SSC CGL 4 Sep 2016 Paper

Let, AB = height of tower = h metre
∠ ACB = 30°,
∠ ADB = 60°
CD = 20 metre ; BC = x metre
In Δ ABC,
tan 30° = $\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{C}}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{\mathit{h}}{\mathit{x}}$
⇒ x = $\sqrt{3}$ h ... (i)
In Δ ABD
an 60° = $\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{D}}$
⇒ $\sqrt{3}$ = $\frac{\mathit{h}}{\mathit{x}-20}$
⇒ h = $\sqrt{3}\mathit{x}-20\sqrt{3}$
= $\sqrt{3}\times \sqrt{3}\mathit{h}$ - 20 $\sqrt{3}$
⇒ h = 3h - 20 $\sqrt{3}$
⇒ 3h - h = $20\sqrt{3}$
⇒ 2h = 20 $\sqrt{3}$
⇒ h = $\frac{20\sqrt{3}}{2}$ = $10\sqrt{3}$ metre