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Question
The length of a shadow of a tower standing on a level plane is found to be 2y meters longer when the seen's altitude is 30° than when it was 45° prove that the height of the tower is y ( √3 + 1 ) meter.
Solution
In the right-angled triangle BCD.
tan 45° = `h/(BC)`
h = BC ....(1)
In right-angled Δ ACD,
tan 30° = `h/(2y + BC)`
⇒ `1/sqrt3 = h/(2y + h)`
⇒ `h(sqrt3 - 1) = 2y`
⇒ h = y ( √3 + 1 ) m
Hence proved.
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