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).
Assume that AB is a telegraph post. When the sun's angle of elevation was 45°, then \(\angle\)ACB = 45° and the shadow length was BC. When the angle of elevation became 60°, then \(\angle\)ADB = 60° and the shadow length was BD.
∴ CD = 4 meters.
When the sun's angle of elevation was \(\angle\)AEB = 30°, the shadow length of the telegraph post became BE.
From \( \triangle \)ABC, with respect to \( \angle \)ACB:
\(\tan 45° = \cfrac{AB}{BC}\)
Or, \(1 = \cfrac{AB}{BC}\)
Or, \(BC = AB\) ----(i)
From \( \triangle \)ABD, with respect to \( \angle \)ADB:
\(\tan 60° = \cfrac{AB}{BD}\)
Or, \( \sqrt{3} = \cfrac{AB}{BD}\)
Or, \( BD = \cfrac{AB}{\sqrt{3}} \) ----(ii)
Again, \(BC - BD = CD = 4\)
Or, \(AB - \cfrac{AB}{\sqrt{3}} = 4\)
Or, \(AB (1 - \cfrac{1}{\sqrt{3}}) = 4\)
Or, \(AB \times \cfrac{\sqrt{3} - 1}{\sqrt{3}} = 4\)
Or, \(AB = 4 \times \cfrac{\sqrt{3}}{\sqrt{3} - 1}\)
Or, \(AB = \cfrac{4\sqrt{3}}{(\sqrt{3} - 1)}\) -----(iii)
From \( \triangle \)ABE, with respect to \( \angle \)AEB:
\(\tan 30° = \cfrac{AB}{BE}\)
Or, \( \cfrac{1}{\sqrt{3}} = \cfrac{AB}{BE}\)
Or, \(BE = \sqrt{3} AB\)
Or, \(BE = \sqrt{3} \times \cfrac{4\sqrt{3}}{(\sqrt{3} - 1)} = \cfrac{12}{(\sqrt{3} - 1)}\)
\(= \cfrac{12(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)
\(= \cfrac{12(\sqrt{3} + 1)}{3 - 1}\)
\(= \cfrac{12(\sqrt{3} + 1)}{2}\)
\(= 6(\sqrt{3} + 1)\)
\(\therefore\) When the sun's angle of elevation was 30°, the shadow length of the telegraph post was \(6(\sqrt{3} + 1)\) meters.
(Proved).