Q.Two poles are placed 120 meters apart, and the height of one pole is double that of the other. If the angles of elevation to the tops of the two poles from the midpoint of the line joining their bases are complementary, find the height of the shorter pole.

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.
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