What is the sum of the internal angles of a polygon?

The sum of the internal angles of a polygon = (n-2) × 180, where n represents the number of sides of the polygon. The triangle formed by the connecting line between the outer angle of any regular polygon and any two adjacent sides of the regular polygon is an isosceles triangle.

Proof of the theorem of sum of interior angles of polygons;

Take any point O in the N-polygon, connect O with each vertex, and divide the N-polygon into N triangles.

Because the sum of the internal angles of these n triangles is equal to N × 180, the sum of the n angles with O as the common vertex is 360.

So the sum of the internal angles of the N-polygon is n×180-2×180 = (n-2)180.

That is, the sum of the internal angles of the N-polygon is equal to (n-2) × 180.

Internal angle indirect:

Internal angle, in mathematical terms, the angle formed by two adjacent sides of a polygon is called Dao as the internal angle of the polygon. Mathematically, the sum of the internal angles of a triangle is 180, and the sum of the internal angles of a quadrilateral (polygon) is 360. And so on, add an edge back, and the sum of internal angles will be added 180.

The formula for the sum of internal angles is: (n? -? 2) The degree of the internal angle of the regular polygon × 180 is: (n-2 )×180 ÷ n.

For example, the sum of the internal angles of a triangle is the sum of three angles inside the triangle, and the internal angle is any one of them.