Question 731560
{{{drawing(300,300,-7,7,-7,7,
circle(0,0,6),green(line(0,0,6,0)),
green(line(0,0,3.74,4.69)),green(line(5,0,3.12,3.91)),
circle(5,0,3/32),circle(3.12,3.91,3/32),
circle(-1.11,4.875,3/32),circle(-4.505,2.17,3/32),
circle(-4.505,-2.17,3/32),circle(-1.09,-4.78,3/32),
circle(3.12,-3.91,3/32),locate(0.5,0.7,A),
locate(1.75,2.19,5),locate(3.4,4.35,1)
)}}} The central angle A measures {{{360^o/7}}} (or {{{2pi/7}}} in radians)
The ceter of the flange and two adjacent holes form an isosceles triangle, like the green one in the drawing.
The legs of such a triangle measure
{{{6-1=5}}} inches.
That is the radius of the flange, {{{6}}} inches, minus the {{{1}}} inch distance from the whole to the edge of the plate.
According to the law of cosines, if the sides of a triangle with lengths {{{b}}} and {{{c}}} form an angle A, the length of the third side, opposite angle A, is given by
{{{a=sqrt(b^2+c^2-2*b*c*cos(A))}}}
In this case {{{b=c=5}}} inches, {{{A=2pi/7=360^o/7}}} and {{{a}}} is the distance between the centers of adjacent holes. Substituting:
{{{a=sqrt(5^2+5^2-2*5*5*cos(A))}}}
{{{a=sqrt(25+25-50*cos(2pi/7))}}}
{{{a=sqrt(50-50*cos(2pi/7))}}}
{{{a}}} = 4.338837.... = {{{highlight(4.3388)}}} (rounded to {{{1/10000}}} inch)