Let the height of the shorter pole be AB \(= x\) meters and the height of the taller pole be CD \(= 2x\) meters. The point O is the midpoint of the line segment BD connecting the bases of the two poles. From O, the angles of elevation to the tops of the poles are \(\angle\)AOB = \(\theta\) and \(\angle\)COD = \(90^\circ - \theta\). Since O is the midpoint of BD, BO = OD = \(\frac{120}{2}\) meters = 60 meters
From \(\triangle\)ABO, \(\cfrac{AB}{BO} = \tan \theta\) i.e., \(\cfrac{x}{60} = \tan \theta\) ——— (i)
Again, from \(\triangle\)COD, \(\cfrac{CD}{OD} = \tan (90^\circ - \theta)\) i.e., \(\cfrac{2x}{60} = \cot \theta\) ——— (ii)
Multiplying equations (i) and (ii), \(\cfrac{x}{60} \times \cfrac{2x}{60} = \tan \theta \times \cot \theta\) i.e., \(\cfrac{2x^2}{60 \times 60} = 1\) i.e., \(x^2 = \cfrac{60 \times 60}{2} = 1800\) i.e., \(x = 30\sqrt{2}\)
∴ The height of the shorter pole is \(30\sqrt{2}\) meters.