In a regular octagon ABCDEFGHIJ what would be the measure of ABC?

A regular octagon is a convex plane shape, having 8 equal sides, 8 vertices , eight angles and 20 diagonals.

Consider a point O inside the octagon. Then Join OA,OB,OC, .....OG and OH. Now you have 8 triangles , OAB,OBC,OCD,....OFG,OGH and OGA and the total angles of these 8 triangles is 180*8 =1440 degree or 8*2=16 right angle. If you remove the 360 degree around the centre O , you get,1440-360= 1080 dgrees or 16-4=12 rigt angles. Therefore, the sum of the angles at the vertices of a regular octagon is 1080 degrees or 12 right angles. Each angle at the vertices of a regular octagon is, therefore, 1080/8= 135 degrees.

Measure of ABC:

We tell about the  angles, sides and  area of the triangle ABC.

Angles:

Since ABC is a triangle formed of the consecutive  of AB and BC of a regular Octogon, angle ABC= 135 degree and angle ACB=angle BAC = (80-135)/2 = 22.5 degree each.

The sides of  the triangle ABC:

From the triangle OAB, angle OAB= 360/8= 45 degrees.

Consider the triangle OAC, the angle AOC = (360/8)*2=90 degrees. Therefore, Triangle OAC is right angled at A. Therefore, by Pythagorus theorem, AC= sqrt (OA^2+OC^2)= (sqrt2)*r, where r is the radius of the circumcircle of the octagon.

AB =r*sqrt(2 - sqrt2) , where r is the radius of the circumcirle of the octagon a derived result for triangle for the isoscelus triangle where AB=AC and BC= sqrt(2-sqrt2) and angle ABC=135 deg using Pythagorus theorem.

Area of the triangle ABC:

ABC is an isosceles triangle bouded by AB=BC=r*sqrt(2-sqrt2)and BC= (sqrt2)*r, Therefore the  area of the triangle is (r^2/4) (2-sqrt2) , using Heron's formula.