SSC CGL 6 Sep 2016 Paper

AB = height of tower = 60 metre
BE = CD = height of pole = h metre
BC = ED = x metre
In Δ AED,
tan45° = $\frac{\mathit{A}\mathit{E}}{\mathit{E}\mathit{D}}$
⇒ 1 = $\frac{60-\mathit{h}}{\mathit{x}}$
⇒ x = 60 - h ... (i)
In Δ ABC
tan 60° = $\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{C}}$
⇒ $\sqrt{3}$ = $\frac{60}{\mathit{x}}$
⇒ $\sqrt{3}$ x = 60
⇒ x = $\frac{60}{\sqrt{3}}$ = 2 $\sqrt{3}$ metre
From equation (i),
20 $\sqrt{3}$ = 60 - h
⇒ h = 60 - 20 $\sqrt{3}$
= 20 $\left(3-\sqrt{3}\right)$ metre