [tex]\(162^\circ \times \frac{10}{9} = 180 \text{ grades}\)[/tex]
### Problem 2: A Hexagon with angles in the ratio 1:2:3:4:5:5
#### Part 1: Degrees
A hexagon has a total interior angle sum of [tex]\( (6 - 2) \times 180^\circ = 720^\circ \)[/tex].
Given the ratio 1:2:3:4:5:5, let’s denote the angles as [tex]\(x\)[/tex], [tex]\(2x\)[/tex], [tex]\(3x\)[/tex], [tex]\(4x\)[/tex], [tex]\(5x\)[/tex], [tex]\(5x\)[/tex]. The sum of these angles is:
[tex]\[ x + 2x + 3x + 4x + 5x + 5x = 720^\circ \][/tex]
[tex]\[ 20x = 720^\circ \][/tex]
[tex]\[ x = 36^\circ \][/tex]
Thus, the angles are:
- [tex]\(x = 36^\circ\)[/tex]
- [tex]\(2x = 2 \times 36^\circ = 72^\circ\)[/tex]
- [tex]\(3x = 3 \times 36^\circ = 108^\circ\)[/tex]
- [tex]\(4x = 4 \times 36^\circ = 144^\circ\)[/tex]
- [tex]\(5x = 5 \times 36^\circ = 180^\circ\)[/tex]
#### Part 2: Grades
- [tex]\(36^\circ \times \frac{10}{9} = 40 \text{ grades}\)[/tex]
- [tex]\(72^\circ \times \frac{10}{9} = 80 \text{ grades}\)[/tex]
- [tex]\(108^\circ \times \frac{10}{9} = 120 \text{ grades}\)[/tex]
- [tex]\(144^\circ \times \frac{10}{9} = 160 \text{ grades}\)[/tex]
- [tex]\(180^\circ \times \frac{10}{9} = 200 \text{ grades}\)[/tex]
### Problem 3: An Octagon with angles in the ratio 1:2:5:6:5
Given ratio needs correcting: Typical ratios sum of octagon isn’t provided.
### Problem 4: Ratio of interior to exterior angles is 2:1
#### Part 1: Number of sides
Let the exterior angle be [tex]\(e\)[/tex]. Since the ratio of the interior to exterior angle is 2:1, the interior angle, [tex]\(i\)[/tex], is [tex]\(2e\)[/tex].
[tex]\[ i + e = 180^\circ \quad \text{(since interior and exterior angles are supplementary)} \][/tex]
Substituting [tex]\(i = 2e\)[/tex],
[tex]\[ 2e + e = 180^\circ \][/tex]
[tex]\[ 3e = 180^\circ \][/tex]
[tex]\[ e = 60^\circ \][/tex]
The number of sides is [tex]\( \frac{360^\circ}{60^\circ} = 6 \)[/tex]. Thus the polygon is a hexagon.
#### Part 2: Express the angles in Degrees and Radians
- Interior angle [tex]\(i = 2 \times 60^\circ = 120^\circ \)[/tex]
- [tex]\( 120^\circ \times \frac{\pi}{180} = \frac{2\pi}{3} \)[/tex] radians.
- Exterior angle [tex]\( e = 60^\circ \)[/tex]
- [tex]\( 60^\circ \times \frac{\pi}{180} = \frac{\pi}{3} \)[/tex] radians.
### Problem 5: Ratio of interior to exterior angles 7:2
#### Part 1: Number of sides
Let the exterior angle be [tex]\(e\)[/tex]. The interior angle [tex]\(i\)[/tex] is [tex]\(7e/2\)[/tex].
[tex]\[ i + e = 180^\circ \quad \text{(since interior and exterior angles are supplementary)} \][/tex]
Substituting [tex]\(i = 7e/2\)[/tex],
[tex]\[ \frac{7e}{2} + e = 180^\circ \][/tex]
[tex]\[ \frac{9e}{2} = 180^\circ \][/tex]
[tex]\[ 9e = 360^\circ \][/tex]
[tex]\[ e = 40^\circ \][/tex]
The number of sides is [tex]\( \frac{360^\circ}{40^\circ} = 9 \)[/tex]. Thus the polygon is a nonagon.
#### Part 2: Express the angles in Degrees and Radians
- Interior angle [tex]\(i = \frac{7}{2} \times 40^\circ = 140^\circ \)[/tex]
- [tex]\( 140^\circ \times \frac{\pi}{180} = \frac{7\pi}{9} \)[/tex] radians.
- Exterior angle [tex]\( e = 40^\circ \)[/tex]
- [tex]\( 40^\circ \times \frac{\pi}{180} = \frac{2\pi}{9} \)[/tex] radians.
