Question 1187815
The way I read the problem, it looks like this:
{{{drawing(350,300,-1.2,5.8,-0.5,5.5,
blue(triangle(0,0,4.47,0,4.47,2)),
blue(triangle(0,0,4.47,0,4.47,5)),
rectangle(4.47,0,4.55,2),rectangle(4.47,2,4.5,5),
line(0,-0.02,4.47,-0.02),green(line(4.47,-0.02,6,-0.02)),
green(line(0,-0.02,-1.5,-0.02)),red(arc(0,0,4,4,-24.1,0))
red(arc(0,0,5,5,-48.2,-24.1)),locate(1.7,0.5,red(A)),
locate(1.85,1.5,red(A)),locate(1.4,-0.1,path),
locate(2.2,-0.1,width),arrow(1.35,-0.22,0,-0.22),
arrow(3,-0.22,4.47,-0.22),locate(4.6,1.3,pedestal),
locate(4.6,3.5,flag),locate(4.6,3.2,staff)
)}}} As seen across the walk path from an ant on the grass at the edge of the path opposite the pedestal,
the pedestal subtends an angle {{{red(A)}}} , and so does the flag staff.
With the path walk surface being horizontal, and the pedestal and flag staff being vertical,
I see two right triangles, sharing the path width as one leg.
The height of the pedestal, {{{2m}}} , is the other leg of one of those triangles, opposite angle {{{A}}} .
The sum of the heights of pedestal and flag staff, {{{2m+3m=5m}}} , is a leg of the other triangle, opposite angle {{{2A)}}} .
If {{{x}}} path width in m, from those right triangles, we get the trigonometric ratios:
{{{tan(A)=2/x}}} and {{{tan(2A)=5/x}}}
Substituting those tangents into the trigonometric identity {{{tan(2a)=2*tan(A)/(1-tan^2(A))}}} , we get
{{{5/x}}}{{{"="}}}{{{2(2/x)/(1-(2/x)^2)}}} --> {{{5/x}}}{{{"="}}}{{{(4/x)/(1-4/x^2)}}}
Multiplying both sides times {{{x<>0}}} , we get and equivalent equation:
{{{5}}}{{{"="}}}{{{(4)/(1-4/x^2)}}}
Assuming that {{{x<>2}}} so that {{{1-4/x^2<>0}}} , we multiply both sides times {{{1-4/x^2}}} to get the equivalent equation
{{{5(1-4/x^2)}}}{{{"="}}}{{{4}}} --> {{{5-20/x^2=4}}} --> {{{5-4=20/x^2}}} --> {{{1=20/x^2}}} --> {{{x^2=20}}} --> {{{x=sqrt(20)}}} --> {{{x=about 4.47}}}
The path is {{{highlight(4.47m)}}} wide.