Q.Two pillars are 120 meters apart, with one pillar's height being twice that of the other. At the midpoint of the line segment connecting their bases, the angles of elevation to the tops of the pillars are complementary. Determine the heights of the pillars.

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).
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