Let AB be the tower and BC be distance between tower and car. Let θ be the angle of depression of the car. According to the given information, In Δ ABC, tanθ=

BC

AB

[Using(1)] and tan30°=1√3 ∴BC=

150

√3

=

150√3

3

=50√3 Hence, distance between the tower and car is50√3 .