UPSC 2016 CDS II Math Paper

Let AB be the tower.
AB = 100 m
Suppose P be the initial position of the car and Q be the position of the car after some time.
In right ΔABQ,
tan‌60°=√3=

AB

BQ

⇒BQ=

100

√3

=

100√3

3

m
In right ΔABP,
tan‌30°=

1

√3

=

AB

BP

⇒

1

√3

=

100

BP

⇒BP=100√3m
∴ Distance travelled by the car = PQ = BP – BQ
=100√3−

100−√3

3

=

200√3

3

m