UPSC NDA 2 2019 Mathematics Solved Paper

Let A ladder = AD = 9 m
and CD = 9 m
∴ CD = AD = 9 am
BC = height of the flagstaff
∵ Δ ACD is isosceles triangle
∠CAB = 60°
∴ ∠ACB = ∠CAD = 30°
∴ ∠ACB = ∠CAD = 30°
∴ ∠BAD = 30° In Δ ABD,
sin‌30°=

BD

AD

⇒

1

2

=

BD

9

⇒BD=

9

2

m=4.5m
Therefore, BC = BD + DC
=4.5+9m=13.5m