Question

the sum of the exterior angles of any polygon is always equal to
exterior angles: non - regular polygons
find the value of x.

1. x=
2. x=
3. x=
4. x=

exterior angles: regular polygons
find the measure of one exterior angle for the regular polygon with n sides.

1. quadrilateral 10) octagon 11) 16 - gon

Explanation:

Step1: Recall the sum of exterior angles of a polygon

The sum of the exterior angles of any polygon is \(360^{\circ}\).

Step2: Solve for \(x\) in problem 1

We know that \(75 + 44+65 + 54+76+x=360\).
First, add the known angles: \(75 + 44+65 + 54+76=(75 + 65)+(44 + 54)+76=140+98 + 76=314\).
Then, \(314+x = 360\), so \(x=360 - 314=46\).

Step3: Solve for \(x\) in problem 2

We have \(71+85 + 44+3x+2x=360\).
Combine like - terms: \(71+85 + 44+5x=360\).
Add the known angles: \(71+85 + 44=200\).
So, \(200+5x=360\).
Subtract 200 from both sides: \(5x=360 - 200 = 160\).
Divide by 5: \(x=\frac{160}{5}=32\).

Step4: Solve for \(x\) in problem 3

We know that \(60+80 + 110+70+x=360\).
Add the known angles: \(60+80 + 110+70=(60 + 80)+(110 + 70)=140+180=320\).
Then, \(320+x=360\), so \(x=360 - 320 = 40\).

Step5: Solve for \(x\) in problem 4

We have \(120+70+70+x=360\).
Add the known angles: \(120+70+70=260\).
Then, \(260+x=360\), so \(x=360 - 260=100\).

Step6: Find the measure of one exterior angle for a quadrilateral (\(n = 4\))

For a regular polygon, the measure of one exterior angle \(=\frac{360}{n}\).
When \(n = 4\), the measure of one exterior angle \(=\frac{360}{4}=90^{\circ}\).

Step7: Find the measure of one exterior angle for an octagon (\(n = 8\))

When \(n = 8\), the measure of one exterior angle \(=\frac{360}{8}=45^{\circ}\).

Step8: Find the measure of one exterior angle for a 16 - gon (\(n = 16\))

When \(n = 16\), the measure of one exterior angle \(=\frac{360}{16}=22.5^{\circ}\).

Answer:

1. \(x = 46\)
2. \(x = 32\)
3. \(x = 40\)
4. \(x = 100\)
5. \(90^{\circ}\)
6. \(45^{\circ}\)
7. \(22.5^{\circ}\)