Exercise 4.5 - Regular pentagon, given a side

By construction, if AB = 1 then BF = [math]\frac{\sqrt[]{5}}{2}[/math], and BG = [math]\frac{\left(1+\sqrt{5}\right)}{2}[/math] , hence the triangle ABI has angles of 72[math]^o[/math] and 36[math]^o[/math]

Information: Exercise 4.5 - Regular pentagon, given a side