Let AB and CD be the poles and O is the point from where the elevation angles are measured.
In ∆ABO,
\(\frac{AB}{BO} = tan 60°\)
\(\frac{AB}{BO} = \sqrt3\)
\(BO = \frac{AB}{ \sqrt3}\)
In ∆CDO,
\(\frac{CD}{ DO} = tan 30°\)
\(\frac{CD }{ 80- BO} =\frac{ 1}{ \sqrt3 }\)
\(CD \sqrt3 = 80 -BO \)
\( CD\sqrt3 = 80 - \frac{AB}{ \sqrt3}\)
\(CD \sqrt3 + \frac{AB}{\sqrt3} = 80\)
Since the poles are of equal heights,
\(CD = AB \)
\(CD [\sqrt3 + \frac{1}{ \sqrt3}\, ] = 80\)
\(CD (\frac{3 +1}{ \sqrt3}) = 80\)
\(CD = 20\sqrt3 m\)
\(BO = \frac{AB}{ \sqrt3} = \frac{CD}{\sqrt3} = (\frac{20 \sqrt3}{\sqrt3} )m = 20m\)
\(DO = BD − BO = (80 − 20) m = 60 m \)
Therefore, the height of poles is \(20\sqrt3 m\) and the point is 20 m and 60 m far from these poles.