Find the sum of the measures of the interior angles of each convex polygon.

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Find the sum of the measures of the interior angles of each convex polygon.

Mathematics
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Here is the question: 16-gon
???

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i dont know
It wants me to find the sum of a convex polygon and the polygon is a 16-gon
& I can't seem to figure it out.
Use the formula (n-2)*180
Ok, So would it be 16 in place of the n?
You will get 2520 degrees
Ok, But what number would I put in place of the n?
16
Alright thanks man, Can ya help me with a couple of more?
ok
My next one is: The measure of the interior angles of a regular polygon is given. Find the number of sides in the polygon.
Here is the question: 1980.
Sum of the angles is 1980?
I just got to find the polygon that has a number of sides of 1980.
Thats basically what it is.
..Sam, Can ya help?
\[\frac{(n-2)\times180}{n}=\text{Each \angle inside the polygon}\] \[(n-2)\times180=\text{Sum of angles inside the polygon}\] ---------------------------------------------------------- So, \[(n-2)\times180=\text{Sum of angles inside the polygon}\] \[(n-2)\times180=1980\] n=13
Alright thanks, Can ya help with 4 more lol? Sorry man
post as a new question
Alright, Give me a min
The equation is (180)(n-2), where n is the number of sides. Here's why: |dw:1334620067825:dw| The number of triangles that I can split the polygon up into, is n-2. Each triangle's interior angles sum to 180. How cool! =D

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