SSC CGL 31 Aug 2016 Paper

Let, AB = height of pole
= h metre
CD = height of building
= 20 metre = BE
∠ADB = 60° ; ∠ACE = 30°
Let, AE = x metre ; BD = EC
= y metre
In Δ ABD,
tan 60° = $\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{D}}$ ⇒ $\sqrt{3}$ = $\frac{\mathit{x}+20}{\mathit{y}}$
⇒ x + 20 = $\sqrt{3}$ y ... (i)
In Δ AEC ,
tan 30° = $\frac{\mathit{A}\mathit{E}}{\mathit{E}\mathit{C}}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{\mathit{x}}{\mathit{y}}$ ⇒ y = $\sqrt{3}$ x ... (ii)
From equation (i),
x + 20 = $\sqrt{3}\times \sqrt{3}\mathit{x}$
⇒ 3x - x = 20
⇒ 2x = 20
⇒ x = $\frac{20}{2}$ = 10 m
∴ Height of pole
= (20 + 10) metre
= 30 metre