1. Let the smaller pillar be AB \(= x\) meters and the larger pillar be CD \(= 2x\) meters.
The midpoint of the base connection BD is O; the angles of elevation from O to the tops of the pillars are \(\angle\)AOB = \(\theta\) and \(\angle\)COD = 90\(^o-\theta\).
Since O is the midpoint of BD, we have BO = OD = \(\frac{120}{2}\) meters = 60 meters.

From \( \triangle \)ABO, we get:
\(\cfrac{AB}{BO} = \tan \theta\)
Or, \(\cfrac{x}{60} = \tan\theta\) --------(i)

From \( \triangle \)COD, we get:
\(\cfrac{CD}{OD} = \tan(90^o - \theta)\)
Or, \(\cfrac{2x}{60} = \cot\theta\) --------(ii)

Multiplying equations (i) and (ii), we get:
\(\cfrac{x}{60} \times \cfrac{2x}{60} = \tan\theta \times \cot\theta\)
Or, \(\cfrac{2x^2}{60 \times 60} = 1\)
Or, \(x^2 = \cfrac{60 \times \cancel{60}30}{\cancel{2}}\)
Or, \(x = 30\sqrt2\)

\(\therefore\) The height of the smaller pillar is \(30\sqrt2\) meters.
And the height of the larger pillar is \(30\sqrt2 \times 2\) meters \(= 60\sqrt2\) meters.
(Proved).