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Question
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30° to 45°.
Solution
Let AB be the building of height h m.
Let the two points be C and D such that CD = 40 m, ∠ADB = 30° and ∠ACB = 45°
In ΔABC,
`(AB)/(BC) = tan 45^circ = 1`
`=>` BC = h
In ΔABD,
`(AB)/(BD) = tan 30^circ`
`=> h/(40 + h) = 1/sqrt(3)`
`=> sqrt(3)h = 40 + h`
∴ `h = 40/(sqrt(3) - 1)`
= `40/(0.732)`
= 54.64 m
Hence, height of the building is 54.64 m.
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