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


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