A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30° and 60° respectively. What is the distance between the two cars and how far is each car from the tower?
In the fig AB is the tower on the highway.

In ∆ABC
tan 30° = ![]()
= ![]()
x + y = 50√3 ………………..(1)
In ∆ABD
tan 60° = ![]()
√3 = ![]()
y = ![]()
On multiplying and dividing by √3, we get
y =
⇒
………………(2)
Therefore the distance between
the first car and tower is 28.87m
On substituting value of y
from eqn (2) in eqn (1)
x +
= 50√3
x = 50√3 - ![]()
⇒ ![]()
⇒
=
= 57.73m
The distance between two cars is 57.73m
The distance between second car and tower is (x + y) = 57.73 + 28.87 = 86.60m