For a polygon to be convex means that all of its interior angles are less than 180 degrees. prove that for all integers n ≥ 3, the angles of any n-sided convex polygon add up to 180(n−2) degrees. you may assume that the interior angles of a triangle add up to 180 degrees.

check explanation

Step-by-step explanation:

Let K(n) be the sum of the interior angles in any n-sided convex polygon which is exactly 180(n −2)

degrees.

CASE: n = 3. A 3-sided polygon is a triangle, whose interior angles were shown always to sum to be 180 degree

INDUCTION HYPOTHESIS: Suppose that K(b) holds for some b ≥ 3. Which means that the interior angles in any b-sided convex

polygon is exactly 180(n −2) degrees.

INDUCTION STEP: We need to show that K(n is greater than or equals to 3) . That is, the interior angles of any b+1-sided convex polygon

is exactly 180(b−2) = 180(b −1) degrees,

Let X be any (b+1)-vertex convex polygon, say with successive vertices x1, x2,..., xb+

Now Y is also a convex polygon , so by the induction hypothesis K(b), the sum of the interior

angles of Y is 180(k −2).

Now let T be the triangle with vertices xk , xb+1, x1. The sum of the interior angles in X is the sum of those

in Y plus the sum of those in T .

So the sum of the interior angles in X is

180(b −2)+180 = 180((b +1)−2) = 180(b −1).

Since X was arbitrary, we conclude that the sum of the interior angles of any (b +1)-sided convex polygon

is 180((b−2)+1) = 180(b−1). That is, P(b +1) holds.

And which means that n is greater or equals to 3

b. astronomers

55/i dont know i do it for the points

step-by-step explanation: