Question 762980
Drawing (not to scale):
{{{drawing(400,200,-80,320,-30,170,
line(-80,-20,320,-20),circle(-70,0,3),
line(-70,-20,-70,-3),blue(line(-67,0,320,0)),
blue(line(-68.5,2.3,-2.3,138.5)),red(circle(0,140,3)),
red(line(0,-20,0,137)),red(circle(300,140,3)),
red(line(300,-20,300,137)),
blue(line(-67,0.5,297,139.5)),
locate(-68,0,P),locate(2,140,Q),locate(302,140,R),
locate(2,0,S),locate(302,0,T)
)}}} The player's eye is at P. The goal posts are red, with crossbars at Q and R.
The blue lines are the player's horizontal line of sight, and the player's lines of sight to the crossbars of the two goal posts.
angle u = angle QPS
angle v = angle RPT
c = PS
QS = RT = height of goal post crossbar
ST = distance between goalposts
{{{tan(u)=QS/PS}}} --> {{{tan(u)=QS/c}}} --> {{{c*tan(u)=QS}}}
{{{cot(v)=PT/RT}}} --> {{{cot(v)=PT/QS}}} --> {{{QS*cot(v)=PT}}} --> {{{PT=c*tan(u)*cot(v)}}}
{{{ST=PT-PS}}} --> {{{ST=PT-c}}} --> {{{ST=c*tan(u)*cot(v)-c}}} --> {{{ST=c(tan(u)*cot(v)-1)}}}
 
EXTRA:
If a regular polygon with {{{n}}} sides is circumscribed around a circle,
connecting the vertices to the center will give you {{{n}}} isosceles triangles whose vertex angles measure {{{360^o/n}}}.
Then, connecting to the center the point of tangency at the midpoint of the sides will split each of those isosceles triangles into two congruent right triangles, with angles measuring {{{360^o/2n}}} at the center of the circle.
{{{drawing(300,300,-1.3,1.3,-1.3,1.3,
blue(circle(0,0,1)),green(rectangle(0,-1,0.1,-0.9)),
red(triangle(-0.727,-1,0.727,-1,0,0)),
green(line(0,0,0,-1)),locate(0.02,-0.45,green(R)),
red(triangle(0,1.236,0,0,1.176,0.382)),
red(triangle(-0.727,-1,-1.176,0.382,0,0)),
red(triangle(0,1.236,0,0,-1.176,0.382)),
red(triangle(1.176,0.382,0.727,-1,0,0)),
locate(0.24,-1,red(side/2)),locate(-0.5,-1,red(side/2))
)}}} In the right triangle {{{side/2=R*tan(360^o/10)}}} (for the pentagon in the drawing)
From this point, you can calculate the perimeter of the polygon, and the area of each triangle. Then, adding the areas of the triangles you get the area of the polygon.
 
For the pentagon in the drawing,
{{{side/2=R*tan(36^o)}}} --> {{{side=2R*tan(36^o)}}}
{{{Perimeter=5*2R*tan(36^o)}}} --> {{{Perimeter=10*R*tan(36^o)}}}
With {{{R=100feet}}}, we calculate the approximate value as {{{Perimeter=10*100feet*tan(36^o)=1000*0.7265feet=726.5feet}}} (or maybe {{{727feet}}}).
For each red isosceles triangle in the pentagon in the drawing,
{{{area = base*height/2=side*R/2=(side/2)*R=(R*tan(36^o))*R=R^2*tan(36^o)}}}
and for the whole pentagon
{{{area=5R^2*tan(36^o)}}} is approximately {{{area=5*(100feet)^2*0.72654=36327feet^2}}}
 
For a decagon, {{{n=10}}}
{{{side/2=R*tan(360^o/20)}}} --> {{{side/2=R*tan(18^o)}}} --> {{{side=2R*tan(18^o)}}}
{{{Perimeter=10*2R*tan(18^o)}}} --> {{{Perimeter=20*R*tan(36^o)}}}
With {{{R=100feet}}}, we calculate the approximate value as {{{Perimeter=20*100feet*tan(18^o)=2000*0.3249feet=649.8feet}}} (or maybe {{{650feet}}}).
For each red isosceles triangle in the decagon,
{{{area = base*height/2=side*R/2=(side/2)*R=(R*tan(18^o))*R=R^2*tan(18^o)}}}
and for the whole decagon
{{{area=10R^2*tan(18^o)}}} is approximately {{{area=10*(100feet)^2*0.32492=32492feet^2}}}