Given regular pentagon ABCDE a circle can be drawn that is tangent to DE at D and to AB at A. What is the ratio of the radius to the circle to side AB?
+128407 Let O be the center of the circle
Then, by symmetry, EO ies on the diameter of the circle
Form triangle EOD
Angle CDO = 90°
Angle EDC =108°
So angle EDO = 108 - 90 = 18 °
And angle DEO = (1/2) 108 = 54°
And angle EOD = 180 - 54 - 18 = 108°
Let the side of the pentagon = S = AB
Let the radius of the circle = R
The by the Law of Sines
AB / sin EOD = R / sin DEO
AB / sin 108 = R / sin 54
R / AB = (sin 54 / sin 108) = ( sqrt [ 5 + sqrt 5 ] / sqrt 10) ≈ .85
