Answers

2016-02-20T09:22:20+05:30
Refer to diagram for this.
Let AB be tower.
BC be original shadow.
BD be final shadow.
Given BD-BC = CD = 50m

∴ In ΔABC,
tan 60° = AB/BC
=> AB/√3 = BC ...(1)

∴ In ΔABD,
tan 30° = AB/BD
=> 1/√3 = AB/(BC+CD)
=> AB√3 = AB/√3+50 (From 1)
=> AB√3-AB/√3 = 50
=> 3AB-AB/√3 = 50
=> AB = 50√3/2 = 25√3 = 25*1.732 = 43.30m
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