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−