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Question
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is
Options
`20, 10sqrt(3)`
`30, 5sqrt(3)`
20, 10
`30, 10sqrt(3)`
Solution
`30, 10sqrt(3)`
Explanation;
Hint:
Let the height of the multistoried building AB be h
AE = h – 20
Let BC be x
In the right ∆ABC, tan 60° = `"AB"/"BC"`
⇒ `sqrt(3) = "h"/x`
x = `"h"/sqrt(3)` ...(1)
In the right ∆ABC, tan 30° = `"AE"/"ED" = ("h" - 20)/x`
`1/sqrt(3) = ("h" - 20)/x`
`1/sqrt(3) = ("h" - 20)/x`
x = `("h" - 20) sqrt(3)` ...(2)
From (1) and (2) we get,
`"h"/sqrt(3) = ("h" - 20) sqrt(3)`
h = 3h – 60
⇒ 60 = 2h
h = `60/2` = 30
Distance between the building (x) =
`"h"/sqrt(3) = 30/sqrt(3)`
= `(30sqrt(3))/3`
= `10sqrt(3)`
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