Given regular pentagon ABCDE a circle can be drawn that is tangent to DE at D and to AB at A. In degrees, what is the measure of minor arc AD?
180(n-2), 180*3=540. So, 540/5=108 and the answer is 108.
Let us let the center of the circle be at F
Drawing radii from F to both D and A will result in right angles CDF and BAF
And ABCDF will form an irregular pentagon.....the sum of the interior angles of this pentagon = 540°
Angles DCB and ABC = 108°
So....we can find angle AFD as
540 - 108 - 108 - 90 - 90 = 144°
But AFD is a central angle in the circle....so....its measure = minor arc AD = 144°