Current location - Training Enrollment Network - Mathematics courses - The problem of finding the number of angles in junior one mathematics
The problem of finding the number of angles in junior one mathematics
Let the number of sides of this polygon be n, and the sum of its internal angles = (n-3) * 180.

Let the vertex of this angle be A, and its two adjacent vertices are B and C, extending slightly from CA to D.

That is, ∠BAC is the inner corner ∠BAD is the outer corner.

Connect B and C to divide the original polygon into a triangle and a polygon with n- 1 sides.

Therefore: (n-1-3) *180+∠ bad+∠ ABC+∠ ACB = 600.

And ∠BAD=∠ABC+∠ACB.

So: (n-4) * 180+2 * ∠ bad = 600.

Because: 0 < 2 * ∠ bad < 360.

Therefore: 600-360 < (n-4) * 180 < 600.

So: 1.3 < n-4 < 3.4.

So: n = 6,7

When n=6, the external angle ∠ bad = 120.

When n = 7°, the external angle ∠ bad = 30.