SSC CGL Tier-1 31-08-2016 Solved 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° =

AB

BD

⇒ √3 =

x+20

y

⇒ x + 20 = √3 y ... (i)
In Δ AEC ,
tan 30° =

AE

EC

⇒

1

√3

=

x

y

⇒ y = √3 x ... (ii)
From equation (i),
x + 20 = √3×√3x
⇒ 3x - x = 20
⇒ 2x = 20
⇒ x =

20

2

= 10 m
∴ Height of pole
= (20 + 10) metre
= 30 metre