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Question
A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of 30° and 60°, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use `sqrt(3)` = 1.73)
Solution
Given, height of tower = AB = 75 m
Let distance between the cars = CD = x m
In ΔABC,
tan 60° = `(AB)/(BC)`
`\implies sqrt(3) = 75/(BC)` ...(∵ tan 60° = `sqrt(3)`)
`\implies` BC = `75/sqrt(3)`
= `(75sqrt(3))/3`
= `25sqrt(3)` m
In ΔABD,
tan 30° = `(AB)/(BD)`
`\implies 1/sqrt(3) = 75/(BC + CD)` ...`(∵ tan 30^circ = 1/sqrt(3))`
`\implies 1/sqrt(3) = 75/(25sqrt(3) + CD)`
`\implies 25sqrt(3) + CD = 75sqrt(3)`
`\implies` CD = `75sqrt(3) - 25sqrt(3)`
= `50sqrt(3)`
= 50 × 1.73
= 86.5 m
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