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.

Answer: check explanationStep-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 degreeINDUCTION HYPOTHESIS: Suppose that K(b) holds for some b ≥ 3. Which means that the interior angles in any b-sided convexpolygon 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 polygonis 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 interiorangles 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 thosein Y plus the sum of those in T . So the sum of the interior angles in X is180(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 polygonis 180((b−2)+1) = 180(b−1). That is, P(b +1) holds. And which means that n is greater or equals to 3