SSC CGL 26 Oct 2014 Shift 1 Paper

AB = Tower = h metre
BC = x metre
In Δ ABC,
tan 60° = $\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{C}}$ ⇒ $\sqrt{3}$ = $\frac{\mathit{h}}{\mathit{x}}$
⇒ h = $\sqrt{3}$ x metre ... (i)
In Δ ABD,
tan 30° = $\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{D}}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{\mathit{h}}{\mathit{x}+40}$
⇒ $\sqrt{3}$ h = x + 40
⇒ $\sqrt{3}\mathit{h}$ = $\frac{\mathit{h}}{\sqrt{3}}$ + 40
⇒ $\sqrt{3}\mathit{h}-\frac{\mathit{h}}{\sqrt{3}}$ = 40
⇒ $\frac{\sqrt{3}\times \sqrt{3}\mathit{h}-\mathit{h}}{\sqrt{3}}$ = 40
⇒ $\frac{3\mathit{h}-\mathit{h}}{\sqrt{3}}$ = 40
⇒ 2h = $40\sqrt{3}$
⇒ h = 20 $\sqrt{3}$ metre