From the figure,
Let the length of football field = l m
Height of the pole = x m
Therefore, In \( \Large \triangle ABC, \tan 60 =\frac{x}{l} \)
\( \Large \sqrt{3}=\frac{x}{l}; x=\sqrt{3}l \)
Now, in \( \Large \triangle ABD \)
\( \Large \tan 30 ^{\circ} =\frac{x}{l+80} => \frac{1}{\sqrt{3}} = \frac{x}{l+80} \)
\( \Large l + 80 = \sqrt{3}x \)
Now, from Eq. (i), we get
\( \Large l + 80 = \sqrt{3} \left(\sqrt{3} l\right) \)
=> \( \Large l + 80 = 3l => 80 = 3l - l \)
\( \Large \therefore\ l = \frac{80}{2} = 40 m \)