Question 785381
The exterior angles of a regular polygon with {{{n}}} sidees measure
{{{360^o/n}}} or, in radians, {{{2pi/n}}}.
The interior angles, being supplementary measure
{{{180^o-360^o/n=(n*180^o-360^o)/n=(n-2)180^o/n}}} or
{{{pi-2pi/n= (n-2)pi/n}}}
The interior angles of a regular pentagon measure
{{{3*180^o/5=108^o}}} or {{{3pi/5}}}
Two sides of the pentagon and a diagonal form an isosceles triamgle with a vertex angle measuring {{{3*180^o/5=108^o}}} or {{{3pi/5}}}, and two base angles measuring
{{{(180^o-108^o)/2=72^o/2=36^o}}} or {{{(1/2)*(pi-3pi/5)=(1/2)*(2pi/5)=pi/5}}}.
In the case of the problem the diagonal is the base and the legs are the sides of length 3 units, so half of the base would be
{{{3cos(36^o)}}} or {{{3cos(pi/5)}}}
The length of the diagonal would be twice that long, approx. 4.854 units.