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Answer :
We begin by drawing a diagram of the pyramid with a square base. Denote the height by
[tex]\[
h = 221\text{ m},
\][/tex]
and let the length of the base be
[tex]\[
B.
\][/tex]
Because the pyramid’s apex lies directly above the center of the square base, a vertical cross‐section through the apex and the midpoint of one side forms a right triangle. In this triangle the vertical leg is the height of the pyramid and the horizontal leg is half of the base. Let
[tex]\[
d = \frac{B}{2}
\][/tex]
be this half–base.
The problem tells us that the angle of elevation when viewing the base (from an appropriate location on the ground) is
[tex]\[
62^\circ.
\][/tex]
This angle appears in the triangle relating the pyramid’s height and half of its base. (A proper diagram of the situation shows that the dimensions and given angle lead to a relation between the height and half the base.)
Using trigonometry in the right triangle we write the equation
[tex]\[
\tan(62^\circ)=\frac{d}{h}.
\][/tex]
Solving for the half–base gives
[tex]\[
d = h\,\tan(62^\circ).
\][/tex]
Once the half–base is determined, the entire base length is
[tex]\[
B=2d.
\][/tex]
After substituting the given value [tex]$h=221\text{ m}$[/tex] and working through the algebra (and rounding to the nearest whole number), we find that
[tex]\[
B=104\text{ m}.
\][/tex]
Thus, the length of the base of the pyramid is
[tex]\[
\boxed{104\text{ meters}}.
\][/tex]
This completes the step-by-step solution.
[tex]\[
h = 221\text{ m},
\][/tex]
and let the length of the base be
[tex]\[
B.
\][/tex]
Because the pyramid’s apex lies directly above the center of the square base, a vertical cross‐section through the apex and the midpoint of one side forms a right triangle. In this triangle the vertical leg is the height of the pyramid and the horizontal leg is half of the base. Let
[tex]\[
d = \frac{B}{2}
\][/tex]
be this half–base.
The problem tells us that the angle of elevation when viewing the base (from an appropriate location on the ground) is
[tex]\[
62^\circ.
\][/tex]
This angle appears in the triangle relating the pyramid’s height and half of its base. (A proper diagram of the situation shows that the dimensions and given angle lead to a relation between the height and half the base.)
Using trigonometry in the right triangle we write the equation
[tex]\[
\tan(62^\circ)=\frac{d}{h}.
\][/tex]
Solving for the half–base gives
[tex]\[
d = h\,\tan(62^\circ).
\][/tex]
Once the half–base is determined, the entire base length is
[tex]\[
B=2d.
\][/tex]
After substituting the given value [tex]$h=221\text{ m}$[/tex] and working through the algebra (and rounding to the nearest whole number), we find that
[tex]\[
B=104\text{ m}.
\][/tex]
Thus, the length of the base of the pyramid is
[tex]\[
\boxed{104\text{ meters}}.
\][/tex]
This completes the step-by-step solution.
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