Question

1. a. predict whether each combination of angles in the following table will make a star. if a combination will make a star, predict the number of points it will have.

measure of the interior angle at each point (a)	measure of the exterior angles between adjacent arms (b)	makes a star? (yes/no)	number of points (n)
25°	70°
20°	100°
15°	60°

b. check your predictions by trying to construct each star.

1. give 3 different combinations for angle measures a and b that will form a nine - pointed star.
2. a. what type of polygon is formed by drawing segments connecting the non - point vertices of an n - pointed star as in the example at the left?

b. what is the measure of an interior angle of this polygon?
c. what type of triangle is formed at each point?
d. what is the measure of each of the base angles in the triangle?
e. what is the sum of the four angles formed at each non - point vertex of the star?
f. use your results from parts b, d, and e to show that the equation you wrote in exercise 2b is true for all stars.

Explanation:

Step1: Recall the condition for star - formation

For an $N$ - pointed star, the relationship between the interior angle at each point $a$ and the exterior angle between adjacent arms $b$ is given by the formula $b=(N - 2)\times\frac{180^{\circ}}{N}-a$. Also, $N=\frac{360^{\circ}}{180^{\circ}-a - b}$. If $N$ is a positive integer greater than 2, then a star can be formed.

Step2: Analyze the first row ($a = 33^{\circ},b = 105^{\circ}$)

Calculate $N=\frac{360^{\circ}}{180^{\circ}-33^{\circ}-105^{\circ}}=\frac{360^{\circ}}{42^{\circ}}
otin\mathbb{Z}$. So, no star is formed.

Step3: Analyze the second row ($a = 25^{\circ},b = 70^{\circ}$)

Calculate $N=\frac{360^{\circ}}{180^{\circ}-25^{\circ}-70^{\circ}}=\frac{360^{\circ}}{85^{\circ}}
otin\mathbb{Z}$. So, no star is formed.

Step4: Analyze the third row ($a = 20^{\circ},b = 100^{\circ}$)

Calculate $N=\frac{360^{\circ}}{180^{\circ}-20^{\circ}-100^{\circ}}=\frac{360^{\circ}}{60^{\circ}} = 6$. A 6 - pointed star is formed.

Step5: Analyze the fourth row ($a = 15^{\circ},b = 60^{\circ}$)

Calculate $N=\frac{360^{\circ}}{180^{\circ}-15^{\circ}-60^{\circ}}=\frac{360^{\circ}}{105^{\circ}}
otin\mathbb{Z}$. So, no star is formed.

1.

a.

The polygon formed by drawing segments connecting the non - point vertices of an $N$ - pointed star is an $N$ - gon.

b.

The measure of an interior angle of an $N$ - gon is given by $\theta=\frac{(N - 2)\times180^{\circ}}{N}$.

c.

The triangle formed at each point of a star is an isosceles triangle.

d.

Let the interior angle at the point of the star be $a$. If the triangle at the point is isosceles, and the sum of angles in a triangle is $180^{\circ}$, the base - angle $\beta=\frac{180^{\circ}-a}{2}$.

e.

At each non - point vertex of the star, the sum of the four angles formed is $360^{\circ}$ (since the angles around a point sum to $360^{\circ}$).

f.

The details of Exercise 2b are not given, but in general, using the angle - sum properties of polygons and triangles formed in a star, we can show the relevant equation.

For question 4:
Let $N = 9$. Using $b=(N - 2)\times\frac{180^{\circ}}{N}-a$, we have $b=\frac{(9 - 2)\times180^{\circ}}{9}-a=140^{\circ}-a$.
Combination 1: $a = 20^{\circ},b = 120^{\circ}$
Combination 2: $a = 30^{\circ},b = 110^{\circ}$
Combination 3: $a = 40^{\circ},b = 100^{\circ}$

Measure of the interior angle at each point ($a$)	Measure of the exterior angles between adjacent arms ($b$)	Makes a star? (yes/no)	Number of points ($N$)
$25^{\circ}$	$70^{\circ}$	no	-
$20^{\circ}$	$100^{\circ}$	yes	6
$15^{\circ}$	$60^{\circ}$	no	-

Answer:

For the table:

Measure of the interior angle at each point ($a$)	Measure of the exterior angles between adjacent arms ($b$)	Makes a star? (yes/no)	Number of points ($N$)
$25^{\circ}$	$70^{\circ}$	no	-
$20^{\circ}$	$100^{\circ}$	yes	6
$15^{\circ}$	$60^{\circ}$	no	-

1.
a. An $N$ - gon
b. $\frac{(N - 2)\times180^{\circ}}{N}$
c. Isosceles triangle
d. $\frac{180^{\circ}-a}{2}$
e. $360^{\circ}$
f. (Details depend on Exercise 2b)

4.
Combination 1: $a = 20^{\circ},b = 120^{\circ}$
Combination 2: $a = 30^{\circ},b = 110^{\circ}$
Combination 3: $a = 40^{\circ},b = 100^{\circ}$