SSC CGL 30 Aug 2015 Paper

OC = Height of plane = h km
(let)
∠DOA = ∠OAC = 60° ;
∠BOE = ∠OBC = 30°
AB = 2 km.
AC = x km (let)
∴ BC = (2 – x) km.
From, ΔOAC
tan 60° = $\frac{\mathit{O}\mathit{C}}{\mathit{A}\mathit{C}}$
⇒ $\sqrt{3}$ = $\frac{\mathit{h}}{\mathit{x}}$
⇒ x = $\frac{\mathit{h}}{\sqrt{3}}$ km ... (i)
From ΔOBC,
tan 30° = $\frac{\mathit{O}\mathit{C}}{\mathit{C}\mathit{B}}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{\mathit{h}}{2-\mathit{x}}$
⇒ $\sqrt{3}$ h = 2 - $\frac{\mathit{h}}{\sqrt{3}}$
⇒ $\sqrt{3}$ h + $\frac{\mathit{h}}{\sqrt{3}}$ = 2
⇒ $\frac{3\mathit{h}+\mathit{h}}{\sqrt{3}}$ = 2
⇒ 4h = $2\sqrt{3}$
⇒ h = $\frac{2\sqrt{3}}{4}$ = $\frac{\sqrt{3}}{2}$ km
= $\frac{1.732}{2}$ = .866 km